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Let $\varphi:(A,\mathfrak{m})\to(B,\mathfrak{n})$ be a local morphism of regular local $\mathbb{C}$-algebras (of the same dimension) which makes $B$ integral over $A$.

Let $\mathfrak{m}=(x_1,\ldots,x_d)$ be a (fixed) regular system of parameters of $A$ and assume that there exist principal prime ideals $\mathfrak{q}_i\subset B$ such that $\mathfrak{q}_i\cap A=(x_i)$. By going up, $\sum_i\mathfrak{q}_i=\mathfrak{n}$. Is it possible to choose a regular system of parameters $\mathfrak{n}=(y_1,\ldots,y_d)$ for $B$ such that $\varphi(x_i)=y_i^{n_i}$ for certain integers $n_i$ and such that $y_i$ generats $\mathfrak{q}_i$? If not, can I add any conditions that will give me such a result?

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I don't understand the "By going up..." part. Doesn't Karl's example show that the sum of the $\mathfrak{q}$'s need not be maximal? – Graham Leuschke Aug 16 '11 at 12:54
$\mathfrak{q}_i$ lies over $(x_i)$, so $\mathfrak{a}:=\sum_i\mathfrak{q}_i$ lies over $\mathfrak{m}$, so by going-up, we know $\dim(\mathfrak{a})=\dim(\mathfrak{m})$. Hence, $\mathfrak{a}$ must be maximal. Oh shoot. Do I need to require that $A$ and $B$ have the same dimension? Maybe I will just add it. – Jesko Hüttenhain Aug 16 '11 at 13:35
Look at Karl's example. $\mathfrak{q}_1$ is $\langle x^2-y^3 \rangle$, $\mathfrak{q}_2$ is $\langle y \rangle$ so $\mathfrak{a}$ is $\langle x^2-y^3, y \rangle = \langle x^2, y\rangle$, which is not maximal. – David Speyer Aug 16 '11 at 13:51
Hm. Yea that sounds about right? Where does my argument fail? – Jesko Hüttenhain Aug 16 '11 at 14:27
You never showed (and it isn't true) that $\mathfrak{a}$ is prime. – David Speyer Aug 16 '11 at 15:02
up vote 3 down vote accepted

Are you assuming that the $x_i$ are fixed? In that case, if one of the $x_i$ happens to be sent to something such that $B/x_i$ is reduced but singular, this seems to be a problem.

For example, consider $\mathbb{C}[[s,t]] \to \mathbb{C}[[x,y]]$ where $s$ is sent to $x^2 - y^3$ and $t$ is sent to $y$.

The only choice for the $n_i$ are to be $1$. But $B / (x^2 - y^3)$ isn't regular, so $(x^2 - y^3, y)$ can't generate the maximal ideal (as is obvious anyways).

In this case, if you are allowed to change $x_i$, then things are better, but I'm not sure if you want that or not.

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You are completely right - I was sloppy when translating from geometry and did not include one very important assumption; I edited the question. Sorry for that. – Jesko Hüttenhain Aug 16 '11 at 11:42

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