How would I show that there is a ring $R$ with identity $1_R$ and a subring $S$ not containing $1_R$, but such that $S$ has its own identity $1_S$ not equal to $1_R$?
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2Your problem is asking for an example of a ring with such a subring. Consider $\mathbb{Z}_6$ and subring ${0,2,4 }$. In the subring $4$ is acting as the multiplicative identity. – Bill Cook Apr 24 '13 at 15:46
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By looking for an example. You may look at quotient rings ${\bf Z}/n{\bf Z}$ in particular for some accessible ones. – anon Apr 24 '13 at 15:46
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@BillCook Please don't clutter up the comments with answers: there is an answer section for that! :) – rschwieb Apr 24 '13 at 15:48
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I answered this in your prior related question. Did you miss it? – Math Gems Apr 24 '13 at 16:43
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The same question was asked already many times on math.SE. – Martin Brandenburg Apr 24 '13 at 20:47
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You can take $R = \mathbb{Z} \times \mathbb{Z}$ and $S = \mathbb{Z} \times \{0\} \subset R$. Then $1_R = (1,1) \neq (1,0) = 1_S$.
Martin Sleziak
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Alex Provost
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what exactly are the elements in Z x {0}, would they be all elements of the form (0,0),(1,0),(2,0)...ect – user72195 Apr 24 '13 at 16:12
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A simple example (related to the one by Alex P.) is given by the rings of matrices $$ \left\{ \begin{bmatrix}a&0\\0&0\end{bmatrix} : a \in F \right\} = S \subseteq R = \left\{ \begin{bmatrix}a&b\\c&d\end{bmatrix} : a, b, c, d \in F \right\} $$ where $F$ might be $\Bbb{Z}, \Bbb{Q}, \Bbb{R}, \dots$.
Here $$ \begin{bmatrix}1&0\\0&0\end{bmatrix} = 1_{S} \ne 1_{R} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. $$
Andreas Caranti
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The rings of the form $(\mathbb Z_n,\oplus,\odot)$ are quite simple, so let us try them.
If $R=\mathbb Z_n$ then $1_R=1$. A candidate for $1_S$ can be only an element such that $1_S\odot 1_S=1_S$.
In $\mathbb Z_2$, $\mathbb Z_3$, $\mathbb Z_5$ we do not have non-trivial subrings.
In $\mathbb Z_4$ the only element such that $a\odot a=a$ is $a=1$.
What about $R=\mathbb Z_6$? Could you find a subring $S$ with the required properties?
Martin Sleziak
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