The contraction of a maximal ideal of $A[[x]]$ is a maximal ideal of $A$?

Question

I am working on the problems in Atiyah and MacDonald's famous Introduction to Commutative Algebra. On p. 11, problem 5 part iv reads:

[Show that] the contraction of a maximal ideal $\mathfrak{m}$ of $A[[x]]$ is a maximal ideal of $A$, and $\mathfrak{m}$ is generated by $\mathfrak{m}^c$ and $x$.

Here $A$ is an arbitrary commutative ring with unity, and $A[[x]]$ (as usual) the ring of formal power series over $A$. By the contraction $\mathfrak{m}^c$, Atiyah and MacDonald mean the pullback of an ideal of $A[[x]]$ under the inclusion $A\hookrightarrow A[[x]]$; thus, if I am understanding correctly, $\mathfrak{m}^c=\mathfrak{m}\cap A$.

What's bugging me is that I am having trouble believing the claimed result, due to the following example:

Let $A=k[t]$, the polynomial ring over a field; so then $A[[x]]$ is $k[t][[x]]$, the ring of formal power series in $x$ with coefficients that are polynomial in $t$. Consider the ideal $(tx-1)$. Then $A[[x]]/(tx-1)$ is the field of finite-tailed Laurent series in $x$ with coefficients in $k$, right? In which case, $(tx-1)$ is a maximal ideal of $A[[x]]$, right? But, $(tx-1)\cap A=(tx-1)\cap k[t]=(0)$, since no element of $k[t]$ is a multiple of $tx-1$. But $(0)$ is not a maximal ideal of $A=k[t]$, and $(tx-1)$ is certainly not generated by $x$ and $(0)$. So doesn't this example violate the conclusion?

In a less legendarily tight book, I would assume there was a typo or an omitted assumption somewhere, but this is Atiyah and MacDonald, and they are never wrong. Conclusion: I must be missing something.

What am I missing? Is $A[[x]]/(tx-1)$ not actually a field? Is $(tx-1)\cap A$ not actually $(0)$? Did I misunderstand the definition of contraction? Or is it something else?

Answer 1

Maybe it's helpful to write your element as $-1 + tx$. Part (i) of the same exercise has you show that the units of $A[[x]]$ are the power series whose constant terms are units in $A$, so your ideal is the whole ring and the authors remain faultless.

The issue seems to be that $k[t][[x]]$ and $k[[x]][t]$ are not the same. My initial justification for this was that, as you noted, elements of the former ring can involve infinitely many powers of $t$. I'd be interested in seeing other reasons.

For the exercise proper, I think things clear up if you first prove that $x \in \mathfrak m$. If this is not the case, then there exists an $f \in A[[x]]$ such that $xf \equiv 1 \bmod\mathfrak m$, i.e., $1 - xf \in \mathfrak m$. Why is that impossible?

Added later: There's a nice discussion of $k[t][[x]]$ in Example 1.2 of Brian Conrad's handout. I'll try to incorporate some of the details here later on.