Dtyjlui

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?

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).

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.

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.

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.