Reduced rings, idempotents and their prime spectrum

Question

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 $Ae$.

Answer 1

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 $A$.

Answer 2

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 $A$.

Answer 3

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]

Answer 4

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