Analogy between the fundamental theorems of arithmetic and algebra
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For the learned mathematician it may be obvious and not worth mentioning: that the fundamental theorems of arithmetic and algebra look very similar and have to do with each other, in abbreviated form:
$$ n = p_1cdot p_2 cdots p_k$$
$$ P(z) = z_0cdot(z_1 -z)cdot (z_2 -z) cdots (z_k -z)$$
which makes obvious that the irreducible polynoms of first degree play the same role in $mathbb{C}[X]$ as do the prime numbers in $mathbb{Z}$ (which both are unitary rings). It also gives - in this special case - the wording "fundamental theorem" a specific meaning: It is stated that and how some irreducible elements build the fundaments of a structure.
Is this analogy helpful, or is it superficial and maybe misleading? If the former, can it be formalised? If the latter, what are the differences that make it merely superficial?
abstract-algebra arithmetic
$endgroup$
|
show 3 more comments
$begingroup$
For the learned mathematician it may be obvious and not worth mentioning: that the fundamental theorems of arithmetic and algebra look very similar and have to do with each other, in abbreviated form:
$$ n = p_1cdot p_2 cdots p_k$$
$$ P(z) = z_0cdot(z_1 -z)cdot (z_2 -z) cdots (z_k -z)$$
which makes obvious that the irreducible polynoms of first degree play the same role in $mathbb{C}[X]$ as do the prime numbers in $mathbb{Z}$ (which both are unitary rings). It also gives - in this special case - the wording "fundamental theorem" a specific meaning: It is stated that and how some irreducible elements build the fundaments of a structure.
Is this analogy helpful, or is it superficial and maybe misleading? If the former, can it be formalised? If the latter, what are the differences that make it merely superficial?
abstract-algebra arithmetic
$endgroup$
5
$begingroup$
The analogy doesn't generalize to other fundamental theorems: the fundamental theorem of calculus, for instance, is that differentiation and integration are inverse operations of each other, which has nothing to do with "how some irreducible elements build the fundaments of a structure" as far as I can see.
$endgroup$
– zwol
Sep 28 '18 at 11:50
1
$begingroup$
I restricted it to "this special case".
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 11:54
1
$begingroup$
On the other hand, some other fundamental theorems, like the one on finite(ly generated) abelian groups, also describe how certain objects factor. Many of those statements describe how general objects decompose into "primes" in that way, but this is not a general rule.
$endgroup$
– Wojowu
Sep 28 '18 at 12:16
1
$begingroup$
Why would the analogy be made more explicit and stressed? I can't think of any reason why you'd do that.
$endgroup$
– Tanner Swett
Sep 28 '18 at 14:35
1
$begingroup$
Because it's a "deep" analogy?
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 14:38
|
show 3 more comments
$begingroup$
For the learned mathematician it may be obvious and not worth mentioning: that the fundamental theorems of arithmetic and algebra look very similar and have to do with each other, in abbreviated form:
$$ n = p_1cdot p_2 cdots p_k$$
$$ P(z) = z_0cdot(z_1 -z)cdot (z_2 -z) cdots (z_k -z)$$
which makes obvious that the irreducible polynoms of first degree play the same role in $mathbb{C}[X]$ as do the prime numbers in $mathbb{Z}$ (which both are unitary rings). It also gives - in this special case - the wording "fundamental theorem" a specific meaning: It is stated that and how some irreducible elements build the fundaments of a structure.
Is this analogy helpful, or is it superficial and maybe misleading? If the former, can it be formalised? If the latter, what are the differences that make it merely superficial?
abstract-algebra arithmetic
$endgroup$
For the learned mathematician it may be obvious and not worth mentioning: that the fundamental theorems of arithmetic and algebra look very similar and have to do with each other, in abbreviated form:
$$ n = p_1cdot p_2 cdots p_k$$
$$ P(z) = z_0cdot(z_1 -z)cdot (z_2 -z) cdots (z_k -z)$$
which makes obvious that the irreducible polynoms of first degree play the same role in $mathbb{C}[X]$ as do the prime numbers in $mathbb{Z}$ (which both are unitary rings). It also gives - in this special case - the wording "fundamental theorem" a specific meaning: It is stated that and how some irreducible elements build the fundaments of a structure.
Is this analogy helpful, or is it superficial and maybe misleading? If the former, can it be formalised? If the latter, what are the differences that make it merely superficial?
abstract-algebra arithmetic
abstract-algebra arithmetic
edited Oct 10 '18 at 10:18
José Carlos Santos
170k23132238
170k23132238
asked Sep 28 '18 at 7:52
Hans-Peter StrickerHans-Peter Stricker
6,63443995
6,63443995
5
$begingroup$
The analogy doesn't generalize to other fundamental theorems: the fundamental theorem of calculus, for instance, is that differentiation and integration are inverse operations of each other, which has nothing to do with "how some irreducible elements build the fundaments of a structure" as far as I can see.
$endgroup$
– zwol
Sep 28 '18 at 11:50
1
$begingroup$
I restricted it to "this special case".
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 11:54
1
$begingroup$
On the other hand, some other fundamental theorems, like the one on finite(ly generated) abelian groups, also describe how certain objects factor. Many of those statements describe how general objects decompose into "primes" in that way, but this is not a general rule.
$endgroup$
– Wojowu
Sep 28 '18 at 12:16
1
$begingroup$
Why would the analogy be made more explicit and stressed? I can't think of any reason why you'd do that.
$endgroup$
– Tanner Swett
Sep 28 '18 at 14:35
1
$begingroup$
Because it's a "deep" analogy?
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 14:38
|
show 3 more comments
5
$begingroup$
The analogy doesn't generalize to other fundamental theorems: the fundamental theorem of calculus, for instance, is that differentiation and integration are inverse operations of each other, which has nothing to do with "how some irreducible elements build the fundaments of a structure" as far as I can see.
$endgroup$
– zwol
Sep 28 '18 at 11:50
1
$begingroup$
I restricted it to "this special case".
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 11:54
1
$begingroup$
On the other hand, some other fundamental theorems, like the one on finite(ly generated) abelian groups, also describe how certain objects factor. Many of those statements describe how general objects decompose into "primes" in that way, but this is not a general rule.
$endgroup$
– Wojowu
Sep 28 '18 at 12:16
1
$begingroup$
Why would the analogy be made more explicit and stressed? I can't think of any reason why you'd do that.
$endgroup$
– Tanner Swett
Sep 28 '18 at 14:35
1
$begingroup$
Because it's a "deep" analogy?
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 14:38
5
5
$begingroup$
The analogy doesn't generalize to other fundamental theorems: the fundamental theorem of calculus, for instance, is that differentiation and integration are inverse operations of each other, which has nothing to do with "how some irreducible elements build the fundaments of a structure" as far as I can see.
$endgroup$
– zwol
Sep 28 '18 at 11:50
$begingroup$
The analogy doesn't generalize to other fundamental theorems: the fundamental theorem of calculus, for instance, is that differentiation and integration are inverse operations of each other, which has nothing to do with "how some irreducible elements build the fundaments of a structure" as far as I can see.
$endgroup$
– zwol
Sep 28 '18 at 11:50
1
1
$begingroup$
I restricted it to "this special case".
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 11:54
$begingroup$
I restricted it to "this special case".
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 11:54
1
1
$begingroup$
On the other hand, some other fundamental theorems, like the one on finite(ly generated) abelian groups, also describe how certain objects factor. Many of those statements describe how general objects decompose into "primes" in that way, but this is not a general rule.
$endgroup$
– Wojowu
Sep 28 '18 at 12:16
$begingroup$
On the other hand, some other fundamental theorems, like the one on finite(ly generated) abelian groups, also describe how certain objects factor. Many of those statements describe how general objects decompose into "primes" in that way, but this is not a general rule.
$endgroup$
– Wojowu
Sep 28 '18 at 12:16
1
1
$begingroup$
Why would the analogy be made more explicit and stressed? I can't think of any reason why you'd do that.
$endgroup$
– Tanner Swett
Sep 28 '18 at 14:35
$begingroup$
Why would the analogy be made more explicit and stressed? I can't think of any reason why you'd do that.
$endgroup$
– Tanner Swett
Sep 28 '18 at 14:35
1
1
$begingroup$
Because it's a "deep" analogy?
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 14:38
$begingroup$
Because it's a "deep" analogy?
$endgroup$
– Hans-Peter Stricker
Sep 28 '18 at 14:38
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
In both cases, the theorem says "This ring is a unique factorisation domain and these are its irreducible elements". So in this sense, they are similar.
However, there are significant differences. In the case of $Bbb Z$, the content is the unique factorization: since in any UFD, irreducible elements are prime, saying "and the irreducible elements are the prime numbers" doesn't add anything.
On the other hand, in the case of $Bbb C[x]$, the content is what the irreducible elements are: given any field $K$, $K[x]$ is a UFD, and yet we know that the fundamental theorem of algebra doesn't hold over most fields (including $Bbb R$, $Bbb Q$, all finite fields, etc). So here the interesting part is that the polynomials of degree $1$ are the only irreducibles.
$endgroup$
$begingroup$
Sorry, I misread. I agree it's not interesting after another reading; deleting.
$endgroup$
– R..
Sep 28 '18 at 16:26
add a comment |
$begingroup$
That is a good analogy. It turns out that both $mathbb Z$ and $mathbb{C}[x]$ are unique factorization domains. In the case of $mathbb{C}[x]$, this fact, together with the fundamental theorem of Algebra, means what you wrote: every $p(x)inmathbb{C}[x]$ can be written as the product of a non-zero complex number and first degree polynomials. The same thing applies to any algebraically closed field, such as the field of algebraic numbers.
$endgroup$
12
$begingroup$
On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific.
$endgroup$
– red_trumpet
Sep 28 '18 at 8:23
add a comment |
$begingroup$
Some other answers already make very good points. I just want to add that I think the truly amazing analogy is
"polynomials are the integers among the functions"; "polynomials behave like integers, and integers behave like polynomials"
which I could formally just state as: both $Bbb Z$ and a polynomial ring $k[x]$ (over any field $k$) are Euclidean domains.
This realisation (that one can do division with remainder, hence has unique factorisation, what this means about the fraction field, localisation, sheaves, ...) is indeed a profound insight, and arguably the analogy (and its generalisations) are a cornerstone of modern algebraic geometry and number theory.
That analogy, I think, should indeed be stressed more often. I sometimes mention it to my undergrad students, saying:
- You learnt factoring numbers, then you learnt factoring polynomials, have you ever wondered what is the relation? Specifically, for numbers you end up with primes which you cannot factor anymore; are there polynomials like that? Which ones?
- Or: In middle school you divided integers with remainder, in high school you divided polynomials with remainder. Well: the integer part of a fraction tells you its size, the polynomial part of a rational function tells you its behaviour for $xto infty$ ...
- Or: Rational functions = quotients of polynomials, like rational numbers = quotients of integers. But they are not complete, limits of them are analytic functions (like $sin, cos$) -- just like rational numbers are not complete, limits of them are possibly transcendental numbers (like $pi$) ...).
But this has nothing to do with $Bbb C$. The Fundamental Theorem of Algebra, completely sidestepping that analogy, rather focusses on the fact (surely remarkable, but unrelated to all that) that in $Bbb C[x]$ the "primes" are very easy.
$endgroup$
add a comment |
$begingroup$
This analogy can be formalise in ring theory. The sets $mathbb C[X]$ and $mathbb Z$ are both rings, and more precisely, noetherian rings.
In noetherian rings, you have a theorem that stated that every ideal can be decomposed as the intersection of finitely many primary ideals.
Which is exactly the meaning of those two theorems.
In general: if you like this kinds of analogies, study group theory, ring theory, and really any kind of algebraic structures.
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14
$begingroup$
This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1.
$endgroup$
– Wojowu
Sep 28 '18 at 10:33
10
$begingroup$
Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it.
$endgroup$
– Wojowu
Sep 28 '18 at 11:58
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In both cases, the theorem says "This ring is a unique factorisation domain and these are its irreducible elements". So in this sense, they are similar.
However, there are significant differences. In the case of $Bbb Z$, the content is the unique factorization: since in any UFD, irreducible elements are prime, saying "and the irreducible elements are the prime numbers" doesn't add anything.
On the other hand, in the case of $Bbb C[x]$, the content is what the irreducible elements are: given any field $K$, $K[x]$ is a UFD, and yet we know that the fundamental theorem of algebra doesn't hold over most fields (including $Bbb R$, $Bbb Q$, all finite fields, etc). So here the interesting part is that the polynomials of degree $1$ are the only irreducibles.
$endgroup$
$begingroup$
Sorry, I misread. I agree it's not interesting after another reading; deleting.
$endgroup$
– R..
Sep 28 '18 at 16:26
add a comment |
$begingroup$
In both cases, the theorem says "This ring is a unique factorisation domain and these are its irreducible elements". So in this sense, they are similar.
However, there are significant differences. In the case of $Bbb Z$, the content is the unique factorization: since in any UFD, irreducible elements are prime, saying "and the irreducible elements are the prime numbers" doesn't add anything.
On the other hand, in the case of $Bbb C[x]$, the content is what the irreducible elements are: given any field $K$, $K[x]$ is a UFD, and yet we know that the fundamental theorem of algebra doesn't hold over most fields (including $Bbb R$, $Bbb Q$, all finite fields, etc). So here the interesting part is that the polynomials of degree $1$ are the only irreducibles.
$endgroup$
$begingroup$
Sorry, I misread. I agree it's not interesting after another reading; deleting.
$endgroup$
– R..
Sep 28 '18 at 16:26
add a comment |
$begingroup$
In both cases, the theorem says "This ring is a unique factorisation domain and these are its irreducible elements". So in this sense, they are similar.
However, there are significant differences. In the case of $Bbb Z$, the content is the unique factorization: since in any UFD, irreducible elements are prime, saying "and the irreducible elements are the prime numbers" doesn't add anything.
On the other hand, in the case of $Bbb C[x]$, the content is what the irreducible elements are: given any field $K$, $K[x]$ is a UFD, and yet we know that the fundamental theorem of algebra doesn't hold over most fields (including $Bbb R$, $Bbb Q$, all finite fields, etc). So here the interesting part is that the polynomials of degree $1$ are the only irreducibles.
$endgroup$
In both cases, the theorem says "This ring is a unique factorisation domain and these are its irreducible elements". So in this sense, they are similar.
However, there are significant differences. In the case of $Bbb Z$, the content is the unique factorization: since in any UFD, irreducible elements are prime, saying "and the irreducible elements are the prime numbers" doesn't add anything.
On the other hand, in the case of $Bbb C[x]$, the content is what the irreducible elements are: given any field $K$, $K[x]$ is a UFD, and yet we know that the fundamental theorem of algebra doesn't hold over most fields (including $Bbb R$, $Bbb Q$, all finite fields, etc). So here the interesting part is that the polynomials of degree $1$ are the only irreducibles.
answered Sep 28 '18 at 9:08
ChristopherChristopher
6,46711628
6,46711628
$begingroup$
Sorry, I misread. I agree it's not interesting after another reading; deleting.
$endgroup$
– R..
Sep 28 '18 at 16:26
add a comment |
$begingroup$
Sorry, I misread. I agree it's not interesting after another reading; deleting.
$endgroup$
– R..
Sep 28 '18 at 16:26
$begingroup$
Sorry, I misread. I agree it's not interesting after another reading; deleting.
$endgroup$
– R..
Sep 28 '18 at 16:26
$begingroup$
Sorry, I misread. I agree it's not interesting after another reading; deleting.
$endgroup$
– R..
Sep 28 '18 at 16:26
add a comment |
$begingroup$
That is a good analogy. It turns out that both $mathbb Z$ and $mathbb{C}[x]$ are unique factorization domains. In the case of $mathbb{C}[x]$, this fact, together with the fundamental theorem of Algebra, means what you wrote: every $p(x)inmathbb{C}[x]$ can be written as the product of a non-zero complex number and first degree polynomials. The same thing applies to any algebraically closed field, such as the field of algebraic numbers.
$endgroup$
12
$begingroup$
On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific.
$endgroup$
– red_trumpet
Sep 28 '18 at 8:23
add a comment |
$begingroup$
That is a good analogy. It turns out that both $mathbb Z$ and $mathbb{C}[x]$ are unique factorization domains. In the case of $mathbb{C}[x]$, this fact, together with the fundamental theorem of Algebra, means what you wrote: every $p(x)inmathbb{C}[x]$ can be written as the product of a non-zero complex number and first degree polynomials. The same thing applies to any algebraically closed field, such as the field of algebraic numbers.
$endgroup$
12
$begingroup$
On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific.
$endgroup$
– red_trumpet
Sep 28 '18 at 8:23
add a comment |
$begingroup$
That is a good analogy. It turns out that both $mathbb Z$ and $mathbb{C}[x]$ are unique factorization domains. In the case of $mathbb{C}[x]$, this fact, together with the fundamental theorem of Algebra, means what you wrote: every $p(x)inmathbb{C}[x]$ can be written as the product of a non-zero complex number and first degree polynomials. The same thing applies to any algebraically closed field, such as the field of algebraic numbers.
$endgroup$
That is a good analogy. It turns out that both $mathbb Z$ and $mathbb{C}[x]$ are unique factorization domains. In the case of $mathbb{C}[x]$, this fact, together with the fundamental theorem of Algebra, means what you wrote: every $p(x)inmathbb{C}[x]$ can be written as the product of a non-zero complex number and first degree polynomials. The same thing applies to any algebraically closed field, such as the field of algebraic numbers.
edited Feb 7 at 8:54
answered Sep 28 '18 at 8:14
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
12
$begingroup$
On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific.
$endgroup$
– red_trumpet
Sep 28 '18 at 8:23
add a comment |
12
$begingroup$
On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific.
$endgroup$
– red_trumpet
Sep 28 '18 at 8:23
12
12
$begingroup$
On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific.
$endgroup$
– red_trumpet
Sep 28 '18 at 8:23
$begingroup$
On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific.
$endgroup$
– red_trumpet
Sep 28 '18 at 8:23
add a comment |
$begingroup$
Some other answers already make very good points. I just want to add that I think the truly amazing analogy is
"polynomials are the integers among the functions"; "polynomials behave like integers, and integers behave like polynomials"
which I could formally just state as: both $Bbb Z$ and a polynomial ring $k[x]$ (over any field $k$) are Euclidean domains.
This realisation (that one can do division with remainder, hence has unique factorisation, what this means about the fraction field, localisation, sheaves, ...) is indeed a profound insight, and arguably the analogy (and its generalisations) are a cornerstone of modern algebraic geometry and number theory.
That analogy, I think, should indeed be stressed more often. I sometimes mention it to my undergrad students, saying:
- You learnt factoring numbers, then you learnt factoring polynomials, have you ever wondered what is the relation? Specifically, for numbers you end up with primes which you cannot factor anymore; are there polynomials like that? Which ones?
- Or: In middle school you divided integers with remainder, in high school you divided polynomials with remainder. Well: the integer part of a fraction tells you its size, the polynomial part of a rational function tells you its behaviour for $xto infty$ ...
- Or: Rational functions = quotients of polynomials, like rational numbers = quotients of integers. But they are not complete, limits of them are analytic functions (like $sin, cos$) -- just like rational numbers are not complete, limits of them are possibly transcendental numbers (like $pi$) ...).
But this has nothing to do with $Bbb C$. The Fundamental Theorem of Algebra, completely sidestepping that analogy, rather focusses on the fact (surely remarkable, but unrelated to all that) that in $Bbb C[x]$ the "primes" are very easy.
$endgroup$
add a comment |
$begingroup$
Some other answers already make very good points. I just want to add that I think the truly amazing analogy is
"polynomials are the integers among the functions"; "polynomials behave like integers, and integers behave like polynomials"
which I could formally just state as: both $Bbb Z$ and a polynomial ring $k[x]$ (over any field $k$) are Euclidean domains.
This realisation (that one can do division with remainder, hence has unique factorisation, what this means about the fraction field, localisation, sheaves, ...) is indeed a profound insight, and arguably the analogy (and its generalisations) are a cornerstone of modern algebraic geometry and number theory.
That analogy, I think, should indeed be stressed more often. I sometimes mention it to my undergrad students, saying:
- You learnt factoring numbers, then you learnt factoring polynomials, have you ever wondered what is the relation? Specifically, for numbers you end up with primes which you cannot factor anymore; are there polynomials like that? Which ones?
- Or: In middle school you divided integers with remainder, in high school you divided polynomials with remainder. Well: the integer part of a fraction tells you its size, the polynomial part of a rational function tells you its behaviour for $xto infty$ ...
- Or: Rational functions = quotients of polynomials, like rational numbers = quotients of integers. But they are not complete, limits of them are analytic functions (like $sin, cos$) -- just like rational numbers are not complete, limits of them are possibly transcendental numbers (like $pi$) ...).
But this has nothing to do with $Bbb C$. The Fundamental Theorem of Algebra, completely sidestepping that analogy, rather focusses on the fact (surely remarkable, but unrelated to all that) that in $Bbb C[x]$ the "primes" are very easy.
$endgroup$
add a comment |
$begingroup$
Some other answers already make very good points. I just want to add that I think the truly amazing analogy is
"polynomials are the integers among the functions"; "polynomials behave like integers, and integers behave like polynomials"
which I could formally just state as: both $Bbb Z$ and a polynomial ring $k[x]$ (over any field $k$) are Euclidean domains.
This realisation (that one can do division with remainder, hence has unique factorisation, what this means about the fraction field, localisation, sheaves, ...) is indeed a profound insight, and arguably the analogy (and its generalisations) are a cornerstone of modern algebraic geometry and number theory.
That analogy, I think, should indeed be stressed more often. I sometimes mention it to my undergrad students, saying:
- You learnt factoring numbers, then you learnt factoring polynomials, have you ever wondered what is the relation? Specifically, for numbers you end up with primes which you cannot factor anymore; are there polynomials like that? Which ones?
- Or: In middle school you divided integers with remainder, in high school you divided polynomials with remainder. Well: the integer part of a fraction tells you its size, the polynomial part of a rational function tells you its behaviour for $xto infty$ ...
- Or: Rational functions = quotients of polynomials, like rational numbers = quotients of integers. But they are not complete, limits of them are analytic functions (like $sin, cos$) -- just like rational numbers are not complete, limits of them are possibly transcendental numbers (like $pi$) ...).
But this has nothing to do with $Bbb C$. The Fundamental Theorem of Algebra, completely sidestepping that analogy, rather focusses on the fact (surely remarkable, but unrelated to all that) that in $Bbb C[x]$ the "primes" are very easy.
$endgroup$
Some other answers already make very good points. I just want to add that I think the truly amazing analogy is
"polynomials are the integers among the functions"; "polynomials behave like integers, and integers behave like polynomials"
which I could formally just state as: both $Bbb Z$ and a polynomial ring $k[x]$ (over any field $k$) are Euclidean domains.
This realisation (that one can do division with remainder, hence has unique factorisation, what this means about the fraction field, localisation, sheaves, ...) is indeed a profound insight, and arguably the analogy (and its generalisations) are a cornerstone of modern algebraic geometry and number theory.
That analogy, I think, should indeed be stressed more often. I sometimes mention it to my undergrad students, saying:
- You learnt factoring numbers, then you learnt factoring polynomials, have you ever wondered what is the relation? Specifically, for numbers you end up with primes which you cannot factor anymore; are there polynomials like that? Which ones?
- Or: In middle school you divided integers with remainder, in high school you divided polynomials with remainder. Well: the integer part of a fraction tells you its size, the polynomial part of a rational function tells you its behaviour for $xto infty$ ...
- Or: Rational functions = quotients of polynomials, like rational numbers = quotients of integers. But they are not complete, limits of them are analytic functions (like $sin, cos$) -- just like rational numbers are not complete, limits of them are possibly transcendental numbers (like $pi$) ...).
But this has nothing to do with $Bbb C$. The Fundamental Theorem of Algebra, completely sidestepping that analogy, rather focusses on the fact (surely remarkable, but unrelated to all that) that in $Bbb C[x]$ the "primes" are very easy.
edited Sep 28 '18 at 20:05
answered Sep 28 '18 at 19:59
Torsten SchoenebergTorsten Schoeneberg
4,3412834
4,3412834
add a comment |
add a comment |
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This analogy can be formalise in ring theory. The sets $mathbb C[X]$ and $mathbb Z$ are both rings, and more precisely, noetherian rings.
In noetherian rings, you have a theorem that stated that every ideal can be decomposed as the intersection of finitely many primary ideals.
Which is exactly the meaning of those two theorems.
In general: if you like this kinds of analogies, study group theory, ring theory, and really any kind of algebraic structures.
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14
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This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1.
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– Wojowu
Sep 28 '18 at 10:33
10
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Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it.
$endgroup$
– Wojowu
Sep 28 '18 at 11:58
add a comment |
$begingroup$
This analogy can be formalise in ring theory. The sets $mathbb C[X]$ and $mathbb Z$ are both rings, and more precisely, noetherian rings.
In noetherian rings, you have a theorem that stated that every ideal can be decomposed as the intersection of finitely many primary ideals.
Which is exactly the meaning of those two theorems.
In general: if you like this kinds of analogies, study group theory, ring theory, and really any kind of algebraic structures.
$endgroup$
14
$begingroup$
This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1.
$endgroup$
– Wojowu
Sep 28 '18 at 10:33
10
$begingroup$
Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it.
$endgroup$
– Wojowu
Sep 28 '18 at 11:58
add a comment |
$begingroup$
This analogy can be formalise in ring theory. The sets $mathbb C[X]$ and $mathbb Z$ are both rings, and more precisely, noetherian rings.
In noetherian rings, you have a theorem that stated that every ideal can be decomposed as the intersection of finitely many primary ideals.
Which is exactly the meaning of those two theorems.
In general: if you like this kinds of analogies, study group theory, ring theory, and really any kind of algebraic structures.
$endgroup$
This analogy can be formalise in ring theory. The sets $mathbb C[X]$ and $mathbb Z$ are both rings, and more precisely, noetherian rings.
In noetherian rings, you have a theorem that stated that every ideal can be decomposed as the intersection of finitely many primary ideals.
Which is exactly the meaning of those two theorems.
In general: if you like this kinds of analogies, study group theory, ring theory, and really any kind of algebraic structures.
answered Sep 28 '18 at 8:00
E. JosephE. Joseph
11.7k82856
11.7k82856
14
$begingroup$
This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1.
$endgroup$
– Wojowu
Sep 28 '18 at 10:33
10
$begingroup$
Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it.
$endgroup$
– Wojowu
Sep 28 '18 at 11:58
add a comment |
14
$begingroup$
This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1.
$endgroup$
– Wojowu
Sep 28 '18 at 10:33
10
$begingroup$
Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it.
$endgroup$
– Wojowu
Sep 28 '18 at 11:58
14
14
$begingroup$
This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1.
$endgroup$
– Wojowu
Sep 28 '18 at 10:33
$begingroup$
This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1.
$endgroup$
– Wojowu
Sep 28 '18 at 10:33
10
10
$begingroup$
Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it.
$endgroup$
– Wojowu
Sep 28 '18 at 11:58
$begingroup$
Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it.
$endgroup$
– Wojowu
Sep 28 '18 at 11:58
add a comment |
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The analogy doesn't generalize to other fundamental theorems: the fundamental theorem of calculus, for instance, is that differentiation and integration are inverse operations of each other, which has nothing to do with "how some irreducible elements build the fundaments of a structure" as far as I can see.
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– zwol
Sep 28 '18 at 11:50
1
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I restricted it to "this special case".
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– Hans-Peter Stricker
Sep 28 '18 at 11:54
1
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On the other hand, some other fundamental theorems, like the one on finite(ly generated) abelian groups, also describe how certain objects factor. Many of those statements describe how general objects decompose into "primes" in that way, but this is not a general rule.
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– Wojowu
Sep 28 '18 at 12:16
1
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Why would the analogy be made more explicit and stressed? I can't think of any reason why you'd do that.
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– Tanner Swett
Sep 28 '18 at 14:35
1
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Because it's a "deep" analogy?
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– Hans-Peter Stricker
Sep 28 '18 at 14:38