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

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?

• $\sin x$ has infinite roots of 0 – Arjang Dec 30 '18 at 7:00
• @Arjang $f(x)=\sin(x)$ is not a polynomial. – Eevee Trainer Dec 30 '18 at 7:03
• @Arjang As a power series, it has infinitely many terms. A polynomial has only finitely many terms. Therefore, still not a polynomial. – Eevee Trainer Dec 30 '18 at 7:05
• @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—and a finite number of indeterminates, raised to nonnegative integer powers." – Eevee Trainer Dec 30 '18 at 7:17
• Actually even a (non-zero) single variable polynomial can have infinitely many roots. The polynomial $x^2+1$ has infinitely many zeros in the ring of quaternions. The proper formulation of the result is that the number of zeros of a non-zero single variable polynomial can have in a field is bounded by its degree. – Jyrki Lahtonen Jan 28 at 19:18

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).

• 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 – user629353 Dec 30 '18 at 7:11
• 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.) – Dirk Dec 30 '18 at 7:13
• 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. – Dirk Dec 30 '18 at 7:18

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.

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.

• But zero can't be classified to have single or more variable, just like it's degree – user629353 Dec 30 '18 at 5:16
• @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. – Kabo Murphy Dec 30 '18 at 5:28
• @ kavi Rama Please show a single web page or a single book page written that zero is a multivariate or univariate polynomial. Have you? – user629353 Dec 30 '18 at 5:30