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

Let $B$ be a commutative unitary reduced ring and let $A$ be a subring of it. Let $e$ be an idempotent of $B$. Then we have a natural surjective ring homomorphism $A\rightarrow Ae$ defined by $a\mapsto ae$. $Ae$ is just the ring with $e$ as unity and multiplication, addition induced from $B$ (i.e. $ae\cdot be = abe$ and $ae+be = (a+b)e$).

The question is.. How much do we know about Spec $Ae$ ?

Edit: Well I realize that the prime ideals of $Ae$ should be of the form $\mathfrak{p}e$ for some prime ideal $\mathfrak{p} \in$ Spec $A$. But how much do we know about the topology of Spec $Ae$?

Edit: Apparently I won't know much about Spec $Ae$ unless I know more about $B$. In my specific problem, I wouldn't mind $B$ to be a von Neumann regular (i.e. zero dimensional) ring of fractions of $A$ (i.e. for all $b\in B\backslash${0} there is an $a$ $\in A$ such that $ab$ $\in A\backslash${0}) with extremally disconnected spectrum (i.e. closure of any open set in Spec $B$ is open). I just didn't wanted to add this extra condition to avoid confusion. My more specific question was whether the ring $A[e]$ could have an extremally disconnected minimal prime spectrum if the minimal prime spectrum of $A$ werent exd, for that it suffices for me to know this for $Ae$.

share|improve this question
    
If you are asking a question about commutative algebra or algebraic geometry, it's fair to say that you don't need to mention that the rings are commutative (and you certainly don't need to mention that they are unital with unital homomorphisms). As far as I'm concerned, Grothendieck's long list of assumptions at the beginning of EGA makes things sufficiently clear (I'd never even heard of nonunital modules before that...). –  Harry Gindi Jul 24 '10 at 12:50
5  
The tags arent really something every reader look at when they read a post. I'd prefer to make things clear. –  Jose Capco Jul 24 '10 at 13:56
4  
You begin with a dominant map ${\rm{Spec}}(B) \rightarrow {\rm{Spec}}(A)$ and are asking what can be about the closure of the image of a chosen open & closed subset of ${\rm{Spec}}(B)$. That sounds quite hopeless without further information, since the dominance could be entirely due to the complementary clopen set in Spec($B$), in which case you're asking to describe the closure of the image of an arbitrary affine mapping to ${\rm{Spec}}(A)$; could be anything at all. –  BCnrd Jul 24 '10 at 15:25
add comment

4 Answers

Let's agree, with Harry, that all rings are commutative and unital, and that ring maps preserve the $1$, and that subrings must contain the $1$. So for example $Be$ and $Ae$ are not subrings of $B$. Thus it might lead to confusion to call them rings.

An idempotent of $B$ makes for an isomorphism between $B$ and a product $(B/Be)\times (B/(B(1-e))$, and conversely every isomorphism $B=B'\times B''$ arises in this way (with $e$ corresponding to $(1,0)$). So let's change notation and say that $A$ is a subring of a product $B\times C$. The question is, what can we say about ($Spec$ of) the image of $A$ in $B$?

That's not a very precise question, but here's a way of looking at the given data: Choosing a subring of the product $B\times C$ of two rings corresponds precisely to choosing (1) a subring $B_0\subset B$ (the image of $A$ under projection), (2) a subring $C_0\subset C$ (the image of $A$ under the other projection), (3) an ideal $I$ of $B_0$ (the elements $b$ such that $(b,0)\in A$), (4) an ideal $J$ of $C_0$ (the elements $c$ such that $(0,c)\in A$), and (5) an isomorphism between $B_0/I$ and $C_0/J$.

Geometrically, where $B$ and $C$ corresponded to disjoint objects, $B_0$ and $C_0$ are stuck together along isomorphic closed subobjects.

So the question, such as it is, is: Given that data what can we say about $Spec$ of $B_0$ in relation to $Spec$ of $A$.

share|improve this answer
add comment

In general, if $f:R\to S$ is a surjective ring homomorphism, then the map Spec $S\to$ Spec $R$ is a closed embedding (in particular, it's injective).

Edit: I missed that $e$ may not necessarily be in $A$; the following assumes that it is.

If $e$ is an idempotent then so is $1-e$, and $A$ decomposes as a product $A = Ae \times A(1-e)$. From the perspective of schemes, this product turns into a coproduct, and Spec $A$ is the disjoint union of Spec $Ae$ and Spec $A(1-e)$. The schemes of the form Spec $Ae$ are hence those obtained as a union of some subset of the connected components of Spec $A$.

share|improve this answer
    
e isnt necessarily in A. In fact it wouldnt be interesting if it were in A. Ae x A(1-e) is then A[e] not A. –  Jose Capco Jul 24 '10 at 16:32
add comment

Edit: Well I realize that the prime ideals of $Ae$ should be of the form $\mathfrak{p}e$ for some prime ideal $\mathfrak{p} \in$ Spec $A$. But how much do we know about the topology of Spec $Ae$?

$A \to B$ induces an isomorphism $A/(A \cap (1-e)B) \to Ae$; the ideal may be also described as $Ann_A(e)=\{a \in A : ea=0\}$. Thus $Spec(Ae)$ is a closed subscheme of $Spec(A)$, namely the closed image of the composition $V(1-e) \subseteq Spec(B) \to Spec(A)$. As Brian Conrad pointed out, nothing much can be said about this closed subscheme. [I just add this because you did not seem to identify the topology]

share|improve this answer
add comment

if R be reduced ring then every idempotent of R is central?

share|improve this answer
    
If you have a new question, please try the button in the upper right corner. –  Julian Kuelshammer Jun 25 '13 at 7:22
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.