Converse of Weak Nullstellensatz I have been trying to come up with an example of a polynomial ring where all maximal ideals are of the form $(x_i-a_i)$. But the ring $k$ on which $k[x_1,...,x_n]$ is based is not algebraically closed (for more than one variable). I need hints to come up with $k$. And will proceed to try to find an example from thereon .

Fix a positive integer $n$.

Let $k$ be a commutative ring with $1 \
e 0$, and let $R=k[x_1,...,x_n]$.

Suppose $k$ is such that

* $k$ is not algebraically closed.$\\\\[4pt]$
* All maximal ideals of $R$ are of the form $(x_1-a_1,...,x_n-a_n)$, for some $a_1,...,a_n\in k$.


Let $p\in k[t]$ be such that $p$ does not have a root in $k$.

Consider the principal ideal $I = (p(x_1))$ of $R$, and let $M$ be a maximal ideal such that $I\subseteq M$.

By assumption, we have $M=(x_1-a_1,...,x_n-a_n)$, for some $a_1,...,a_n\in k$.

Then, since $p \in M$, we can write $$p(x_1) = (f_1)(x_1-a_1) + \cdots + (f_n)(x_n-a_n)$$ for some $f_1,...,f_n \in R$.

Substituting $x_1=a_1,...,x_n=a_n$ in the above equation, we get $p(a_1)= 0$, contrary to our choice of $p$.

It follows that there is no such ring $k$.