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I have an ideal in $k[[x,y]]$, and I know that it contains $x$ but isn't $\langle x \rangle$ (here $k$ is a field, maybe not alg closed). This means that my ideal must be of the form $\langle x ,y^n \rangle$ for some $n$.

This is not hard to prove, but a proof isn't what I'm looking for. What I want is the suitable commutative algebra lingo: how do I explain this fact in as few words as possible? What more general principle is at work?

Tim
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    The ideals of $k\left[\left[x,y\right]\right]$ containing $x$ are in 1-to-1 correspondence with ideals of $k\left[\left[x,y\right]\right] / \left(x\right)$. But $k\left[\left[x,y\right]\right] / \left(x\right) \cong k\left[\left[y\right]\right]$ (this follows from $k\left[\left[x,y\right]\right] = \left(k\left[\left[y\right]\right]\right)\left[\left[x\right]\right]$), so we are looking for the ideals of $k\left[\left[y\right]\right]$. And that should be well-known. – darij grinberg Feb 19 '15 at 23:41

1 Answers1

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The general principles are:
(1) the correspondence theorem for ideals, and
(2) the isomorphism $R[[X]]/(X)\simeq R$ for any commutative ring $R$.

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