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This is probably a very silly question:

If $R$ is an arbitrary commutative ring with unit and $f\in R[X]$ a polynomial, then for any element $a\in R$ we have $$f(a)=0 \Longleftrightarrow X-a ~\mbox{ divides }~ f \Longleftrightarrow f\in (X-a)$$ where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.

Now let $f\in R[X_1,\ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,\ldots,a_n)\in R^n$ that $$f(a_1,\ldots,a_n)=0 \Longleftrightarrow f\in (X_1-a_1,\ldots,X_n-a_n) ?$$ If this does not work in general, let $R=K$ be a field.

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Yes. Another way to see this is by the following:

We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=\varphi_0+\varphi_1X_n+\cdots + \varphi_kX_n^k$$ for some $\varphi\in R[X_1,...,X_{n-1}]$ with $\varphi_k(a_1,..,a_{n-1})\neq 0$. This reduces to the case you are okay with!

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Consider the case of $(a_1,\ldots,a_n)=(0,\ldots,0)$ first, where it's obvious. Then observe that $$f(a_1,\ldots,a_n)=0\iff g(0,\ldots,0)=0$$ and $$f\in (x_1-a_1,\ldots,x_n-a_n)\iff g\in (x_1,\ldots,x_n)$$ where $$g(x_1,\ldots,x_n)=f(x_1+a_1,\ldots,x_n+a_n)$$

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