Suppose that $k$ is an algebraically closed field. Then, we know from Hilbert's (weak) Nullstellensatz that the maximal ideals of $k[x_1, \ldots, x_n]$ are of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for some $a_i \in k$. Now consider the ring $A = k[x_1, \ldots, x_n]/(f_1, \ldots, f_r)$ for some $f_i \in k[x_1, \ldots, x_n]$.

Since the maximal ideals of the original polynomial ring correspond to points of $k^n$ and and maximal ideals of $A$ correspond to the maximal ideals of the original polynomial ring containing $(f_1, \ldots, f_r)$, it seems to make sense that the maximal ideals of $A$ correspond to points of $k^n$ such that $f_1 = \cdots = f_r = 0$.

It's not hard to show that $f_i(a_1, \ldots, a_n) = 0$ if $f_i \in (x_1 - a_1, \ldots, x_n - a_n)$. But I'm having trouble proving the converse directly.

So far, I've tried to use an induction argument on the number of variables and using the fact that $x_n - a_n | f_i(a_1, a_2, \ldots, a_{n - 1}, x_n)$, but I haven't been able to finish showing this.

Are there any suggestions for next steps to take? Is what I'm trying to prove even true?

Note: Didn't see that question just boils down to what was pointed out as a possible duplicate as mentioned in the comment and seen in the answer. But I don't think this question is completely identical (though it may be tautological/repetitive in retrospect).


This is probably easier than you think. Consider the map $k[x_{1}, \ldots, x_{n}] \rightarrow k$ given by evaluating at $(a_{1}, \ldots, a_{n})$. The kernel of this map contains $(x_{1}-a_{1}, \ldots, x_{n}-a_{n})$ and is hence equal to the latter as it is maximal. So, what you want to show is that $(f_{1}, \ldots, f_{r})$ is contained in the kernel if and only if $f_{1}(a_{1}, \ldots, a_{n}) = \cdots = f_{r}(a_{1}, \ldots, a_{n}) = 0$. But this is tautological.


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