Prove that two multivariate polynomials that are equal at every point over the reals are identical as formal polynomials [duplicate]

Question

This is a basic question about polynomial rings, but I can't find the precise proof of the following fact.

Theorem: Let $f, g $ be two polynomials in $\mathbb{R}[\overline x]$, where $\mathbb{R}$ are the reals and $\overline x = (x_1,\dots,x_n)$. Assume that $f(\overline a) = g(\overline a)$, for all $\overline a\in\mathbb{R}^n$. Prove that $f = g $ as formal polynomials (i.e., the coefficient vector of their monomials is the same).

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Comment: over finite fields this is not necessarily the case. E.g., $x^2-x$ is not the zero polynomial but computes the zero function over $GF(2)$. See If two polynomials are equal as functions, are they necessarily equal as polynomials?.

Attempt 1: Use Schwartz-Zippel Lemma: Let $S\subseteq \mathbb{F}^n$. If $f-g$ is not always zero over $S$, then it has at most $|S|^{n-1}\cdot d~ $ roots from $S^n$ , with $d$ being the maximal (total) degree of $f-g$. But this doesn't seem to work, because even if we knew that $f-g$ is always zero over $S$, how can we conclude that $f-g$ is the zero polynomial (even for $S$ big enough)?

Comment 2: Note that if we change the formulation of Schwartz-Zippel Lemma to: If $f-g$ is not the zero polynomial, then it has at most $|S|^{n-1}\cdot d~ $ roots from $S^n$, then we finish the proof of the statement I'm after at. But this is not the formulation of Schwartz-Zippel Theorem I've seen.

Answer 1

Pick a polynomial $f\in \mathbb{R}[x_1, \dots, x_n]$ such that $f(\overline{a})=0$ for all $a\in \mathbb{R}^n$. We want to prove that $f$ is the zero polynomial. We can write $f=(a_\alpha)_{\alpha\in \mathbb{N}^n}$ (these are the coefficients of $f$ and this is really the definition of a formal multivariate polynomial).

Denote by $G: \mathbb{R}^n \rightarrow \mathbb{R}$ the polynomial function corresponding to $f$. We can write $$ G(\overline{x})= \sum_{\alpha\in \mathbb{N}^n \ : \ \vert \alpha \vert\leq d } a_\alpha \prod_{j=1}^n x_j^{\alpha_j},$$ where $d$ is the total degree of $f$ if $f$ is not the zero polynomial (in any case all $a_\alpha$ vanish for $\vert \alpha\vert >d$).

In order to show that $f$ is the zero polynomial, we just need to prove that $a_\alpha=0$ for all $\vert \alpha\vert = d$ (by definition of the total degree at least on of these coefficients needs to be nonzero).

Let $\vert \alpha\vert =d$, then we have, using that $d$ is the total degree of $f$, $$ \partial_1^{\alpha_1} \dots \partial_n^{\alpha_n} G(\overline{x})= a_\alpha \prod_{j=1}^n (\alpha_j!).$$ We could also just use Taylor's theorem to get this.

However, $G$ is the zero function by assumption and hence all partial derivatives are the zero function too. This yields $$ a_\alpha =\frac{0}{\prod_{j=1}^n (\alpha_j !)} =0.$$ This completes the proof.

Answer 2

If $f(x) = \sum_{\alpha} c_{\alpha} x^{\alpha}$, where $\alpha$ denotes a multiindex $(\alpha_1,...,\alpha_n)$, then $c_{\alpha}$ is given by the formula $$c_{\alpha} = {1 \over \alpha_1 ! ... \alpha_n !}{\partial^{\alpha_1 + ... + \alpha_n} f \over \partial_{x_1}^{\alpha_1}...\partial_{x_n}^{\alpha_n}}(0) $$ This can be proved rigorously using induction for example. Since this formula is determined by the values of $f(x)$, if two polynomials take the same value at every point, each coefficient $c_{\alpha}$ will be the same for the two polynomials.