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Let $F$ be a field and let $I = \{f(x) \in F[x]\mid f(a) = 0 ~~ \forall a \in F\}$. Prove that $I = \{0\}$ when $F$ is infinite.

I have already shown that $I$ is an ideal and that $I$ is infinite when $F$ is finite. I have also found a monic polynomial to generate $I$ when $F$ is finite.

As to this part of the question, it appears to me that we should investigate what happens if there is a non-zero polynomial, $g(x) \in I$. Well, $\deg(g(x)) = n$ where $n$ is a natural number. Then, we know $g(x)$ has at most $n$ zeros. My intuition tells me this is impossible--that we need it to have infinitely many zeros--but I don't have the theory to support that idea. Help would be appreciated.

Michael Hardy
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Balthazar
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2 Answers2

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As you are aware, a non-zero $f \in F[x]$ has a degree $n \geq 0$ and at most $n$ zeroes. So, if $F$ is infinite, no nonzero $f$ can be in $I$, by essentially the reasoning you sketched.

This question tells how to prove a degree $n$ poly has at most $n$ zeroes, if you want it: How to prove that a polynomial of degree $n$ has at most $n$ roots?

Community
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user131331
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Hint $\ $ If $\ 0\ne f(x)\,$ is the least degree polynomial with infinitely many roots $\,r_1,r_2,r_3,\ldots\,$ then $\,f(x)/(x-r_1)\,$ also has infinitely many roots $\,r_2,r_3,\ldots$ and has smaller degree, contradiction.

Alternatively $\ f(r_i) = 0\color{#c00}\iff x-r_i\mid f(x).\, $ But the $\,x-r_i$ are primes in the UFD $\,F[x],\,$ hence, by existence and uniqueness of prime factorizations, $\,f(x)\,$ has finitely many prime factors $\,x-r_i\,\,$ $\color{#c00}{\rm therefore}$ $f(x)\,$ has finitely many roots (bounded by its degree).

Bill Dubuque
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