Is a polynomial $f$ zero at $(a_1,\ldots,a_n)$ iff $f$ lies in the ideal $(X_1-a_1,\ldots,X_n-a_n)$?

Solution 1:

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 $$g(x_1,\ldots,x_n)=f(x_1+a_1,\ldots,x_n+a_n)$$

Solution 2:

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 $\varphi_j(a_1,..,a_{n-1})\neq 0$ for some $j\leq k$. This reduces to the case you are okay with!