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Something similar was stated in Atiyah's Exercise 16, pg 69, Introduction to Commutative Algebra.

Let $F \in k[x_1, \ldots, x_n]$ be a non-zero polynomial, where $k$ is an infinite field. Then there exists $\lambda_1, \ldots, \lambda_n \in k$, $F(\lambda_1, \ldots, \lambda_n) \not= 0$.

Is this obvious? Is my proof (below) correct - and is there a nicer way to prove it?

Lemma: Let $f \in R[x]$ for a commutative ring $R$. If $f(\alpha)=0$, then $f(x)=g(x)(x-\alpha)$, where $\deg g < \deg f$.

Proof: Prove by induction on degree of $f$. If $\deg f =1$, it is clear. Suppose statement true for $\deg f < n$. Let $f=r_0x^n+ \cdots + r_n$. Then $f(x)=r_0x^{n-1} (x-\alpha)+h(x)$ where $\deg h< \deg f$. We are done by inductive hypothesis.

Corollary: If $R$ is an integral domain, $f \in R[x]$ has at most $\deg f $ roots.

Proof: By induction. $f(x)=g(x)(x-\alpha)$. Now use: if $\beta \not= \alpha$ is a root of $f$, it is a root of $g(x)$.

With these two results:

Proof of statement: By induction on number of variables.

  1. When $n=0$ it is clear. Suppose the statement holds $<n$. Let $F=F'(x_n) \in k[x_1, \ldots, x_{n-1}][x_n]=R[x_n]$.

  2. As $F'$ has coefficients in $R$, and $\deg F' < \infty$ exists $F'(x_n)$ has at most $\deg F'$ roots. As $k$ is infinite, exists $\lambda_n \in k$, $F'(\lambda_n) \not= 0$.

Step 2. is where I used the hypothesis that $k$ is infinite and lemma.

  1. Now $F'(\lambda_n) = F_1 \in k[x_1, \ldots, x_{n-1}]$. Then exists $\lambda_1, \ldots, \lambda_{n-1} \in k$, $F_1(\lambda_1, \ldots, \lambda_{n-1}) \not= 0$. Inductive step is complete.
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  • $\begingroup$ Take $n=1$, $k=\mathbb{F}_2$ and $F=x(x-1)$. Then in this scenario the proposed statement is not true. So I think it needs some more assumptions. $\endgroup$ – asdq Apr 26 '18 at 8:19
  • $\begingroup$ Is $k$ algebraically closed? We know that $x^{2}+1\in \mathbb{R}[x]$ doesn't have any solution in $\mathbb{R}$. $\endgroup$ – Seewoo Lee Apr 26 '18 at 8:57
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    $\begingroup$ @asdq Yes, I forgot to add that $k$ is an infinite field. See-Woo Lee, I am not proving existence of solution, but rather something which doesn't annihilate the polynomial. $\endgroup$ – CL. Apr 26 '18 at 9:15
  • $\begingroup$ It seems you forgot to assume $F\ne0$. $\endgroup$ – user26857 Apr 28 '18 at 8:45

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