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For no real reason, I did a classification of (associative, with a 1) rings with less than 8 elements (they all happen to be commutative). Most of the rings I got were of a type I knew - namely: cyclic, field, or Boolean - but one wasn't.

$R \hspace{.04 in} := \hspace{.04 in} \Bigg\langle \bigg\{\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\bigg\} \text{ , matrix operations over the field } \mathbb{Z}/2\mathbb{Z} \Bigg\rangle$

Does $R$ arise in any remotely natural situations?

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try removing the single quotes you have round the latex expression. Also, double dollar signs for displayed equations. –  George Lowther Jan 12 '11 at 0:18
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And, writing $F_2\equiv\mathbb{Z}/2\mathbb{Z}$, that ring is (isomorphic to) the quotient $F_2[X]/(X^2)$. i.e. add a single element $x$ to $F_2$ subject the relation $x^2=0$, –  George Lowther Jan 12 '11 at 0:23
    
What is a cyclic ring? –  Qiaochu Yuan Jan 12 '11 at 0:23
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@George: Thanks, that worked, and good point about $F_2[\sqrt0]$. @Qiaochu: a ring whose underlying additive group is a cyclic group –  Ricky Demer Jan 12 '11 at 0:31
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Also, you might be interested in mathoverflow.net/questions/7133/… . –  Qiaochu Yuan Jan 12 '11 at 0:40
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4 Answers 4

up vote 9 down vote accepted

I'm not sure what would be natural in this context, but I would find it natural to denote $0 = \begin{pmatrix}0&0\\0&0\end{pmatrix}$, $1 = \begin{pmatrix}1&0\\ 0&1\end{pmatrix}$, $\epsilon = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ , $1+\epsilon = \begin{pmatrix}1&1\\0&1\end{pmatrix}$ Where 0 is a additive unit, $1$ is a multiplicative unit and $\epsilon^2 = 0$.

So this ring arises as $\mathbb Z[\epsilon]/(2,\epsilon^2)$.

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These are "rings of dual numbers"; see Bill's answer. –  Arturo Magidin Jan 12 '11 at 2:41
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Let $R$ be an associative ring with $1 \neq 0$, and let $M$ be an $(R,R)$-bimodule. Then the following set of matrices forms a ring with respect to the usual operations for matrix addition and matrix multiplication. \begin{equation} T(R,M) = \left\{\begin{pmatrix}r & x \\ 0 & r\end{pmatrix} \mid r \in R,~ x \in M\right\} \end{equation} Any such ring is known as a trivial extension. The ring described in the question has the form $T(R,R)$ for $R = \mathbb{Z}/2\mathbb{Z}$.

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It's the case $\rm\ R = \mathbb F_2$ of $\rm R[t]/t^2\:.\ $ This is known as the algebra of dual numbers over the ring $\rm R$. Such rings and higher order analogs $\:\rm R[t]/t^n \;$ prove quite useful when studying (higher) derivations algebraically since such rings provide very convenient algebraic models of tangent / jet spaces. For example, they permit easy transfer of properties of homomorphisms to derivations -- see for example section 8.15 in Jacobson, Basic Algebra II. See this post for further discussion and links.

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$R$ is a certain subring of the ring $B$ of upper triangular $2 \times 2$ matrices over $\mathbb{F}_2$. $B$ arises as the endomorphism ring of a complete flag in $(\mathbb{F}_2)^2$, so in that sense it's a natural geometric object to look at. The construction generalizes to a vector space over any field.

I'm not sure exactly what you mean by "natural situation," though, or why you should expect that every finite ring arises in such a situation. For any finite field $\mathbb{F}_q$, every subring of $\mathcal{M}_n(\mathbb{F}_q)$ is a finite ring, for example. As George Lowther says, for finite commutative rings we can also consider quotients $\mathbb{F}_q[x]/p(x)$ for arbitrary polynomials $p(x)$. These break up into products of rings of the form $\mathbb{F}_q[x]/r(x)^n$ for irreducible $r$, which is a form of the Chinese remainder theorem and is related to the structure theorem for finitely generated modules over a principal ideal domain (in this case, $\mathbb{F}_q[x]$). Is this "natural"?

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Qiaochu: The Jordan normal form only works over algebraically closed fields. Here, the quotients break up into rings of the form $\mathbb{F}_q[x]/(f^n)$ for irreducible $f\in\mathbb{F}_q[x]$. –  George Lowther Jan 12 '11 at 0:38
    
ok, I see you were on to it already. –  George Lowther Jan 12 '11 at 0:39
    
@George: yes, sorry, I realized my mistake. –  Qiaochu Yuan Jan 12 '11 at 0:39
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