Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve a problem in the book Introduction to commutative algebra from Atiyah.

Page 11, exercise 5 (iv). Let $A[\![x]\!]$ be the ring of all formal power series of the form $\sum_{i=0}^{\infty} a_ix^i$. It is said that the contraction $m^c$ of a maximal ideal $m$ of $A[\![x]\!]$ is a maximal ideal of $A$ and $m$ is generated by $m^c$ and $x$.

There is a counter example. Let $A=\mathbb{Z}$, $p$ be a prime and $$m = \left\{ a_0+\sum_{i=1}^{\infty}a_ix^i \biggm| a_0 = kp, k\in \mathbb{Z}, a_i \in \mathbb{Z}\right\}.$$ Then $m$ is a maximal ideal of $\mathbb{Z}[\![x]\!]$. The contraction $m^c$ of $m$ is $\{kp \mid k\in \mathbb{Z} \}$. The smallest ideal containing $m^c$ and $x$ is $$K=\left\{\sum_{i=0}^{\infty}a_ix^i \biggm| a_i=k_i p, k_i \in \mathbb{Z}\right\}$$ which is smaller than $m$. It seems that $m$ is not generated by $m^c$ and $x$. I don't know where is the problem. Thank you very much.

share|cite|improve this question
up vote 2 down vote accepted

The ideal $$ I = \left\{\sum_{i=0}^{\infty}a_ix^i \biggm| a_i=k_i p, k_i \in \mathbb{Z}\right\} $$ doesn't contain $x$. Let $J$ be an ideal of $\mathbb Z[\![x]\!]$ containing $x$ and $m^c = p\mathbb Z$. Let $k \in \mathbb Z$ and $a_i \in \mathbb Z$ for $i \ge 1$. Then $$ kp + \sum_{i\ge 1} a_i x^i = kp + x \cdot \sum_{i\ge 0} a_i x^{i-1} \in m^c + x \cdot \mathbb Z[\![x]\!] \subseteq J $$ Hence $m \subseteq J$, by maximality $m = J$ and $m$ is the only ideal of $\mathbb Z[\![x]\!]$ containing $x$ and $m^c$.

share|cite|improve this answer
thank you very much. But the ideal K also contains $m^c$ and $x$. K is smaller than $m$. – LJR Nov 13 '12 at 8:15
@user9791 Your $K$ is my $I$. And it doesn't contain $x$! It contains $px$, and $2px$, but not $x$ (as $a_1$ has to be a multiple of $p$ and $1$ isn't). – martini Nov 13 '12 at 8:17
thank you very much. – LJR Nov 13 '12 at 8:24

The ideal you say is the smallest containing $x$ and $m^c$ do not contain $x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.