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Can you suggest a good name for a local homomorphism $(R,\mathfrak{m})\stackrel{\varphi}{\rightarrow}(S,\mathfrak{n})$ of local rings with the property that $\varphi(m)S$ is $\mathfrak{n}$-primary?

EDIT: Would you use 'map of finite length'? (because $S/f(\mathfrak{m})S$ has finite length over $S$?) Assume $R$ and $S$ are noetherian.

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Why not just call the morphism $n$-primary? – David White Sep 20 '11 at 22:31
@David: because it's not an ideal to be $n$-primary. It would be confused with an $n$-primary ideal, I think. – Mahdi Majidi-Zolbanin Sep 20 '11 at 23:04
$\mathfrak{n}$-secondary ? $\;\;$ – Ricky Demer Sep 21 '11 at 4:09
Primal? $\,\,\!$ – Dan Petersen Sep 21 '11 at 6:14
I don't think $\mathfrak n$-primary is too bad. If $\psi:A\to B$ is a morphism, and $\mathfrak a\subset A$ an ideal and $\mathfrak p\subset B$ prime, we can think of an ideal $\mathfrak a$ as $\mathfrak p$-primary relative to $\psi$ if the extension of $\mathfrak a$ along $\psi$ is $\mathfrak p$-primary. Now since $(A,\mathfrak m)$ and $(B,\mathfrak n)$ are local, if a subideal of $\mathfrak m$ is $\mathfrak n$-primary relative to $\psi$ then so is $\mathfrak m$ (uuh, right?). Perhaps $(\mathfrak m,\mathfrak n)$-primary? – Eivind Dahl Sep 21 '11 at 9:40
up vote 4 down vote accepted

One can notice that your hypothesis is equivalent to say that the closed fiber of $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ has dimension $0$ because this closed fibers is equal to $V(\varphi(\mathfrak m)S)$, and $V(I)$ of an ideal $I$ in $S$ is reduced to the closed point if and only the nilradical of $I$ is equal to $\mathfrak n$ which is equivalent to $I$ being $\mathfrak n$-primary as $\mathfrak n$ is maximal.

So one could call your homomorphism ''of relative dimension $0$'', except that I am not sure whether the same condition holds over the other prime ideals of $R$.

Don't you need some noetherian condition to get the finiteness of the length over $S$ ?

Update The dimension condition on the closed fiber doesn't propagate to other fibers. Let us consider the following example. Let $k$ be a field, $R=k[[x,y]]$, $S=k(z)[[x,y]]$ et let $\phi$ be the inclusion $R\subset S$. Then $\mathfrak mS=\mathfrak n$. Now I claim that the generic fiber of $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ has positive dimension. Otherwise, $S\otimes_R \mathrm{Frac}(R)$ would be a field. But this tensor product is the localization of $k(z)[[x,y]]$ with respect to the multiplicative subset $k[[x,y]]\setminus \lbrace 0 \rbrace$. It is easy to see that there is not enough denominators to make the localization into a field (e.g. $x+zy$ is not invertible in the localization).

So, one can only say "of relative dimension $0$ at the closed fiber". And yes this sounds too long.

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"of relative dimension zero" makes sense. Would you say "of relative dimension zero at the closed point", or does that sound too long? In my mind everything is noetherian :)) but you are right, I should have said that (I'll add it now.) – Mahdi Majidi-Zolbanin Sep 22 '11 at 19:32

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