0

Let $F$ be an algebraically closed (hence infinite) field and $n$ a positive integer. For a subset $S$ of $F^n$, let us use the following notation:

$J(S):=\{f \in F[x_1,...,x_n] : f(a_1,...,a_n)=0$ for all $(a_1,...,a_n) \in S \}$

There is a theorem in Hungerford's Algebra, Section VIII.7, which asserts

$J(F^n)=\emptyset$,

with proof omitted. Is this an typo? I think, at least, the zero polynomial $0$ belongs to $J(F^n)$. On the other hand, how do I have to prove the inclusion

$J(F^n) \subset 0$?

It seems obvious, but I cannot write down rigorously.

user302934
  • 1,698
  • 1
    It is a typo. Obviously $0 \in J(F^n)$. For the other inclusion, proceed by induction. For $n = 1$ you simply use that $F$ is infinite. For the inductive step, write $F[X_1, \ldots, X_n, X_{n+1}] \cong F[X_1, \ldots, X_n][X_{n+1}]$. – juan diego rojas Nov 13 '19 at 16:40

1 Answers1

1

Here's an inductive proof that if $f \neq 0$ (in the sense that the sum has a non-zero coefficient), then there exists some $(a_1,\dots,a_n) \in F^n$ such that $f(a_1,\dots,a_n) \neq 0$.

The case of $n=1$ is straightforward (but uses the fact that $F$ is algebraically complete).

For the inductive step: suppose that $f \neq 0$. Write $f$ as a polynomial in $x_n$. That is, regard $f(x_1,\dots,x_{n-1},x_n)$ as a function of $x_n$ with coefficients that are expressions in $F[x_1,\dots,x_{n-1}]$. One of these coefficients must be non-zero. By our inductive hypothesis, there exists some $v = (a_1,\dots,a_{n-1})$ such that this coefficient evaluated at $v$ is non-zero. Now, consider the function $f(a_1,\dots,a_{n-1},x)$, which is a non-zero element of $F[x]$.

Ben Grossmann
  • 225,327