MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R$ be an associative ring with 1 and suppose that $Q$ is a central extension of $R.$ I'd like to know how the ring structure of $Q$ and $R$ are related. For example, it's easy to see that if $Q$ is a prime ring, then $R$ is prime too. Now, suppose that $Q = I \oplus J$ for some ideals $I, J$ of $Q$ where $I$ is a central prime ideal of $Q.$ Would this imply $R=(R \cap I) \oplus (R \cap J)$? What if I also add this condition that $Q$ is the classical (left) quotient ring of $R$? Thanks.

share|cite|improve this question
At least without the extra assumption that $Q$ is the quotient ring, the answer is "no". Consider $Q=\mathbb{Z}/p\mathbb{Z}\oplus\mathbb{Z}/p\mathbb{Z}$ with $R=\mathbb{Z}/p\mathbb{Z}$ diagonally embedded into $Q$, so that both intersections are trivial. – Alex B. Jan 5 '11 at 7:25

In general, the answer is No. Certainly Alex's example above shows that the condition that $Q$ is the classical left quotient ring of $R$ is necessary for your property to hold. Here is a (commutative) example which shows that it is not sufficient.

Let $k$ be any field, let $R = k[x,y] / \langle xy\rangle$, and let $k(a)$ be the field of rational functions in one variable. Let $Q = k(a) \oplus k(a)$ be the direct sum of two copies of these, and consider the map $\varphi$ $: R \to Q$ which sends $x$ to $(a,0)$ and $y$ to $(0,a)$.

Now $Q = I \oplus J$ where $I = k(a) \oplus 0 = Q\varphi(x)$ and $J = 0 \oplus k(b) = Q\varphi(y)$. This shows that $xR \subseteq \varphi^{-1}(I)$. Since $R$ is spanned over $k$ by powers of $x$ and powers of $y$, $R = xR \oplus k[y]$ as a $k$-vector space. If $\varphi( f(y) ) = (0, f(a)) \in I$ for some polynomial $f[t] \in k[t]$, then $f = 0$ because $k[a] \hookrightarrow k(a)$. This shows that $\varphi^{-1}(I) = xR$. Similarly $\varphi^{-1}(J) = yR$.

Next, $\ker \varphi = \varphi^{-1}(I \cap J) = xR \cap yR = 0$, so we can view $\varphi$ as an embedding of $R$ into $Q$. If we identify $R$ with its image in $Q$ using $\varphi$, then $I \cap R = xR$ and $J \cap R = yR$ by the above paragraph, but $xR + yR$ is not the whole of $R$ (since the ideal $\langle x,y \rangle$ of $k[x,y]$ is proper). Thus $R$ properly contains the direct sum $(I \cap R) \oplus (J \cap R)$.

Let $S$ be the set of regular elements of $R$, that is, the set of non-zero-divisors. We will show that $Q$ is the classical (left and right) quotient ring of $R$. To do this, we have to prove that (a) $\varphi(S)$ consists of units in $Q$, and (b) every element of $Q$ has the form $\varphi(r) \varphi(s)^{-1}$ for some $r \in R$, $s \in S$.

For (a), note that $S$ is precisely $R \backslash (xR \cup yR)$; so if $s \in S$ then $\varphi(s) = (v,w)$ for some non-zero $v,w \in k(a)$. But then $(1/v,1/w) \in Q$ is an inverse for $\varphi(s)$.

For (b), let $(v,w) \in Q$ and write $v = f(a)/h(a), w = g(a)/h(a)$ for some polynomials $f,g,h \in k[t]$ with $h \neq 0$ (you can choose the same denominator $h$ by passing to a common denominator). Then

$(v,w) = \frac{ \varphi(xf(x) + yg(y)) }{ \varphi((x+y) h(x+y))} = \varphi(r) \varphi(s)^{-1}$

with $r = xf(x) + yg(y) \in R$ and $s = (x+y) h(x+y) \in S$.

What's happened here is that the classical ring of quotients $Q$ of $R$ (also known in the commutative literature as the "total ring of quotients") has orthogonal idempotents $\frac{x}{x+y}$ and $\frac{y}{x+y}$ which don't lie in $R$; this stops $R$ from breaking up into a direct sum of two ideals.

Geometrically, $Spec(Q) = \{xQ, yQ\}$ is disconnected, but $Spec(R)$ is not (it's the union of two intersecting lines in $\mathbb{A}^2_k$). This is a fairly common phenomenon.

There are more complicated, but similar, non-commutative examples, too.

Chapter 2 of the book "Noncommutative Noetherian Rings" by McConnell and Robson has information that you may find useful. See in particular Proposition 2.1.16.

share|cite|improve this answer
Nice! Welcome to MO, Костя! – Alex B. Jan 5 '11 at 15:07

Your Answer


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.