Pullback of maximal ideal in $k[y]$ is not maximal in $k[x]$.

Let $k$ be a field and let $k[x]=k[x_1,\ldots,k_m]$ and $k[y]=k[y_1,\ldots,y_n]$ be $k$-algebras. Let $\varphi:k[x]\to k[y]$ be a $k$-algebra homomorphism. It can be shown (pretty readily) that $\varphi$ is determined by polynomials $f_1,\ldots,f_m\in k[y]$ which tell us where to send the $x_i$.

In the case where $k$ is algebraically closed, as Hilbert's Nullstellensatz and the associated algebraic geometry thereabouts, the pullback of every maximal ideal in $k[y]$ is maximal in $k[x]$. However, if $k$ is not algebraically closed, then there should be a counterexample, i.e. some maximal ideal in $k[y]$ that does not pull back maximally along $\varphi$.

It's pretty clear that the case of $k[x]=k[t]=k[y]$ for one variable $t$ is not very fruitful, so I've been trying to come up with an example in $\mathbb R[x,y]\to\mathbb R[x,y]$ using $(x^2+1,y)$ or some other ideal. Also suggested online (in weird contexts) was the ideal $(x^2+y^2+1)$, but I can't see how that's going to help.

In theory I'd like $\mathbb R[x,y]/\mathfrak m\cong\mathbb C$ to pull back to $\mathbb R[x,y]/\varphi^{-1}(\mathfrak m)\cong\mathbb R[t]$, but I've been unable to succeed. Thanks for the read.

PS: this is a homework question, and in theory the answer could be 'the pullback is always maximal', but I don't think so.

• In general, if $\varphi:A\to B$ is a $k$-algebra homomorphism, $A$ and $B$ f.g. $k$-algebras, then maximals pull back to maximal. So $A/\varphi^{-1}(\mathfrak{m})\hookrightarrow B/\mathfrak{m}$. But, by Zariski's lemma $B/\mathfrak{m}$ is a finite dimensional $k$-space, and so $A/\varphi^{-1}(\mathfrak{m})$ is f.g. over $k$, but since it's an integral domain this implies it's a field. I hope I've understood your question correctly. – Alex Youcis Oct 8 '13 at 7:57
• I meant "and so $A/\varphi^{-1}(\mathfrak{m})$ is a finite $k$-space". @JeskoHüttenhain $t^2+1\mapsto 1-y$, right? – Alex Youcis Oct 8 '13 at 8:09
• @AlexYoucis: Thanks ;) – Jesko Hüttenhain Oct 8 '13 at 8:10

You're in for a big surprise:

The pull back of any maximal ideal $\mathfrak n\subset k[y_1,\ldots,y_n]$ under a $k$-algebra morphism $\phi: k[x]\to k[y]$ is a maximal ideal $\mathfrak m=\phi^{-1}(\mathfrak n)\subset k[x]$ for $k$ a completely arbitrary field, possibly not algebraically closed.
This is a non trivial theorem (using Zariski's version of the Nullstellensatz) , best seen in the following context:

A ring $A$ is said to be a Jacobson ring if every prime ideal of $A$ is the intersection of the maximal ideals which contain it.
For example a field $k$ or $\mathbb Z$ are trivially Jacobson rings and, much less trivially, a finitely generated algebra over a Jacobson ring is Jacobson: this implies that your rings of polynomials $k[x],k[y]$ are Jacobson.
The property you need to conclude is the following theorem:
Given a morphism of finite type $\phi:A\to B$ between Jacobson rings, the inverse image of any maximal ideal $\mathfrak n\subset B$ is a maximal ideal $\phi ^{-1} (\mathfrak n)=\mathfrak m\subset A$

Bibliography
Jacobson rings are not very present in the standard textbooks.
You can find all of the above and more in Bourbaki's Commutative Algebra, chapter V, $\S$3.4 or in the Stacks Project Chapter 9, §32.

• (+1) They are also in Pete L Clark's notes, I believe. – Alex Youcis Oct 8 '13 at 8:12
• In particular, check on page 187 here: math.uga.edu/~pete/integral.pdf – Alex Youcis Oct 8 '13 at 9:02
• @Alex: Yes; thanks for the plug. I looked last night to see whether this particular application is in my notes, and it seems to me that it isn't. It would make a good exercise. – Pete L. Clark Oct 8 '13 at 14:40
• Jacobson rings and the generalised Nullstellensatz are also discussed in Eisenbud's textbook. – Zhen Lin Oct 8 '13 at 14:59
• I'll be damned. Thanks a lot. – Ian Coley Oct 8 '13 at 19:25