Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
You never showed (and it isn't true) that $\mathfrak{a}$ is prime. –  David Speyer Aug 16 '11 at 15:02
show 2 more comments

1 Answer

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.

share|improve this answer
    
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
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.