Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
add comment

2 Answers 2

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|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
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

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.