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Let $k$ be a field and $A^{n}$ denote the affine $n$ space over $k$, consisting of all $n-$tuples with entires from $k$ that is, $A^{n}:=\{(a_{1},\ldots,a_{n})|a_{i}\in k\}$

We know every polynomial $f\in k[x_{1},\ldots,x_{n}]$ defines a map $$f:A^{n}\longrightarrow k$$ given by $$(a_{1},...,a_{n})\longmapsto f(a_{1},...,a_{n})$$

My question is the following,

When several polynomials define the same mapping ?

My attempt:

Suppose our base field has infinite elements. Let $P$ and $Q$ be two polynomial which define the same map. This is equivalent to saying that $P(a_{1},\ldots,a_{n})=Q(a_{1},\ldots,a_{n})$ for all $(a_{1},\ldots,a_{n})\in A^{n}$

Consider the polynomial $H=P-Q$, since $P$ and $Q$ take same value for each point in affine space $H$ has infinitely many roots (as there are infinitely many elements in $k$, so infinitely many points in $A^{n}$). But every polynomial of given degree $n$ has atmost $n$ roots. So the only choice for $H$ is $0$ hence we conclude (by comparing the coefficient ) that $P$ and $Q$ are same.

But what about the case when there are only finitely many elements in our field.

I am pretty much sure that we can find several polynomials that define the same mapping.

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    Absolutely. In F_p x^p=x. – 2017-02-12
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    @Tensor_Product: Your argument for the infinite case requires a little more care. When you said "every polynomial of degree $n$ has at most $n$ roots in $k$", that claim relates to _univariate_ polynomials. But your polynomial $f$ is multivariate. – 2017-02-12
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    @quasi, Oh thanks, I did not noticed it. What will be the correct and more precise argument ? – 2017-02-12
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    @Tensor_Product: Induction on the number of variables. – 2017-02-12
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    In the book *Elementary algebraic geometry, Klaus Hulek* page 2, has written "Unless $k$ has infinitely many elements, for example, when $k$ is algebraically closed, several polynomial can define the same map " – 2017-02-12
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    If you use scheme-theoretic language (ie $\mathbb{A}^n_k=\operatorname{Spec} k[x_1,\cdots,x_n]$) then the answer is no even for $k$ a finite field- $\operatorname{Spec}$ can see points of your variety that belong to field extensions of your base field. – 2017-02-13

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