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Let $k$ be an algebraically closed field (of characteristic zero), and let $F_1,\ldots,F_m \in k[x_1,\ldots,x_n]$, for some $m,n$.

Let $I$ be the ideal in $k[x_1,\ldots,x_n]$ generated by $F_1,\ldots,F_m$, $I=\langle F_1,\ldots,F_m \rangle$ and assume that $I$ is a proper ideal of $k[x_1,\ldots,x_n]$.

Further assume that $\bar{a}=(a_1,\ldots,a_n) \in k^n$ is such that $F_1(\bar{a})=\cdots=F_m(\bar{a})=0$, namely, $\bar{a}$ is a common zero for $F_1,\ldots,F_m$.

As a proper ideal, $I$ is contained in some maximal ideal $M$ of $k[x_1,\ldots,x_n]$. Since $k$ is algebraically closed, Hilbert's Nullstellensatz implies that $M= \langle x_1-c_1,\ldots,x_n-c_n \rangle$, for some $c_1,\ldots,c_n \in k$.

Question: Is it true that $I \subseteq \langle x_1-a_1,\ldots,x_n-a_n \rangle$?

Perhaps this is related to one of the versions of Hilbert's Nullstellensatz. See this (page 2): I guess I am looking for a result of the following form: Let $I,J$ be two ideals of $k[x_1,\ldots,x_n]$. Then, $V(J) \subseteq V(I)$ implies that $I \subseteq J$ -- applynig this to our above $I$ and $J=\langle x_1-a_1,\ldots,x_n-a_n \rangle$, for which $V(J)=\{\bar{a}\}$.

Thank you very much!

user237522
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1 Answers1

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Yes, this is true. There is no need for your hypotheses on $k$, either. In fact, we have the following: for any ring $R$ and any $a_1,\dots,a_n\in R$, if $f\in R[x_1,\dots,x_n]$ satisfies $f(a_1,\dots,a_n)=0$, then $f$ lies in the ideal $I=\langle x_1-a_1,\dots,x_n-a_n\rangle$ of $R[\overline{x}]$. To see this, consider the quotient ring $R[\overline{x}]\big/I$. To show $f\in I$, it suffices to show $f\equiv_I 0$, where $\equiv_I$ denotes equivalence modulo $I$. But $x_i\equiv_I a_i$ for each $i\leqslant n$, and so, since addition and multiplication commute with $\equiv_I$, we have $$f=f(x_1,\dots,x_n)\equiv_I f(a_1,\dots,a_n)=0$$ as desired. (If this seems surprising, then it is perhaps more intuitive to think specifically of the case $a_1=\dots=a_n=0$. Then this fact amounts merely to saying that $\langle x_1,\dots,x_n\rangle$ consists precisely of those elements of $R[\overline{x}]$ with constant term $0$, which is hopefully not too surprising. One can recover the general case from this case by considering the unique $R$-algebra endomorphism of $R[\overline{x}]$ that takes $x_i$ to $x_i-a_i$ for each $i\leqslant n$; can you see how?)

Atticus Stonestrom
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