# M_2(k) as a central extension

Does there exist a field $k$ and a subring $R$ of $S = M_2(k)$ such that $R$ is not finitely generated over its center, $S=kR$ and $1_R = 1_S$? ($S$ is the algebra of $2 \times 2$ matrices over $k$.)

-

## 2 Answers

I think the answer is "yes". Let $A$ be a non-Noetherian integral domain (for example a polynomial ring in infinitely many variables over a field), let $I$ denote a non-finitely-generated ideal, and let $k$ be the field of fractions of $A$. Let $R$ denote the ring of $2\times 2$ matrices with coefficients in $A$ and with bottom left hand entry in $I$.

I think this ticks all the boxes. For example $kR=M_2(k)$ because I can scale any element of $M_2(k)$ until it's in $M_2(R)$ and then again so that all entries are in $I$.

However, I don't think $R$ can be finitely-generated over its centre (which is easily checked to be $A$). For if $r_1,r_2,\ldots,r_n$ are finitely many elements of $R$ then the ring they generate over $A$ will be contained in the $2\times 2$ matrices with coefficients in $A$ and bottom left hand entry in $J$, the finitely-generated ideal generated by the bottom left hand entries of the $r_i$, and this is a proper subset of $I$.

-
Nice example, thanks. –  iravan Nov 20 '10 at 0:59

This question has been explored in the context of polynomial identity rings (PI-rings). Your hypothesis implies that $R$ is a prime PI-ring of PI-degree 2 (see below). However, the main structure theorems of PI-theory, Kaplansky's Theorem, Posner's Theorem, Artin-Procesi Theorem and central polynomials, are in the opposite direction. They show that if a ring $R$ satisfies a polynomial identity plus a suitable further hypothesis, then $R$ is, or almost is, a finite module over its center, or at least has a large center.

One example is the strong form of Posner's theorem using central polynomials: Let $R$ be a prime PI-ring with center $C$, and let $T = C - \{0\}$. Then for some integer $n$, $T^{-1}R$ is a central simple algebra of finite dimension $n^2$ over its center $T^{-1}C$.

The theory of central simple algebras says that if $A$ is finite dimensional central simple over a field $L$, then its dimension over $L$ is a square $n^2$, and there is a unit preserving $L$-algebra embedding into $M_n(k)$ for some extension field $k$ of $L$.

Combining this theory with Posner's theorem gives: A ring $R$ (with unit) is a prime PI-ring if and only if there is a field $k$, an integer $n$, and a unit preserving ring embedding $R \to M_n(k)$ such that $kR = M_n(k)$, where $R$ is identified with its image in $M_n(k)$.

The field $k$ is not unique, but the integer $n$ is. It is called the PI-degree of $R$. Thus your hypothesis: "$R$ is a subring of $S = M_2(k)$, where $k$ is a field, $S = kR$, and $1_R = 1_S$", implies that $R$ is a prime PI-ring of PI-degree 2.

One result showing that a prime PI-ring is close to being a finite module over its center is a theorem of mine (p. 174 in Drensky-Formanek, "Polynomial Identity Rings"): If $R$ is a prime PI-ring of PI-degree $n$ with center $C$, then there is a $C$-module embedding of $R$ into a free $C$-module of rank $n^2$.

As for your question, there is an example due to Cauchon (p. 228 in Rowen, "Polynomial Identities in Ring Theory") of a Noetherian prime PI-ring of PI-degree 2 which is not a finite module over its center. Cauchon also proved that if a prime PI-ring has the ascending chain condition on two-sided ideals, then it has the ascending chain condition on left and right ideals. In other words, left Noetherian, right Noetherian, and ACC on two-sided ideals are equivalent for prime PI-rings.

-