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

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?

• Is it clear that $k[t][[x]] \simeq k[[x]][t]$? Jun 14, 2012 at 23:08
• @DylanMoreland hmmm... I think they are not isomorphic. In $k[t][[x]]$, arbitrarily high powers of $t$ could occur. Is this the problem? Perhaps $k[t][[x]]/(tx-1)$ isn't a field after all? Jun 14, 2012 at 23:16
• ...of finite-tailed Laurent series in x with coefficients in k...Why? Jun 14, 2012 at 23:18
• @H.Kabayakawa I was thinking that the elements were only polynomial in $t$, so in the quotient the powers of $x^{-1}$ in a given element are bounded, but in light of Dylan's comment this isn't true, so I'm now thinking $k[t][[x]]/(tx-1)$ isn't a field... Jun 14, 2012 at 23:22
• @BenBlum-Smith One can see easily that a polynomial $f \in A[[x]]$ is a unit iff its constant term is a unit. In fact this is exercise 5(i) of AM on page 11. In your case, -1 is a unit so that $tx - 1$ is a unit in $k[t][[x]]$, so your example does not hold.
– user38268
Jun 15, 2012 at 4:08

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.

• Seems like it's because of your same point about $tx-1$. $xf\in (x)$, so $1-xf$ has unit constant term, thus is a unit, and $1-xf\in \mathfrak{m}$ implies $\mathfrak{m}=(1)$, a contradiction. Thanks! Jun 15, 2012 at 4:04
• @BenBlum-Smith Looks good. Do things work out from there? Jun 15, 2012 at 4:07
• @DylanMoreland Ben is looking at exercise 5(iv) of AM on page 11. The exercie you mention is 5(i) on the same page.
– user38268
Jun 15, 2012 at 4:10
• @BenjaminLim Ah, thanks. I'll add that (I'm away from my office for a few weeks, otherwise I'd check it myself!) Jun 16, 2012 at 8:02
• @DylanMoreland - yes, it's all worked out now. Jun 19, 2012 at 12:54