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Suppose I have two polynomials of equal degree $f, g \in \mathbb{C}[x_1, \ldots, x_n]$ that satisfy $f(\mathbf{a}) = g(\mathbf{a})$ for all $\mathbf{a} \in \mathbb{R}^n$. Does this imply $f = g$? I know this is true if the values are equal for all $\mathbb{C}^n$, but was not sure about this case. Thank you!

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  • $\begingroup$ Standard trick: Polynomials being equal means their difference is zero. Polynomials being zero is much easier to work with, generally. $\endgroup$
    – Arthur
    Jul 27 '20 at 11:51
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    $\begingroup$ If there are infinite sets $S_1,S_2,\ldots,S_n$ of $\mathbb{C}$ such that $f(\textbf{a})=g(\textbf{a})$ for all $\textbf{a}\in S_1\times S_2\times\ldots\times S_n$, then it already follows that $f\equiv g$. If you know that the degree of $x_i$ in both $f$ and $g$ is less than a certain positive integer $d_i$, you can even require that $|S_i|\geq d_i$ for each $i=1,2,\ldots,n$. $\endgroup$ Jul 27 '20 at 11:52
  • $\begingroup$ @Batominovski How do you prove this? $\endgroup$
    – Johnny T.
    Jul 27 '20 at 11:59
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    $\begingroup$ @JohnnyT. See Theorem 1.2 and Lemma 2.1 of this article: cs.tau.ac.il/~nogaa/PDFS/null2.pdf. $\endgroup$ Jul 27 '20 at 12:01
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Yes, this is true. For simplicity, let's study the polynomial $h=f-g$, as a function $\Bbb R^n\to\Bbb C$. It is constantly equal to $0$.

The function $h$ may be Taylor expanded at, say, the origin. As it is the zero function, the Taylor expansion is trivial. But any polynomial is equal to its finite Taylor expansion. Thus $h$ must be given by the zero polynomial.

One might object and say "But expanding $h$ to a polynomial function $\Bbb C^n\to \Bbb C$ might give you more room to manoeuvre and make other polynomials". To address this, note that the real partial derivatives of $h$ as a function $\Bbb R^n\to\Bbb C$ are always equal to the partial derivatives of $h$ as a function $\Bbb C^n\to\Bbb C$. So expanding the domain of $h$ in this way cannot change the Taylor expansion.

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I realized this after reading Arthur's answer.

Let us write $f = f_1 + i f_2$ and $g = g_1 + i g_2$ where $f_1, f_2, g_1, g_2 \in \mathbb{R}[x_1, \ldots, x_n]$. Then for any $\mathbf{a} \in \mathbb{R}^n$ the real part of $f(\mathbf{a})$ is $f_1(\mathbf{a})$ and the imaginary part $f_2(\mathbf{a})$ and similarly for $g$. Therefore $f_1 - g_1$ and $f_2 - g_2$ are real polynomials which vanish at all values of $\mathbb{R}^n$, hence they are the zero polynomials. Thus $f = g$.

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