Does the strong form of Hilbert Nullstellensatz say how the maximal ideals look like?

Question

Let $k$ be a field such that for every proper ideal $J$ of $k[x_1;...;x_n]$ ; the ideal set of $V(J)$ is $I(V(J))=rad(J)$ (i.e. Hilbert's Nullstellansatz strong form holds ) . Then ; without using the fact $k$ is algebraically closed ; from the expression of the above mentioned ideal set ; can we conclude that any maximal ideal of $k[x_1;...;x_n]$ is of the form $(x_1-a_1 ; ... ; x_n - a_n )$ ?

Definitely ; if $m$ is a maximal ideal of $k[x_1;...;x_n]$ then $m=rad(m)=I(V(m))$ ; so it will be enough to prove that $I(V(m))= (x_1-a_1 ; ... ; x_n-a_n )$ for some $a_1;...;a_n \in k$ . But I am unable to prove that . Please help

Answer 1

I am not sure what you want here. Your assumption is never satisfied if $k$ is not algebraically closed:

• For $n = 1$ and $x := x_1$, it suffices to choose $f \in k[x]$ irreducible with no zeros in $k$. Then $I(V(f)) = I(\emptyset) = k[x]$, but $rad((f)) \neq k[x]$ (else $1 \in (f)$).
• For $n > 1$, choose $f$ as above, and consider $g(x_1,...,x_n) := f(x_1)\cdot ... \cdot f(x_n)$. Then $V(g)$ is again empty, so $I(V(g)) = k[x_1,...,x_n]$, but $rad(g)$ is again not equal to $k[x_1,...,x_n]$.

So it necessarily follows that $k$ is algebraically closed; in this case it is ok, but the logic is usually reversed: first you show the weak form of the Nullstellensatz, stating that maximal ideals of $k[x_1,...,x_n]$ correspond to points of $\mathbb A^n_k$, (here $k = \overline{k})$, and afterwards you prove the strong version using the weak one.