I am trying to prove 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 $x$.
I have made the following progress:
Suppose $x\not\in \mathfrak{m}$. Then $1-xg \in \mathfrak{m}$ for some $g\in A[[x]]$, and this is a contradiction because $1-xg$ is a unit. Therefore, $x\in \mathfrak{m}$. Consider any nonzero element $a\in A$. We want to show that there is $b \in A$ such that $ab \equiv 1 \ (\mathfrak{m}^c)$. Since $A[[x]]/\mathfrak{m}$ is a field, we know there is $f = \sum a_nx^n\in A[[x]]$ such that $af \equiv 1 \ (\mathfrak{m})$, and since $x\in \mathfrak{m}$, $aa_0 \equiv 1 \ (\mathfrak{m})$. How do I conclude from this that $aa_0 \equiv 1 \ (\mathfrak{m}^c)$ ?
Thanks.
Best Answer
Just remember what the symbols mean! By definition, $aa_0 \equiv 1 \ (\mathfrak{m})$ means $aa_0-1\in\mathfrak{m}$. But since $aa_0-1\in A$, this means $aa_0-1\in\mathfrak{m}^c$ as well.