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.

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 nonzerodivisors. 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 nonzero $v,w \in k(a)$. 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, noncommutative 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. 

