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.

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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 $m^c$.

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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 $x$.

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