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Let $p$ be a prime and let $F=\mathbb{Z}/p\mathbb{Z}$. Can a nonzero multivariate polynomial $f\in F[x_1,....,x_n]$ such that $\mathrm{deg}_if< p$ for all $i=1,\ldots,n$ be identically zero as a polynomial function $F^n\to F$, where $\mathrm{deg}_if$ stands for the degree of $f$ in $x_i$?

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No. Induct on n. – Qiaochu Yuan Feb 22 '11 at 17:18
up vote 7 down vote accepted

No. The phenomenon of polynomials over fields which are not the zero polynomial but evaluate to the zero function is studied in loving detail in $\S 2.1$ of these notes: in particular, the result you want is Theorem 8.

I also agree with Qiaochu's comment: it would be possible to make much less of a production of this by simply inducting on the number of variables, beginning with the basic fact that no polynomial with coefficients in an integral domain can have more roots than its degree (except the polynomial which has all coefficients zero, if you take the convention that this polynomial has degree $-\infty$). I have my reasons for the more elaborate treatment given in my notes: in particular, an at least somewhat careful analysis of the situation leads directly to a proof of the Chevalley-Warning Theorem.

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Great notes! If I may pose a question (even though the original one was answered four years ago): in the proof of Proposition 6, why is it that $Q(t_n)$ has infinitely many roots in $R$? I suppose you could write $Q$ out like $Q(t_n)=\sum_kP_k(t_1,\ldots,t_{n-1})t_n^k=P(t_1,\ldots,t_n)$, but if $P(\alpha)=0$ for all $\alpha\in R^n$, how does that imply that $Q(x)=0$ for all $x\in R$? – Bryder Apr 3 '15 at 14:37

Hint $\ $ By induction the leading coefficient of $\rm\ f(x_n) \ $ is nonzero for some $\rm (a_1,\cdots,a_{n-1})\in F^{\,n-1}.\,$ Evaluating $\rm\:f\:$ at this point yields a univariate polynomial $\rm\ 0 \ne \bar f(x_n)\in \mathbb F_p[x_n]\ $ of degree $\rm< p.\, $ Some $\rm\,a\in \Bbb F_p\,$ isn't a root of $\rm\,\bar f,\,$ by a poly $\ne 0\,$ over a domain has no more roots than its degree, and $\rm\:\deg \bar f < p =|\Bbb F_p|.\,$ Therefore $\rm\, 0\ne \bar f(a) = f(a_1,\ldots,a_{n-1},a),\,$ hence $\rm\, f\neq 0.\ \ $ QED

Remark $\ $ In fact the above root-bound property completely characterizes integral domains: the ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in $\rm\:D.\:$ For the simple proof see my post here, where I illustrate it constructively in $\rm\: \mathbb Z/m\: $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd.\,$ The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m,\:$ for example, a square root of $\,1\,$ that is not $\:\pm 1$.

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