# Nontrivial subring with unity different from the whole ring?

Is there an example of a ring $R$ with unity and a nontrivial subring $J$, such that $1_J \ne 1_R$?

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The example I give in a class, if this issue arises, is the ring ${\mathbf Z}/6{\mathbf Z}$ and its subset $\{0,3\}$, which is a "subset with ring structure" and multiplicative identity 3. – KCd Jul 15 '12 at 5:40
For a non trivial idempotent $e$, $eRe$ is a ring with diferente unit. – Frank Murphy Jul 15 '12 at 6:11

If to you, "ring" means "ring with unity", then the definition of "subring" requires that the unity be the same as that of the larger ring, just like "submonoid" requres that the identity be the same as that of the larger monoid. So under this definition, the answer is "no", because the definition of "subring" requires that if $R$ is a subring of $S$, then $1_S\in R$, and so $1_R=1_S$.

If by "ring" you don't require unity, and you are asking if it is possible to have rings $R$ and $S$, with $R\subseteq S$, and where both $R$ and $S$ happen to have a unity and $1_R\neq 1_S$, then yes: take $S=\mathbb{Z}\times\mathbb{Z}$, and $R=\mathbb{Z}\times\{0\}$. Then $1_S=(1,1)$ and $1_R=(1,0)$. In fact, every time you write a ring with unity $S$ as $S=R_1\times R_2$, you have that $1_S=(1_{R_1},1_{R_2})$.

The converse holds in part; this is the notion of "central idempotents", which is connected with the decomposition of rings into direct products:

Proposition. Let $S$ be a ring with unity, and suppose that $R\subseteq S$ by a subgroup that is closed under multiplication, and such that there exists $e_R\in R$ that is central in $S$ ($e_Rs=se_R$ for all $s\in S$) such that $e_Rr=re_R=r$ for all $r\in R$. Then $S\cong R\times T$, where $T$ is a ring with unity.

Proof. Let $T=S(1_S-e_R)$. Then $T$ is an ideal of $S$: it is trivially a left ideal; and since $1_S-e_R$ is central, $S(1_S-e_R) = (1_S-e_R)S$ which is trivially a right ideal. In particular, $T$ is a subgroup that is closed under multiplication. Moreover, $1_S-e_R$ is idempotent: note that $(1_S-e_R)(1_S-e_R) = 1_S - e_R - e_R+e_Re_R$. But $e_Re_R=e_R$ since $e_R$ is an identity for $R$, so $(1_S-e_R)^2=1_S-e_R$. Thus, for every $t\in T$, there exists $s\in S$ such that $t=s(1_S-e_R)$, so $$t(1_S-e_R) = s(1_S-e_R)^2 = s(1_S-e_R) = t$$ and since $1_S-e_R$ is central, this proves $1_S-e_R$ is a unity for $T$.

Note also that \begin{align*} e_R(1_S-e_R) &= e_R-e_Re_R = e_R-e_R = 0,\\ \text{and}\qquad (1_S-e_R)e_R &= e_R-e_Re_R = e_R-e_R=0. \end{align*}

Now consider the map $S\to R\times T$ given by $s\mapsto (se_R,s(1_S-e_R)$. Note that the map is one-to-one: if $se_R= te_R$ and $s(1_S-e_R) = t(1_S-e_R)$, then $$s = s(e_R+1_S-e_R) = se_R + s(1_S-e_R) = te_R+t(1_S-e_R) = t(e_R+1_S-e_R) = t.$$ And the map is onto: given $r\in R$ and $t\in T$, there exist $s,s'\in S$ such that $r=se_R$ and $t=s'(1_S-e_R)$. Let $u=se_R + s'(1_S-e_R)$. Then \begin{align*} ue_R &= se_Re_R + s'(1_S-e_R)e_R = se_R + s'0 = se_R = r\\ u(1_S-e_R) &= se_R(1_S-e_R) + s'(1_S-e_R)(1_S-e_R) = s0+s'(1_S-e_R) = s'(1_S-e_R)=t. \end{align*} Thus, the image of $u$ is $(ue_R,u(1_S-e_R)) = (r,t)$. SO the map is onto. It is easy to verify that it is both additive and multiplicative, so we get an isomorphism of rings. $\Box$

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The Proposition is a special case of the Peirce decomposition. – Bill Dubuque Jul 15 '12 at 15:14
+1 for pointing out that for rings (with unity), that "subring" means, among other things, that $1_S = 1_R$. – Hurkyl Jul 15 '12 at 16:00
I don't see how the non-trivial example you gave doesn't involve a ring with unity. A ring with unity S is defined as having a multiplicative identity element $1_{S}$, and a subset R of S is a subring of R if it is itself a ring under the same binary operations (not necessarily identity elements, though). I don't think any of those definitions requires the unity to carry over to subrings if one wishes to speak of a ring S with unity $1_{S}$. And to make it clear, the only reason I'm commenting on this is because the reply confused the hell out of me when I first read it. – Ryker Aug 13 '13 at 1:48

If $R$ is a ring with such a subring $S$, then $1_S$ must have the property that $s 1_S = 1_S s = s$ for all $s \in S$. In particular, $1_S$ is an idempotent in $R$. This motivates the following construction: given a ring $R$ and an idempotent $e \in R$, consider $$\{ r \in R : er = re = r \}.$$

This is a subring of $R$ which is by definition the maximal subring with respect to which $e$ is an identity (hence any subring in which $e$ is an identity is a unital subring of this subring). If $r$ is in this subring, then $ere = r$, and conversely any element of the form $ere$ lies in this ring. Consequently this is just $eRe$, which was mentioned by Frank Murphy in the comments.

Here is an example which doesn't come from direct products (equivalently where the idempotent $e$ is not central). Let $R = \mathcal{M}_{\infty}(k)$ be the ring of matrices with entries in another ring $k$ with countably many rows and columns but with only finitely many nonzero entries. Then there is a sequence of idempotents $e_n$, none of which are central (in fact $R$ has no center), defined as the diagonal matrices with first $n$ entries equal to $1$ and remaining entries equal to $0$. The corresponding subrings are the rings $\mathcal{M}_n(k)$. These idempotents give an approximate identity in $\mathcal{M}_{\infty}(k)$.

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Or for that matter, we can embed the $n\times n$ matrix ring over $R$ into the $n+k\times n+k$ matrix ring by adding rows and columns of zeros. This embedding is additive and multiplicative, but the identity of these upper-left $n\times n$ block matrices is different from that of the full larger matrix ring. They don't form an ideal, though. – Arturo Magidin Jul 15 '12 at 19:48
The example in the third paragraph doesn't have an identity, and so it doesn't fit the OP's request totally. However, there isn't any reason against expanding it to be the column-finite matrices over $k$ (each column only contains finitely many nonzero entries, but there may be infinitely many nonzero columns) and require $k$ to be a field. The resulting ring has identity and still has infinitely many nontrivial idempotents, none of which are central (since the ring is prime.) – rschwieb Jun 22 '13 at 17:05
@rschwieb: you misunderstand me. $R$ isn't the example; its finite-dimensional subrings are. – Qiaochu Yuan Jun 22 '13 at 19:10
@QiaochuYuan OK, sorry about that then. It just looked that way to me because the OP also asked for the superring to have identity, and $R$ looked like the superring. It's still a good example! – rschwieb Jun 22 '13 at 19:21

Sure. Let $S,T$ be any nontrivial rings and consider $R=S\times T$. Then the unity of $R$ is $(1_S,1_T)$ but the subring $S\times \{0\}$ has unity $(1_S,0_T)$.

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Universally: $R[x]/(x^2\!-\!x)$ has subring $xR \cong R\:$ and unit $x.$ Indeed, $r \mapsto x\,r\:$ is an isomorphism.