I saw this question: Is a polynomial $f$ zero at $(a_1,\ldots,a_n)$ iff $f$ lies in the ideal $(X_1-a_1,\ldots,X_n-a_n)$?
I was wondering if a polynomial $ f $ of a ring $ K [x_1, ..., x_n] $, where $ K $ is a field, is zero at $(a_1, ..., a_k, x_{k+1}, ..., x_n)$ then $f$ lies in the ideal $(x_1-a_1, ..., x_n-a_n)?$ I tried to adapt the answer given there, but no success. If this is not true, what can we say about the relationship between $f$ and the ideal $(x_1-a_1, ..., x_n-a_n)$?
One has a homomorphism $\phi:K[x_1,\ldots,x_n]\to K[x_{k+1},\ldots,x_n]$ given by $$\phi:f(x_1,\ldots,x_n)\mapsto f(a_1,\ldots,a_k,x_{k+1},\ldots,x_n).$$ You are asking its kernel is $I=(x_1-a_1,\ldots,x_k-a_k)$.
Yes it is. Each $x_i-a_i\in \ker\phi$, Conversely suppose $f\in\ker\phi$. Then $$f(x_1,\ldots,x_n)=f(x_1,\ldots,x_n)-f(a_1,\ldots,a_k,x_{k+1},\ldots,x_n).$$ We claim that $g(x_1,\ldots,x_n)-g(a_1,\ldots,a_k,x_{k+1},\ldots,x_n)\in I$ for any polynomial $g$. It suffices to prove this for $g=x_1^{i_1}\cdots x_n^{i_n}$. This is just a calculation: $$g(x_1,\ldots,x_n)-g(a_1,\ldots,a_k,x_{k+1},\ldots,x_n) =(x_1^{i_1}\cdots x_k^{i_k}-a_1^{i_1}\cdots a_k^{i_k})x_{k+1}^{i_{k+1}}\cdots x_n^{i_n}$$ and the bracketed expression is in $I$. To see this set $y_i=x_i-a_i$ and in terms of $y_i$, the bracket has zero constant term.