Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do one usually define the general linear group over a ring $R$, denoted by $\text{GL}(n,R)$. I was told in a paper that $\text{GL}(n,R)$ is a group, and I presumed that $$\text{GL}(n,R)=\{A\in M_{n\times n}(R)|\text{det}(A)~\mbox{is a unit in}~R\}.$$However, I tried google it and found $$\text{GL}(n,R)=\{A\in M_{n\times n}(R)|\text{det}(A)\neq0\}.$$See for example, http://gmcninch.math.tufts.edu/Math215-Fall-2012/storage/HW4.pdf. As $R$ is not necessarily a unital ring, so it would happen that $\text{GL}(n,R)$ is not a group. Could any expert tell me which understanding is correct? And also, could you recommend any textbook which provides detailed discussion about this kind of group? I need to learn this more, thank you very much!

share|improve this question
We want inverses, so the first definition is definitely the right one. –  Zhen Lin Feb 6 '13 at 12:05
If so, how do explain the definition in the website? Why he still calls it a group? –  Easy Feb 6 '13 at 12:12
The homework sheet you link to after "See for example" does not appear to contain any definition of the general linear group over an arbitrary ring. It defines $GL_2(\mathbb Z/n\mathbb Z)$, but its definition is the non-unit determinant one. –  Henning Makholm Feb 6 '13 at 12:18
Whatever, could you recommend any textbook which introduces the general linear groups over a unital commutative ring? I am struggling to find a proper reference. Thanks –  Easy Feb 6 '13 at 12:20
sorry, I think I should drop the "unital" condition.. –  Easy Feb 6 '13 at 12:22

1 Answer 1

If $R$ is a commutative ring with identity, then for an integer $n\geq 1$, $\mathrm{GL}_n(R)$ is the set of $n\times n$ matrices in $g\in\mathrm{M}_n(R)$ (coefficients in $R$) such that $\mathrm{det}(g)\in R^\times$. This is precisely the group of invertible elements of the ring $\mathrm{M}_n(R)$ of matrices.

If $R$ is a field, then $R^\times=R\setminus\{0\}$, so in this case, the condition $\det(g)\in R^\times$ is equivalent to $\det(g)\neq 0$, but this is not true for (non-zero) commutative rings which aren't fields.

share|improve this answer
why $det(g)\in R^\times$ implies $g$ invertible? For example, suppose $R=\mathbb{Z}_4$ and $det(g)=2$, then $det(g^{-1})=2^{-1}$, which doesn't exist. –  Easy Feb 6 '13 at 12:11
@user60079: Here $R^\times$ is the group of units in the ring, so $2\notin R^\times$ in your example. If the determinant of a matrix is a unit, then Cramer's rule produces an inverse for it. –  Henning Makholm Feb 6 '13 at 12:13

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.