I'm self-learning commutative algebra following "Introduction to Commutative Algebra". When dealing with concepts like "contraction" and "extension", some exercises in this book don't specify which homomorphism it uses and make it hard to understand, like this problem:

Let $A$ be a ring and let $A[[x]]$ be the ring of formal power series with coefficients in $A$. Show that the contraction of a maximal ideal $m$ of $A[[x]]$ is maximal ideal of $A$, and $m$ is generated by $m^c$ and $x$. (Here $m^c$ is the contraction of $m$.)

In this problem, the author doesn't specify the homomorphism he uses. Is it the inclusion mapping $f(a) = a$??? And please help me to solve this problem, too. Thanks. I really appreciate.

  • 2
    $\begingroup$ 1) You should name the authors of your book 2) What you highlight is your condensation of the problem. The authors split their exercise in five parts and what you quote is only the fourth. If you want users to help you, don't make it more difficult for them than it already is. $\endgroup$ – Georges Elencwajg Apr 18 '14 at 7:04
  • $\begingroup$ Ok, sorry for not mentioning the author. I will try to be careful later. Thanks $\endgroup$ – le duc quang Apr 18 '14 at 14:01

The statement of the problem talks about the contraction to $A$ of a maximal ideal $m$ of $A[[x]]$. So it is understood that the underlying ring homomorphism is the inclusion $f: A \rightarrow A[[x]]$. To solve the problem, try to understand the structure of $m$. In particular, show that the units of $A[[x]]$ are precisely the power series with a constant term $a_0$ that is a unit of $A$, i.e. $a_0 \in A^*$. So an element of $m$ must have constant term that is not a unit of $A$ and thus it must belong to some maximal ideal of $A$.

  • $\begingroup$ Sorry for asking a silly question but there's some point still unclear for me. With the inclusion $f$, so is it true that image of $f$ is the power series having only constant term. So is it possible for $m$ not containing power series having only constant term, and then the $m^c$ is null??? $\endgroup$ – le duc quang Apr 19 '14 at 17:14
  • $\begingroup$ If $m$ contains only power series that have zero constant term, then $m$ is not maximal. So if i understand your question, the answer is no. $\endgroup$ – Manos Apr 20 '14 at 17:00

The map is the inclusion map $f:A \mapsto A[[x]]$ which simply sends $a$ to the formal series $a$. I'll use the following Lemma which is easy to prove.

Lemma(Criterion for maximal ideal):If $m$ is a maximal ideal of $R$, then for every $a \notin m$, there exists some $r \in R$ such that $1+ra \in m$.

We have to show that if $M \subset A[[x]]$ is a maximal ideal, then the pullback $f^{-1}(M)$ is a maximal ideal of $A$. Following the note of Manos' answer, note that $x \in M$ because if not, then by the Lemma, for some $g\in A[[x]]$, we have $1+xg \in M$. Since $1$ is a unit, $1+xg$ is a unit, a contradiction. So, every element of $M$ is of the form $\sum a_{n}x^{n}$ where $a_{0}$ is not a unit. Now, assume that $f^{-1}(M)$ is not a maximal ideal.Pulling back the maximal ideal, we find that there must be some $a \notin f^{-1}(M)$ such that $f^{-1}(M)+(a) \subset A$.

$\Rightarrow M+(a)=A[[x]] \Rightarrow 1=m+ra$ for some $r\in A[[x]]$ and $m \in M$. $r$ must be degree 0(why?) i.e in the image of the inclusion. So, $m$ is the image of the inclusion $f$ too. But this means that $f^{-1}(M)+(a)$ contains $1$, a contradiction.


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