Why is the zero polynomial the only one to have infinite roots? [closed]












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How can it be that the zero polynomial ($f(x)=0$) is the only polynomial which has an infinite number of roots? As stated on wikipedia:




The polynomial $0$, which may be considered to have no terms at all, is called the zero polynomial. Unlike other constant polynomials, its degree is not zero. Rather the degree of the zero polynomial is either left explicitly undefined, or defined as negative (either $−1$ or $−∞$). These conventions are useful when defining Euclidean division of polynomials. The zero polynomial is also unique in that it is the only polynomial having an infinite number of roots. The graph of the zero polynomial, $f(x) = 0$, is the $x$-axis.




We can have the polynomial $x-y$ to have infinitely many roots: $x=y=text{all real numbers}$.



Where is my misunderstanding?










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closed as off-topic by RRL, user91500, Holo, Shailesh, clathratus Dec 30 '18 at 9:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, user91500, Holo, Shailesh, clathratus

If this question can be reworded to fit the rules in the help center, please edit the question.













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    $sin x$ has infinite roots of 0
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    – Arjang
    Dec 30 '18 at 7:00






  • 1




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    @Arjang $f(x)=sin(x)$ is not a polynomial.
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    – Eevee Trainer
    Dec 30 '18 at 7:03










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    @EeveeTrainer if you write it as taylor series then it is.
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    – Arjang
    Dec 30 '18 at 7:05






  • 4




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    @Arjang As a power series, it has infinitely many terms. A polynomial has only finitely many terms. Therefore, still not a polynomial.
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    – Eevee Trainer
    Dec 30 '18 at 7:05








  • 1




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    @Arjang en.wikipedia.org/wiki/Polynomial "a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms. Each term consists of the product of a number—called the coefficient of the term[2]—and a finite number of indeterminates, raised to nonnegative integer powers."
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:17
















-5












$begingroup$


How can it be that the zero polynomial ($f(x)=0$) is the only polynomial which has an infinite number of roots? As stated on wikipedia:




The polynomial $0$, which may be considered to have no terms at all, is called the zero polynomial. Unlike other constant polynomials, its degree is not zero. Rather the degree of the zero polynomial is either left explicitly undefined, or defined as negative (either $−1$ or $−∞$). These conventions are useful when defining Euclidean division of polynomials. The zero polynomial is also unique in that it is the only polynomial having an infinite number of roots. The graph of the zero polynomial, $f(x) = 0$, is the $x$-axis.




We can have the polynomial $x-y$ to have infinitely many roots: $x=y=text{all real numbers}$.



Where is my misunderstanding?










share|cite|improve this question











$endgroup$



closed as off-topic by RRL, user91500, Holo, Shailesh, clathratus Dec 30 '18 at 9:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, user91500, Holo, Shailesh, clathratus

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    $sin x$ has infinite roots of 0
    $endgroup$
    – Arjang
    Dec 30 '18 at 7:00






  • 1




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    @Arjang $f(x)=sin(x)$ is not a polynomial.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:03










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    @EeveeTrainer if you write it as taylor series then it is.
    $endgroup$
    – Arjang
    Dec 30 '18 at 7:05






  • 4




    $begingroup$
    @Arjang As a power series, it has infinitely many terms. A polynomial has only finitely many terms. Therefore, still not a polynomial.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:05








  • 1




    $begingroup$
    @Arjang en.wikipedia.org/wiki/Polynomial "a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms. Each term consists of the product of a number—called the coefficient of the term[2]—and a finite number of indeterminates, raised to nonnegative integer powers."
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:17














-5












-5








-5


0



$begingroup$


How can it be that the zero polynomial ($f(x)=0$) is the only polynomial which has an infinite number of roots? As stated on wikipedia:




The polynomial $0$, which may be considered to have no terms at all, is called the zero polynomial. Unlike other constant polynomials, its degree is not zero. Rather the degree of the zero polynomial is either left explicitly undefined, or defined as negative (either $−1$ or $−∞$). These conventions are useful when defining Euclidean division of polynomials. The zero polynomial is also unique in that it is the only polynomial having an infinite number of roots. The graph of the zero polynomial, $f(x) = 0$, is the $x$-axis.




We can have the polynomial $x-y$ to have infinitely many roots: $x=y=text{all real numbers}$.



Where is my misunderstanding?










share|cite|improve this question











$endgroup$




How can it be that the zero polynomial ($f(x)=0$) is the only polynomial which has an infinite number of roots? As stated on wikipedia:




The polynomial $0$, which may be considered to have no terms at all, is called the zero polynomial. Unlike other constant polynomials, its degree is not zero. Rather the degree of the zero polynomial is either left explicitly undefined, or defined as negative (either $−1$ or $−∞$). These conventions are useful when defining Euclidean division of polynomials. The zero polynomial is also unique in that it is the only polynomial having an infinite number of roots. The graph of the zero polynomial, $f(x) = 0$, is the $x$-axis.




We can have the polynomial $x-y$ to have infinitely many roots: $x=y=text{all real numbers}$.



Where is my misunderstanding?







polynomials roots






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday







user629353

















asked Dec 30 '18 at 5:10









user629353user629353

1147




1147




closed as off-topic by RRL, user91500, Holo, Shailesh, clathratus Dec 30 '18 at 9:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, user91500, Holo, Shailesh, clathratus

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by RRL, user91500, Holo, Shailesh, clathratus Dec 30 '18 at 9:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, user91500, Holo, Shailesh, clathratus

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    $sin x$ has infinite roots of 0
    $endgroup$
    – Arjang
    Dec 30 '18 at 7:00






  • 1




    $begingroup$
    @Arjang $f(x)=sin(x)$ is not a polynomial.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:03










  • $begingroup$
    @EeveeTrainer if you write it as taylor series then it is.
    $endgroup$
    – Arjang
    Dec 30 '18 at 7:05






  • 4




    $begingroup$
    @Arjang As a power series, it has infinitely many terms. A polynomial has only finitely many terms. Therefore, still not a polynomial.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:05








  • 1




    $begingroup$
    @Arjang en.wikipedia.org/wiki/Polynomial "a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms. Each term consists of the product of a number—called the coefficient of the term[2]—and a finite number of indeterminates, raised to nonnegative integer powers."
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:17


















  • $begingroup$
    $sin x$ has infinite roots of 0
    $endgroup$
    – Arjang
    Dec 30 '18 at 7:00






  • 1




    $begingroup$
    @Arjang $f(x)=sin(x)$ is not a polynomial.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:03










  • $begingroup$
    @EeveeTrainer if you write it as taylor series then it is.
    $endgroup$
    – Arjang
    Dec 30 '18 at 7:05






  • 4




    $begingroup$
    @Arjang As a power series, it has infinitely many terms. A polynomial has only finitely many terms. Therefore, still not a polynomial.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:05








  • 1




    $begingroup$
    @Arjang en.wikipedia.org/wiki/Polynomial "a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms. Each term consists of the product of a number—called the coefficient of the term[2]—and a finite number of indeterminates, raised to nonnegative integer powers."
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:17
















$begingroup$
$sin x$ has infinite roots of 0
$endgroup$
– Arjang
Dec 30 '18 at 7:00




$begingroup$
$sin x$ has infinite roots of 0
$endgroup$
– Arjang
Dec 30 '18 at 7:00




1




1




$begingroup$
@Arjang $f(x)=sin(x)$ is not a polynomial.
$endgroup$
– Eevee Trainer
Dec 30 '18 at 7:03




$begingroup$
@Arjang $f(x)=sin(x)$ is not a polynomial.
$endgroup$
– Eevee Trainer
Dec 30 '18 at 7:03












$begingroup$
@EeveeTrainer if you write it as taylor series then it is.
$endgroup$
– Arjang
Dec 30 '18 at 7:05




$begingroup$
@EeveeTrainer if you write it as taylor series then it is.
$endgroup$
– Arjang
Dec 30 '18 at 7:05




4




4




$begingroup$
@Arjang As a power series, it has infinitely many terms. A polynomial has only finitely many terms. Therefore, still not a polynomial.
$endgroup$
– Eevee Trainer
Dec 30 '18 at 7:05






$begingroup$
@Arjang As a power series, it has infinitely many terms. A polynomial has only finitely many terms. Therefore, still not a polynomial.
$endgroup$
– Eevee Trainer
Dec 30 '18 at 7:05






1




1




$begingroup$
@Arjang en.wikipedia.org/wiki/Polynomial "a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms. Each term consists of the product of a number—called the coefficient of the term[2]—and a finite number of indeterminates, raised to nonnegative integer powers."
$endgroup$
– Eevee Trainer
Dec 30 '18 at 7:17




$begingroup$
@Arjang en.wikipedia.org/wiki/Polynomial "a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms. Each term consists of the product of a number—called the coefficient of the term[2]—and a finite number of indeterminates, raised to nonnegative integer powers."
$endgroup$
– Eevee Trainer
Dec 30 '18 at 7:17










4 Answers
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I would say that the statement you quote is ok, but it omits that it is only about polynomials of a single variable. For polynomials of several variables the statement is not true and you gave a good counterexample $P(x,y) = x-y$.



Actually, the study of the "roots" of polynomials of several variables is a huge field (the respective sets of roots are called "varieties" and they are studied in algebraic geometry).






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    First of all for a zero polynomial there is no such thing defined as in single variable or in multiple variable just like it's degree, a zero polynomial is a zero polynomial
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    – user629353
    Dec 30 '18 at 7:11










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    I'd say that "zero polynomial" is ambiguous as is does not specify the number of variables. (In the same way, terms like "identity" or "zero function" are ambiguous.)
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    – Dirk
    Dec 30 '18 at 7:13










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    Moreover, I find the Wikipedia article about polynomials quite confusing. As the article treats polynomials of several variables from the start, the quote is plain wrong in the context.
    $endgroup$
    – Dirk
    Dec 30 '18 at 7:18



















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Your confusion lies in definitions, to a degree, but after discussing things in the comments of this question with you, I'm coming to the realization that a lot also has to do with implicit assumptions in definitions. So this answer might be a bit long and messy but hopefully I'll have edited it to include every little confusing nuance by the time I'm done.





Roots of Functions:



A "root" of a function $f(x)$ is any $x$ such that $f(x) = 0$.



The function $f(x) = 0$ has infinitely many roots, since for all real numbers $x$, $f(x) =0$. Plug in whatever $x$ you desire, it is always zero.



In general, a polynomial of degree $n$,



$$f(x) = sum_{k=0}^n a_kx^k = a_0 + a_1x + a_2x^2 + ... + a_nx^n tag 1$$



has at most $n$ real roots, and exactly $n$ roots if we extend this to complex values. This is known as "The Fundamental Theorem of Algebra." The only exception to this rule is $f(x) = 0$, for reasons outlined previously. The fundamental theorem of algebra is usually stated in such a way as to explicitly state $f(x)=0$ (or constant polynomials in general) is excepted.





Definition of Polynomial:




Note: throughout any discussion of the roots of, say, polynomials, we implicitly assume the polynomial to be a function. By itself, $x^2 + 3x$ or stuff of the sort is just an expression. (Expressions differ from equations in that they lack any sort of equal sign.) It might be a polynomial, but it is just an expression. Without being written as a function, say $f(x) = x^2 + 3x$, there is no inherent notion of roots. Thus in any discussion of the notion of "roots of a polynomial," pretty much any time you see "polynomial," you can replace it with "polynomial function."




In particular, note that the related tenets only apply to polynomials. For example, $x-y$ is not a polynomial as you suggest. A polynomial will be of the form



$$y = text{a linear combination of } 1, x, x^2, ..., x^n$$



That is, a polynomial will have a "$y$" on the left side of the equals sign, and on the right side you only have $x$ terms to positive integer powers, each multiplied by some constant as applicable, and a constant term that has no $x$ attached. See equation $(1)$ above: that is the general form for a polynomial, where $a_0, a_1, a_2, ..., a_n$ are all constants.



So why is $x-y$ not a polynomial? Well, it is just an expression. There is no equality, for once. $y$ and $x$ are together, and not set equal to anything.



Since $x-y$ is not a polynomial, it is definitely not a "polynomial of infinitely many roots" as you posit.





"But Wait! What About Functions of Multiple Variables?:"



A fair question. We can look at functions of two variables - indeed, let's turn the $x-y$ expression from before into a function of two variables!



$$f(x,y) = x-y tag 2$$



Okay, so what about this function? You could say it is a polynomial function of multiple variables, and has zeroes for all real numbers when $x=y$. That yields an infinite number of zeroes/roots, doesn't it?



Well, yes, but then we invoke another subtler statement of the fundamental theorem of algebra. Namely, it is considered with "univariate" polynomial functions: that is to say, those of one variable. As in equation $(1)$, for example, we have the solve variable $x$. In function $(2)$ above, however, we have two variables: $x$ and $y$.



That takes us outside of the realm of univariate polynomials - what we expect to hold for univariate polynomials need not hold for polynomials of multiple variables. In that sense, $f(x,y)=x-y$ does not present a polynomial other than the $0$ polynomial with infinite roots - because in the discussion of $f(x) =0$ being the only such polynomial, we assume we are talking about univariate polynomials, polynomials of one variable. Taking one of two variables takes all of that off the table, in that bivariate polynomials can have infinitely many zeroes with absolutely no trouble.






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    Actually you took my counterexample as a function i.e.,y= f(x)=x, but this is not the case I am talking about polynomial, i.e., x-y,
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    – user629353
    Dec 30 '18 at 6:56










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    $x-y$ is not a polynomial. A polynomial is as defined. I will of course edit that into my post to address that point of confusion, but, essentially, a polynomial is in the form $y=text{something}$. So in that sense, you don't have a counterexample because that's not a polynomial
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    – Eevee Trainer
    Dec 30 '18 at 6:57










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    Polynomial has no equality sign, if it has then there would be no distinction from function or equation
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    – user629353
    Dec 30 '18 at 7:05






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    Well, $x-y$ is certainly a polynomial in the variables $x$ and $ y$.
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    – Dirk
    Dec 30 '18 at 7:09








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    Literally right in the opening of the article it starts using definition of polynomials in the functional context by defining $$p(x) = a_0 + a_1x + ... + a_nx^n$$ This is a function. And a function has roots.
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    – Eevee Trainer
    Dec 30 '18 at 7:10



















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Polynomial in a single variable can have only finite number of roots unless it is identically $0$.
This does not extend to polynomials in several variables as your example shows.






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    But zero can't be classified to have single or more variable, just like it's degree
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    – user629353
    Dec 30 '18 at 5:16












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    @user629353 You are simply misquoting a result without verifying the definitions and posting too many comments. You will not a better answer than the one's already available.
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    – Kavi Rama Murthy
    Dec 30 '18 at 5:28










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    @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you?
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    – user629353
    Dec 30 '18 at 5:30





















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The result in question is only about single-variable polynomials (hence the reference to "$f(x)$"). As $p(x,y)=x-y$ shows, a polynomial in more than one variable can indeed have infinitely many zeroes without being the zero polynomial.






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    4 Answers
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    4 Answers
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    I would say that the statement you quote is ok, but it omits that it is only about polynomials of a single variable. For polynomials of several variables the statement is not true and you gave a good counterexample $P(x,y) = x-y$.



    Actually, the study of the "roots" of polynomials of several variables is a huge field (the respective sets of roots are called "varieties" and they are studied in algebraic geometry).






    share|cite|improve this answer









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    • $begingroup$
      First of all for a zero polynomial there is no such thing defined as in single variable or in multiple variable just like it's degree, a zero polynomial is a zero polynomial
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      – user629353
      Dec 30 '18 at 7:11










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      I'd say that "zero polynomial" is ambiguous as is does not specify the number of variables. (In the same way, terms like "identity" or "zero function" are ambiguous.)
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:13










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      Moreover, I find the Wikipedia article about polynomials quite confusing. As the article treats polynomials of several variables from the start, the quote is plain wrong in the context.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:18
















    3












    $begingroup$

    I would say that the statement you quote is ok, but it omits that it is only about polynomials of a single variable. For polynomials of several variables the statement is not true and you gave a good counterexample $P(x,y) = x-y$.



    Actually, the study of the "roots" of polynomials of several variables is a huge field (the respective sets of roots are called "varieties" and they are studied in algebraic geometry).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      First of all for a zero polynomial there is no such thing defined as in single variable or in multiple variable just like it's degree, a zero polynomial is a zero polynomial
      $endgroup$
      – user629353
      Dec 30 '18 at 7:11










    • $begingroup$
      I'd say that "zero polynomial" is ambiguous as is does not specify the number of variables. (In the same way, terms like "identity" or "zero function" are ambiguous.)
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:13










    • $begingroup$
      Moreover, I find the Wikipedia article about polynomials quite confusing. As the article treats polynomials of several variables from the start, the quote is plain wrong in the context.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:18














    3












    3








    3





    $begingroup$

    I would say that the statement you quote is ok, but it omits that it is only about polynomials of a single variable. For polynomials of several variables the statement is not true and you gave a good counterexample $P(x,y) = x-y$.



    Actually, the study of the "roots" of polynomials of several variables is a huge field (the respective sets of roots are called "varieties" and they are studied in algebraic geometry).






    share|cite|improve this answer









    $endgroup$



    I would say that the statement you quote is ok, but it omits that it is only about polynomials of a single variable. For polynomials of several variables the statement is not true and you gave a good counterexample $P(x,y) = x-y$.



    Actually, the study of the "roots" of polynomials of several variables is a huge field (the respective sets of roots are called "varieties" and they are studied in algebraic geometry).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 30 '18 at 7:08









    DirkDirk

    8,8302446




    8,8302446












    • $begingroup$
      First of all for a zero polynomial there is no such thing defined as in single variable or in multiple variable just like it's degree, a zero polynomial is a zero polynomial
      $endgroup$
      – user629353
      Dec 30 '18 at 7:11










    • $begingroup$
      I'd say that "zero polynomial" is ambiguous as is does not specify the number of variables. (In the same way, terms like "identity" or "zero function" are ambiguous.)
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:13










    • $begingroup$
      Moreover, I find the Wikipedia article about polynomials quite confusing. As the article treats polynomials of several variables from the start, the quote is plain wrong in the context.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:18


















    • $begingroup$
      First of all for a zero polynomial there is no such thing defined as in single variable or in multiple variable just like it's degree, a zero polynomial is a zero polynomial
      $endgroup$
      – user629353
      Dec 30 '18 at 7:11










    • $begingroup$
      I'd say that "zero polynomial" is ambiguous as is does not specify the number of variables. (In the same way, terms like "identity" or "zero function" are ambiguous.)
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:13










    • $begingroup$
      Moreover, I find the Wikipedia article about polynomials quite confusing. As the article treats polynomials of several variables from the start, the quote is plain wrong in the context.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:18
















    $begingroup$
    First of all for a zero polynomial there is no such thing defined as in single variable or in multiple variable just like it's degree, a zero polynomial is a zero polynomial
    $endgroup$
    – user629353
    Dec 30 '18 at 7:11




    $begingroup$
    First of all for a zero polynomial there is no such thing defined as in single variable or in multiple variable just like it's degree, a zero polynomial is a zero polynomial
    $endgroup$
    – user629353
    Dec 30 '18 at 7:11












    $begingroup$
    I'd say that "zero polynomial" is ambiguous as is does not specify the number of variables. (In the same way, terms like "identity" or "zero function" are ambiguous.)
    $endgroup$
    – Dirk
    Dec 30 '18 at 7:13




    $begingroup$
    I'd say that "zero polynomial" is ambiguous as is does not specify the number of variables. (In the same way, terms like "identity" or "zero function" are ambiguous.)
    $endgroup$
    – Dirk
    Dec 30 '18 at 7:13












    $begingroup$
    Moreover, I find the Wikipedia article about polynomials quite confusing. As the article treats polynomials of several variables from the start, the quote is plain wrong in the context.
    $endgroup$
    – Dirk
    Dec 30 '18 at 7:18




    $begingroup$
    Moreover, I find the Wikipedia article about polynomials quite confusing. As the article treats polynomials of several variables from the start, the quote is plain wrong in the context.
    $endgroup$
    – Dirk
    Dec 30 '18 at 7:18











    2












    $begingroup$

    Your confusion lies in definitions, to a degree, but after discussing things in the comments of this question with you, I'm coming to the realization that a lot also has to do with implicit assumptions in definitions. So this answer might be a bit long and messy but hopefully I'll have edited it to include every little confusing nuance by the time I'm done.





    Roots of Functions:



    A "root" of a function $f(x)$ is any $x$ such that $f(x) = 0$.



    The function $f(x) = 0$ has infinitely many roots, since for all real numbers $x$, $f(x) =0$. Plug in whatever $x$ you desire, it is always zero.



    In general, a polynomial of degree $n$,



    $$f(x) = sum_{k=0}^n a_kx^k = a_0 + a_1x + a_2x^2 + ... + a_nx^n tag 1$$



    has at most $n$ real roots, and exactly $n$ roots if we extend this to complex values. This is known as "The Fundamental Theorem of Algebra." The only exception to this rule is $f(x) = 0$, for reasons outlined previously. The fundamental theorem of algebra is usually stated in such a way as to explicitly state $f(x)=0$ (or constant polynomials in general) is excepted.





    Definition of Polynomial:




    Note: throughout any discussion of the roots of, say, polynomials, we implicitly assume the polynomial to be a function. By itself, $x^2 + 3x$ or stuff of the sort is just an expression. (Expressions differ from equations in that they lack any sort of equal sign.) It might be a polynomial, but it is just an expression. Without being written as a function, say $f(x) = x^2 + 3x$, there is no inherent notion of roots. Thus in any discussion of the notion of "roots of a polynomial," pretty much any time you see "polynomial," you can replace it with "polynomial function."




    In particular, note that the related tenets only apply to polynomials. For example, $x-y$ is not a polynomial as you suggest. A polynomial will be of the form



    $$y = text{a linear combination of } 1, x, x^2, ..., x^n$$



    That is, a polynomial will have a "$y$" on the left side of the equals sign, and on the right side you only have $x$ terms to positive integer powers, each multiplied by some constant as applicable, and a constant term that has no $x$ attached. See equation $(1)$ above: that is the general form for a polynomial, where $a_0, a_1, a_2, ..., a_n$ are all constants.



    So why is $x-y$ not a polynomial? Well, it is just an expression. There is no equality, for once. $y$ and $x$ are together, and not set equal to anything.



    Since $x-y$ is not a polynomial, it is definitely not a "polynomial of infinitely many roots" as you posit.





    "But Wait! What About Functions of Multiple Variables?:"



    A fair question. We can look at functions of two variables - indeed, let's turn the $x-y$ expression from before into a function of two variables!



    $$f(x,y) = x-y tag 2$$



    Okay, so what about this function? You could say it is a polynomial function of multiple variables, and has zeroes for all real numbers when $x=y$. That yields an infinite number of zeroes/roots, doesn't it?



    Well, yes, but then we invoke another subtler statement of the fundamental theorem of algebra. Namely, it is considered with "univariate" polynomial functions: that is to say, those of one variable. As in equation $(1)$, for example, we have the solve variable $x$. In function $(2)$ above, however, we have two variables: $x$ and $y$.



    That takes us outside of the realm of univariate polynomials - what we expect to hold for univariate polynomials need not hold for polynomials of multiple variables. In that sense, $f(x,y)=x-y$ does not present a polynomial other than the $0$ polynomial with infinite roots - because in the discussion of $f(x) =0$ being the only such polynomial, we assume we are talking about univariate polynomials, polynomials of one variable. Taking one of two variables takes all of that off the table, in that bivariate polynomials can have infinitely many zeroes with absolutely no trouble.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Actually you took my counterexample as a function i.e.,y= f(x)=x, but this is not the case I am talking about polynomial, i.e., x-y,
      $endgroup$
      – user629353
      Dec 30 '18 at 6:56










    • $begingroup$
      $x-y$ is not a polynomial. A polynomial is as defined. I will of course edit that into my post to address that point of confusion, but, essentially, a polynomial is in the form $y=text{something}$. So in that sense, you don't have a counterexample because that's not a polynomial
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 6:57










    • $begingroup$
      Polynomial has no equality sign, if it has then there would be no distinction from function or equation
      $endgroup$
      – user629353
      Dec 30 '18 at 7:05






    • 2




      $begingroup$
      Well, $x-y$ is certainly a polynomial in the variables $x$ and $ y$.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:09








    • 1




      $begingroup$
      Literally right in the opening of the article it starts using definition of polynomials in the functional context by defining $$p(x) = a_0 + a_1x + ... + a_nx^n$$ This is a function. And a function has roots.
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 7:10
















    2












    $begingroup$

    Your confusion lies in definitions, to a degree, but after discussing things in the comments of this question with you, I'm coming to the realization that a lot also has to do with implicit assumptions in definitions. So this answer might be a bit long and messy but hopefully I'll have edited it to include every little confusing nuance by the time I'm done.





    Roots of Functions:



    A "root" of a function $f(x)$ is any $x$ such that $f(x) = 0$.



    The function $f(x) = 0$ has infinitely many roots, since for all real numbers $x$, $f(x) =0$. Plug in whatever $x$ you desire, it is always zero.



    In general, a polynomial of degree $n$,



    $$f(x) = sum_{k=0}^n a_kx^k = a_0 + a_1x + a_2x^2 + ... + a_nx^n tag 1$$



    has at most $n$ real roots, and exactly $n$ roots if we extend this to complex values. This is known as "The Fundamental Theorem of Algebra." The only exception to this rule is $f(x) = 0$, for reasons outlined previously. The fundamental theorem of algebra is usually stated in such a way as to explicitly state $f(x)=0$ (or constant polynomials in general) is excepted.





    Definition of Polynomial:




    Note: throughout any discussion of the roots of, say, polynomials, we implicitly assume the polynomial to be a function. By itself, $x^2 + 3x$ or stuff of the sort is just an expression. (Expressions differ from equations in that they lack any sort of equal sign.) It might be a polynomial, but it is just an expression. Without being written as a function, say $f(x) = x^2 + 3x$, there is no inherent notion of roots. Thus in any discussion of the notion of "roots of a polynomial," pretty much any time you see "polynomial," you can replace it with "polynomial function."




    In particular, note that the related tenets only apply to polynomials. For example, $x-y$ is not a polynomial as you suggest. A polynomial will be of the form



    $$y = text{a linear combination of } 1, x, x^2, ..., x^n$$



    That is, a polynomial will have a "$y$" on the left side of the equals sign, and on the right side you only have $x$ terms to positive integer powers, each multiplied by some constant as applicable, and a constant term that has no $x$ attached. See equation $(1)$ above: that is the general form for a polynomial, where $a_0, a_1, a_2, ..., a_n$ are all constants.



    So why is $x-y$ not a polynomial? Well, it is just an expression. There is no equality, for once. $y$ and $x$ are together, and not set equal to anything.



    Since $x-y$ is not a polynomial, it is definitely not a "polynomial of infinitely many roots" as you posit.





    "But Wait! What About Functions of Multiple Variables?:"



    A fair question. We can look at functions of two variables - indeed, let's turn the $x-y$ expression from before into a function of two variables!



    $$f(x,y) = x-y tag 2$$



    Okay, so what about this function? You could say it is a polynomial function of multiple variables, and has zeroes for all real numbers when $x=y$. That yields an infinite number of zeroes/roots, doesn't it?



    Well, yes, but then we invoke another subtler statement of the fundamental theorem of algebra. Namely, it is considered with "univariate" polynomial functions: that is to say, those of one variable. As in equation $(1)$, for example, we have the solve variable $x$. In function $(2)$ above, however, we have two variables: $x$ and $y$.



    That takes us outside of the realm of univariate polynomials - what we expect to hold for univariate polynomials need not hold for polynomials of multiple variables. In that sense, $f(x,y)=x-y$ does not present a polynomial other than the $0$ polynomial with infinite roots - because in the discussion of $f(x) =0$ being the only such polynomial, we assume we are talking about univariate polynomials, polynomials of one variable. Taking one of two variables takes all of that off the table, in that bivariate polynomials can have infinitely many zeroes with absolutely no trouble.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Actually you took my counterexample as a function i.e.,y= f(x)=x, but this is not the case I am talking about polynomial, i.e., x-y,
      $endgroup$
      – user629353
      Dec 30 '18 at 6:56










    • $begingroup$
      $x-y$ is not a polynomial. A polynomial is as defined. I will of course edit that into my post to address that point of confusion, but, essentially, a polynomial is in the form $y=text{something}$. So in that sense, you don't have a counterexample because that's not a polynomial
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 6:57










    • $begingroup$
      Polynomial has no equality sign, if it has then there would be no distinction from function or equation
      $endgroup$
      – user629353
      Dec 30 '18 at 7:05






    • 2




      $begingroup$
      Well, $x-y$ is certainly a polynomial in the variables $x$ and $ y$.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:09








    • 1




      $begingroup$
      Literally right in the opening of the article it starts using definition of polynomials in the functional context by defining $$p(x) = a_0 + a_1x + ... + a_nx^n$$ This is a function. And a function has roots.
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 7:10














    2












    2








    2





    $begingroup$

    Your confusion lies in definitions, to a degree, but after discussing things in the comments of this question with you, I'm coming to the realization that a lot also has to do with implicit assumptions in definitions. So this answer might be a bit long and messy but hopefully I'll have edited it to include every little confusing nuance by the time I'm done.





    Roots of Functions:



    A "root" of a function $f(x)$ is any $x$ such that $f(x) = 0$.



    The function $f(x) = 0$ has infinitely many roots, since for all real numbers $x$, $f(x) =0$. Plug in whatever $x$ you desire, it is always zero.



    In general, a polynomial of degree $n$,



    $$f(x) = sum_{k=0}^n a_kx^k = a_0 + a_1x + a_2x^2 + ... + a_nx^n tag 1$$



    has at most $n$ real roots, and exactly $n$ roots if we extend this to complex values. This is known as "The Fundamental Theorem of Algebra." The only exception to this rule is $f(x) = 0$, for reasons outlined previously. The fundamental theorem of algebra is usually stated in such a way as to explicitly state $f(x)=0$ (or constant polynomials in general) is excepted.





    Definition of Polynomial:




    Note: throughout any discussion of the roots of, say, polynomials, we implicitly assume the polynomial to be a function. By itself, $x^2 + 3x$ or stuff of the sort is just an expression. (Expressions differ from equations in that they lack any sort of equal sign.) It might be a polynomial, but it is just an expression. Without being written as a function, say $f(x) = x^2 + 3x$, there is no inherent notion of roots. Thus in any discussion of the notion of "roots of a polynomial," pretty much any time you see "polynomial," you can replace it with "polynomial function."




    In particular, note that the related tenets only apply to polynomials. For example, $x-y$ is not a polynomial as you suggest. A polynomial will be of the form



    $$y = text{a linear combination of } 1, x, x^2, ..., x^n$$



    That is, a polynomial will have a "$y$" on the left side of the equals sign, and on the right side you only have $x$ terms to positive integer powers, each multiplied by some constant as applicable, and a constant term that has no $x$ attached. See equation $(1)$ above: that is the general form for a polynomial, where $a_0, a_1, a_2, ..., a_n$ are all constants.



    So why is $x-y$ not a polynomial? Well, it is just an expression. There is no equality, for once. $y$ and $x$ are together, and not set equal to anything.



    Since $x-y$ is not a polynomial, it is definitely not a "polynomial of infinitely many roots" as you posit.





    "But Wait! What About Functions of Multiple Variables?:"



    A fair question. We can look at functions of two variables - indeed, let's turn the $x-y$ expression from before into a function of two variables!



    $$f(x,y) = x-y tag 2$$



    Okay, so what about this function? You could say it is a polynomial function of multiple variables, and has zeroes for all real numbers when $x=y$. That yields an infinite number of zeroes/roots, doesn't it?



    Well, yes, but then we invoke another subtler statement of the fundamental theorem of algebra. Namely, it is considered with "univariate" polynomial functions: that is to say, those of one variable. As in equation $(1)$, for example, we have the solve variable $x$. In function $(2)$ above, however, we have two variables: $x$ and $y$.



    That takes us outside of the realm of univariate polynomials - what we expect to hold for univariate polynomials need not hold for polynomials of multiple variables. In that sense, $f(x,y)=x-y$ does not present a polynomial other than the $0$ polynomial with infinite roots - because in the discussion of $f(x) =0$ being the only such polynomial, we assume we are talking about univariate polynomials, polynomials of one variable. Taking one of two variables takes all of that off the table, in that bivariate polynomials can have infinitely many zeroes with absolutely no trouble.






    share|cite|improve this answer











    $endgroup$



    Your confusion lies in definitions, to a degree, but after discussing things in the comments of this question with you, I'm coming to the realization that a lot also has to do with implicit assumptions in definitions. So this answer might be a bit long and messy but hopefully I'll have edited it to include every little confusing nuance by the time I'm done.





    Roots of Functions:



    A "root" of a function $f(x)$ is any $x$ such that $f(x) = 0$.



    The function $f(x) = 0$ has infinitely many roots, since for all real numbers $x$, $f(x) =0$. Plug in whatever $x$ you desire, it is always zero.



    In general, a polynomial of degree $n$,



    $$f(x) = sum_{k=0}^n a_kx^k = a_0 + a_1x + a_2x^2 + ... + a_nx^n tag 1$$



    has at most $n$ real roots, and exactly $n$ roots if we extend this to complex values. This is known as "The Fundamental Theorem of Algebra." The only exception to this rule is $f(x) = 0$, for reasons outlined previously. The fundamental theorem of algebra is usually stated in such a way as to explicitly state $f(x)=0$ (or constant polynomials in general) is excepted.





    Definition of Polynomial:




    Note: throughout any discussion of the roots of, say, polynomials, we implicitly assume the polynomial to be a function. By itself, $x^2 + 3x$ or stuff of the sort is just an expression. (Expressions differ from equations in that they lack any sort of equal sign.) It might be a polynomial, but it is just an expression. Without being written as a function, say $f(x) = x^2 + 3x$, there is no inherent notion of roots. Thus in any discussion of the notion of "roots of a polynomial," pretty much any time you see "polynomial," you can replace it with "polynomial function."




    In particular, note that the related tenets only apply to polynomials. For example, $x-y$ is not a polynomial as you suggest. A polynomial will be of the form



    $$y = text{a linear combination of } 1, x, x^2, ..., x^n$$



    That is, a polynomial will have a "$y$" on the left side of the equals sign, and on the right side you only have $x$ terms to positive integer powers, each multiplied by some constant as applicable, and a constant term that has no $x$ attached. See equation $(1)$ above: that is the general form for a polynomial, where $a_0, a_1, a_2, ..., a_n$ are all constants.



    So why is $x-y$ not a polynomial? Well, it is just an expression. There is no equality, for once. $y$ and $x$ are together, and not set equal to anything.



    Since $x-y$ is not a polynomial, it is definitely not a "polynomial of infinitely many roots" as you posit.





    "But Wait! What About Functions of Multiple Variables?:"



    A fair question. We can look at functions of two variables - indeed, let's turn the $x-y$ expression from before into a function of two variables!



    $$f(x,y) = x-y tag 2$$



    Okay, so what about this function? You could say it is a polynomial function of multiple variables, and has zeroes for all real numbers when $x=y$. That yields an infinite number of zeroes/roots, doesn't it?



    Well, yes, but then we invoke another subtler statement of the fundamental theorem of algebra. Namely, it is considered with "univariate" polynomial functions: that is to say, those of one variable. As in equation $(1)$, for example, we have the solve variable $x$. In function $(2)$ above, however, we have two variables: $x$ and $y$.



    That takes us outside of the realm of univariate polynomials - what we expect to hold for univariate polynomials need not hold for polynomials of multiple variables. In that sense, $f(x,y)=x-y$ does not present a polynomial other than the $0$ polynomial with infinite roots - because in the discussion of $f(x) =0$ being the only such polynomial, we assume we are talking about univariate polynomials, polynomials of one variable. Taking one of two variables takes all of that off the table, in that bivariate polynomials can have infinitely many zeroes with absolutely no trouble.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 30 '18 at 7:37

























    answered Dec 30 '18 at 6:50









    Eevee TrainerEevee Trainer

    5,4071936




    5,4071936












    • $begingroup$
      Actually you took my counterexample as a function i.e.,y= f(x)=x, but this is not the case I am talking about polynomial, i.e., x-y,
      $endgroup$
      – user629353
      Dec 30 '18 at 6:56










    • $begingroup$
      $x-y$ is not a polynomial. A polynomial is as defined. I will of course edit that into my post to address that point of confusion, but, essentially, a polynomial is in the form $y=text{something}$. So in that sense, you don't have a counterexample because that's not a polynomial
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 6:57










    • $begingroup$
      Polynomial has no equality sign, if it has then there would be no distinction from function or equation
      $endgroup$
      – user629353
      Dec 30 '18 at 7:05






    • 2




      $begingroup$
      Well, $x-y$ is certainly a polynomial in the variables $x$ and $ y$.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:09








    • 1




      $begingroup$
      Literally right in the opening of the article it starts using definition of polynomials in the functional context by defining $$p(x) = a_0 + a_1x + ... + a_nx^n$$ This is a function. And a function has roots.
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 7:10


















    • $begingroup$
      Actually you took my counterexample as a function i.e.,y= f(x)=x, but this is not the case I am talking about polynomial, i.e., x-y,
      $endgroup$
      – user629353
      Dec 30 '18 at 6:56










    • $begingroup$
      $x-y$ is not a polynomial. A polynomial is as defined. I will of course edit that into my post to address that point of confusion, but, essentially, a polynomial is in the form $y=text{something}$. So in that sense, you don't have a counterexample because that's not a polynomial
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 6:57










    • $begingroup$
      Polynomial has no equality sign, if it has then there would be no distinction from function or equation
      $endgroup$
      – user629353
      Dec 30 '18 at 7:05






    • 2




      $begingroup$
      Well, $x-y$ is certainly a polynomial in the variables $x$ and $ y$.
      $endgroup$
      – Dirk
      Dec 30 '18 at 7:09








    • 1




      $begingroup$
      Literally right in the opening of the article it starts using definition of polynomials in the functional context by defining $$p(x) = a_0 + a_1x + ... + a_nx^n$$ This is a function. And a function has roots.
      $endgroup$
      – Eevee Trainer
      Dec 30 '18 at 7:10
















    $begingroup$
    Actually you took my counterexample as a function i.e.,y= f(x)=x, but this is not the case I am talking about polynomial, i.e., x-y,
    $endgroup$
    – user629353
    Dec 30 '18 at 6:56




    $begingroup$
    Actually you took my counterexample as a function i.e.,y= f(x)=x, but this is not the case I am talking about polynomial, i.e., x-y,
    $endgroup$
    – user629353
    Dec 30 '18 at 6:56












    $begingroup$
    $x-y$ is not a polynomial. A polynomial is as defined. I will of course edit that into my post to address that point of confusion, but, essentially, a polynomial is in the form $y=text{something}$. So in that sense, you don't have a counterexample because that's not a polynomial
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 6:57




    $begingroup$
    $x-y$ is not a polynomial. A polynomial is as defined. I will of course edit that into my post to address that point of confusion, but, essentially, a polynomial is in the form $y=text{something}$. So in that sense, you don't have a counterexample because that's not a polynomial
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 6:57












    $begingroup$
    Polynomial has no equality sign, if it has then there would be no distinction from function or equation
    $endgroup$
    – user629353
    Dec 30 '18 at 7:05




    $begingroup$
    Polynomial has no equality sign, if it has then there would be no distinction from function or equation
    $endgroup$
    – user629353
    Dec 30 '18 at 7:05




    2




    2




    $begingroup$
    Well, $x-y$ is certainly a polynomial in the variables $x$ and $ y$.
    $endgroup$
    – Dirk
    Dec 30 '18 at 7:09






    $begingroup$
    Well, $x-y$ is certainly a polynomial in the variables $x$ and $ y$.
    $endgroup$
    – Dirk
    Dec 30 '18 at 7:09






    1




    1




    $begingroup$
    Literally right in the opening of the article it starts using definition of polynomials in the functional context by defining $$p(x) = a_0 + a_1x + ... + a_nx^n$$ This is a function. And a function has roots.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:10




    $begingroup$
    Literally right in the opening of the article it starts using definition of polynomials in the functional context by defining $$p(x) = a_0 + a_1x + ... + a_nx^n$$ This is a function. And a function has roots.
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 7:10











    1












    $begingroup$

    Polynomial in a single variable can have only finite number of roots unless it is identically $0$.
    This does not extend to polynomials in several variables as your example shows.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But zero can't be classified to have single or more variable, just like it's degree
      $endgroup$
      – user629353
      Dec 30 '18 at 5:16












    • $begingroup$
      @user629353 You are simply misquoting a result without verifying the definitions and posting too many comments. You will not a better answer than the one's already available.
      $endgroup$
      – Kavi Rama Murthy
      Dec 30 '18 at 5:28










    • $begingroup$
      @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you?
      $endgroup$
      – user629353
      Dec 30 '18 at 5:30


















    1












    $begingroup$

    Polynomial in a single variable can have only finite number of roots unless it is identically $0$.
    This does not extend to polynomials in several variables as your example shows.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But zero can't be classified to have single or more variable, just like it's degree
      $endgroup$
      – user629353
      Dec 30 '18 at 5:16












    • $begingroup$
      @user629353 You are simply misquoting a result without verifying the definitions and posting too many comments. You will not a better answer than the one's already available.
      $endgroup$
      – Kavi Rama Murthy
      Dec 30 '18 at 5:28










    • $begingroup$
      @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you?
      $endgroup$
      – user629353
      Dec 30 '18 at 5:30
















    1












    1








    1





    $begingroup$

    Polynomial in a single variable can have only finite number of roots unless it is identically $0$.
    This does not extend to polynomials in several variables as your example shows.






    share|cite|improve this answer









    $endgroup$



    Polynomial in a single variable can have only finite number of roots unless it is identically $0$.
    This does not extend to polynomials in several variables as your example shows.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 30 '18 at 5:15









    Kavi Rama MurthyKavi Rama Murthy

    53k32055




    53k32055












    • $begingroup$
      But zero can't be classified to have single or more variable, just like it's degree
      $endgroup$
      – user629353
      Dec 30 '18 at 5:16












    • $begingroup$
      @user629353 You are simply misquoting a result without verifying the definitions and posting too many comments. You will not a better answer than the one's already available.
      $endgroup$
      – Kavi Rama Murthy
      Dec 30 '18 at 5:28










    • $begingroup$
      @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you?
      $endgroup$
      – user629353
      Dec 30 '18 at 5:30




















    • $begingroup$
      But zero can't be classified to have single or more variable, just like it's degree
      $endgroup$
      – user629353
      Dec 30 '18 at 5:16












    • $begingroup$
      @user629353 You are simply misquoting a result without verifying the definitions and posting too many comments. You will not a better answer than the one's already available.
      $endgroup$
      – Kavi Rama Murthy
      Dec 30 '18 at 5:28










    • $begingroup$
      @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you?
      $endgroup$
      – user629353
      Dec 30 '18 at 5:30


















    $begingroup$
    But zero can't be classified to have single or more variable, just like it's degree
    $endgroup$
    – user629353
    Dec 30 '18 at 5:16






    $begingroup$
    But zero can't be classified to have single or more variable, just like it's degree
    $endgroup$
    – user629353
    Dec 30 '18 at 5:16














    $begingroup$
    @user629353 You are simply misquoting a result without verifying the definitions and posting too many comments. You will not a better answer than the one's already available.
    $endgroup$
    – Kavi Rama Murthy
    Dec 30 '18 at 5:28




    $begingroup$
    @user629353 You are simply misquoting a result without verifying the definitions and posting too many comments. You will not a better answer than the one's already available.
    $endgroup$
    – Kavi Rama Murthy
    Dec 30 '18 at 5:28












    $begingroup$
    @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you?
    $endgroup$
    – user629353
    Dec 30 '18 at 5:30






    $begingroup$
    @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you?
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    – user629353
    Dec 30 '18 at 5:30













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    The result in question is only about single-variable polynomials (hence the reference to "$f(x)$"). As $p(x,y)=x-y$ shows, a polynomial in more than one variable can indeed have infinitely many zeroes without being the zero polynomial.






    share|cite|improve this answer









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      0












      $begingroup$

      The result in question is only about single-variable polynomials (hence the reference to "$f(x)$"). As $p(x,y)=x-y$ shows, a polynomial in more than one variable can indeed have infinitely many zeroes without being the zero polynomial.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The result in question is only about single-variable polynomials (hence the reference to "$f(x)$"). As $p(x,y)=x-y$ shows, a polynomial in more than one variable can indeed have infinitely many zeroes without being the zero polynomial.






        share|cite|improve this answer









        $endgroup$



        The result in question is only about single-variable polynomials (hence the reference to "$f(x)$"). As $p(x,y)=x-y$ shows, a polynomial in more than one variable can indeed have infinitely many zeroes without being the zero polynomial.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 5:13









        Eric WofseyEric Wofsey

        181k12208336




        181k12208336















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