When does a ring map $R\to S$ produce a group epimorphism $GL_n(R)\to GL_n(S)$?

Question

Let $R$ and $S$ be rings with $1$ (not necessarily commutative) and $f:R\to S$ a ring homomorphism preserving $1$. Let $\bar{f}$ be the ring map $M_n(R)\to M_n(S)$ given by $f$ acting on the matrix elements.

My question is, what is the most general condition on $f$ for $\bar{f}$ to be a surjective group homomorphism $GL_n(R)\twoheadrightarrow GL_n(S)$? I think this is true if $f$ is an isomorphism but I can't be sure.

Many thanks for your help.

Answer 1

It is not true that if $f : R \to S$ is surjective, then so is $\text{GL}_n(f) : \text{GL}_n(R) \to \text{GL}_n(S)$. In fact this is already false for $n = 1$: for example, the quotient map $f : \mathbb{Z} \to \mathbb{Z}_5$ has the property that the image contains units which don't lift to units in $f$, namely $2$ and $3$. For higher $n$, no matrix in $\text{GL}_n(\mathbb{Z}_5)$ with determinant $2$ or $3$ can lift to a matrix in $\text{GL}_n(\mathbb{Z})$, which must have determinant $1$ or $-1$.

Fortunately, you don't need this fact to solve exercise I.1.11. You can use the identity given in the exercise instead.

Answer 2

Partial answer

If we have an homomorphism $g:S\to R$ such that $f\circ g:R\to S\to R$ is the identity, $g$ induces $\bar{g}$ such that $\bar{f}\circ\bar{g}:\mathrm{GL}_n(R)\to\mathrm{GL}_n(R)$ is the identity, and so $\bar{f}$ is surjective. For example, the map $k[x]\to k$ sending $x$ to $0$ clearly has a section $k\to k[x]$. I suspect that this is also necessary when $R$ is reduced. Surely, there are counterexamples when $R$ is not reduced.

In fact, let $R$ be $k[\sigma]$ with $\sigma^3=0$ and $S$ be $k[\epsilon]$ with $\epsilon^2=0$, and $f:R\to S$ the homomorphism sending $\sigma$ to $\epsilon$. Moreover, call $h:S\to R$ the function (not an homomorphism!) sending $a+b\epsilon$ to $a+b\sigma$. An element $a+b\epsilon$ (or $a+b\sigma+c\sigma^2$) is invertible if and only if $a$ is invertible, and hence if $A$ is an invertible matrix in $\mathrm{GL}_n(S)$, also $\bar{h}(A)$ is invertible. This proves that $\mathrm{GL}_n(R)\to\mathrm{GL}_n(S)$ is surjective. Still, there are no sections $g:S\to R$ of $f$.

Let us suppose that $g$ exists. Call $g(\epsilon)=\alpha+\beta\sigma+\gamma\sigma^2$. Since $fg(\epsilon)=\epsilon$, we have $\alpha=0$ and $\beta=1$, but then $g(\epsilon)^2=g(\epsilon^2)=0$ implies $\beta^2=0$, absurd.

Answer 3

Here are some standard examples: Consider $a$, $b$ integers $>0$ so that $b \mid a \ $. We have the canonical surjective morphism of rings $$\mathbb{Z}/(a) \to \mathbb{Z}/(b)$$

and the induced morphism of groups

$$GL_n(\mathbb{Z}/(a)) \to GL_n(\mathbb{Z}/(b))$$

is surjective.

Even for $n=1$ this is a good exercise.

The general case follows from the surjectivity of

$$SL_n(\mathbb{Z}) \to SL_n(\mathbb{Z}/(b))$$

and the $GL_1 (\cdot )$ case.

Also see http://en.wikipedia.org/wiki/Approximation_in_algebraic_groups