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.

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

This question already has an answer here:

Is there an example of a ring $S$ with identity $1_S$ containing a non-trivial subring $R$ which itself has an identity $1_R$, but $1_R\neq 1_S$ (or equivalently $1_S\notin R$). I'd also like to know under what conditions the identities have to be equal. I know they must be equal if $S$ has no zero divisors, since for every $r\in R$ we have $(1_S-1_R)r=0$

In other words: Is the category of unital rings with identity-preserving morphisms a full subcategory of the category of rings (where morphism are only required to respect addition and multiplication)?

share|cite|improve this question

marked as duplicate by M Turgeon, Amzoti, Lord_Farin, Martin, Dan Rust Jun 22 '13 at 16:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Whoops! I see it now. – Cameron Buie Jun 22 '13 at 15:56
up vote 1 down vote accepted


$$S:=\left\{\;\begin{pmatrix}x&y\\z&w\end{pmatrix}\;;\;\;x,y,z,w\in\Bbb R\;\right\}\;,\;R:=\left\{\;\begin{pmatrix}a&0\\0&0\end{pmatrix}\;;\;\;a\in\Bbb R\;\right\}$$


share|cite|improve this answer
@stefan It's worth drawing attention to the construction of which this is a special case: the ring $eSe$ where $e^2=e\notin \{0,1_S\}$. This relies on the existence of a nontrivial idempotent. Since the identity element of any subring has to be idempotent, you can see that the identity of subrings would have to coincide with the superring's identity if there are only trivial idempotents (like in local rings and in domains). – rschwieb Jun 22 '13 at 16:58

Another example is the ring made up of $\begin{pmatrix} a & b\\ c & d\end{pmatrix}$, where $a,b,c,d\in\Bbb{R}$. The identity is $\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$.

It has a subring: $\begin{pmatrix} a & a\\ a & a\end{pmatrix}$, where $a\in\Bbb{R}$. The identity of this subring is $\begin{pmatrix} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}$.

share|cite|improve this answer
Thanks for your answer. But isn't $GL$ the group of matrices with non-vanishing determinant? I think you meant the ring of all matrices. – Stefan Hamcke Jun 22 '13 at 16:52
You're quite right. The relevant edits have been made. – Ayush Khaitan Jun 22 '13 at 16:56
@dimension10 That's a pretty misleading thing to say without specifying that you mean there is more than one element $x\in R$ such that $xs=s$ for all $s\in S$. Of course, a ring can only contain at most one element acting like an identity. – rschwieb Jun 22 '13 at 18:43
@dimension10- what is the other identity? Let $\begin{pmatrix} b & b\\ b & b\end{pmatrix}$ be the identity. Then on multiplying with $\begin{pmatrix} a & a\\ a & a\end{pmatrix}$, we get $\begin{pmatrix} 2ab & 2ab\\ 2ab & 2ab\end{pmatrix}$. $2ab=a$ iff $b=\frac{1}{2}$, unless $a$ or $b=0$. – Ayush Khaitan Jun 23 '13 at 1:44
@AyushKhaitan Well $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ has all the properties of the identity except that of being in $S$, so I thought that might be what he meant :) – rschwieb Jun 24 '13 at 10:58

Let $D_4=\langle x,y\mid x^4=y^2=xyxy=1\rangle,$ and consider the group ring $S=\Bbb C[D_4],$ a vector space of dimension $8$ over $\Bbb C$ with canonical standard basis $\{\mathbf{e}_{1},\mathbf{e}_{x},\mathbf{e}_{x^2},\mathbf{e}_{x^3},\mathbf{e}_{y},\mathbf{e}_{xy},\mathbf{e}_{x^2y},\mathbf{e}_{x^3y}\}.$ Multiplication in this vector space is given by $\mathbf{e}_{g}\mathbf{e}_{h}=\mathbf{e}_{gh}$ for any $g,h\in D_4,$ with scalars multiplying as usual, and distributivity as expected. The multiplicative identity element of $S$ is $\mathbf{e}_{1}.$

Since the center of $D_4$ is $\{1,x^2\},$ then $\mathbf{v}_{1}:=\frac12(\mathbf{e}_{1}-\mathbf{e}_{x^2})$ commutes multiplicatively with everything in $S.$ Note further that $\mathbf{v}_{1}^2=\mathbf{v}_{1}$. For each $g\in D_4,$ let $\mathbf{v}_{g}:=\mathbf{v}_{1}\mathbf{e}_{g}.$ It can be shown that $\mathbf{v}_{g}\mathbf{v}_{h}=\mathbf{v}_{gh}$ for all $g,h\in D_4,$ and that $\mathbf{v}_{x^2}=-\mathbf{v}_{1},$ so $$\{\mathbf{v}_{g}\mid g\in D_4\}=\{\pm \mathbf{v}_{1},\pm \mathbf{v}_{x},\pm\mathbf{v}_{y},\pm\mathbf{v}_{x}\mathbf{v}_{y}\},$$ whence $$R:=\{a\mathbf{v}_{1}+b\mathbf{v}_{x}+c\mathbf{v}_{y}+d\mathbf{v}_{x}\mathbf{v}_{y}\mid a,b,c,d\in\Bbb C\}$$ is a subset of $S$ that is closed under the addition and multiplication operations of $R$--a subring of $S$--whose multiplicative identity element is $\mathbf{v}_{1}\ne\mathbf{e}_{1}.$

See Arturo's fine answer here for more on this subject.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.