In a lecture, our professor gave an example for a ring. He took it out of another source and mentioned that he does not know the motivation for the chosen operation.

Of course, it's likely that somebody just invented an arbitrary operation satisfying ring axioms. I'd still like to try my luck whether anyone here can decipher the operation and give any kind of motivation for that example.

On $\mathbb{R}^3$ define the operations $+$ and $\cdot$ by $$ \begin{aligned} (a_1, a_2, a_3) + (b_1,b_2,b_3) &= (a_1+b_1,a_2+b_2,a_3+b_3) \\ (a_1, a_2, a_3) \cdot (b_1, b_2, b_3) &= (a_1 \cdot b_1, a_2 \cdot b_2, a_1 \cdot b_3 + a_3 \cdot b_2). \end{aligned} $$ (The $+$ and $\cdot$ operations on the right side are the usual addition and multiplication from $\mathbb{R}$.) With those operations, one can confirm that $\left(\mathbb{R}^3, +, \cdot \right)$ is a ring.

  • $\begingroup$ In case anyone is wondering, the multiplicative identity is $(1,1,0)$. $\endgroup$ – Alex Provost Dec 8 '17 at 19:00
  • $\begingroup$ I think it's just an arbitrary thing. I can't see any motivating pattern here. If someone else does see something, that will be surprising and very interesting. Also, in case anyone is wondering, the multiplication is associative (I checked). $\endgroup$ – Zach Teitler Dec 8 '17 at 19:04
  • $\begingroup$ I can't check at the moment, but is multiplication here commutative? Perhaps someone wanted to construct a noncommutative ring without referencing matricies? $\endgroup$ – Andrew Tawfeek Dec 8 '17 at 19:14
  • $\begingroup$ The multiplication is not commutative, take $(1,0,0) \cdot (0,0,1) \neq (0,0,1) \cdot (1,0,0)$. $\endgroup$ – Kezer Dec 8 '17 at 19:16
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    $\begingroup$ mentioned that he does not know the motivation for the chosen operation Seriously?! If you show him the upper triangular matrix ring he will be rather abashed, then :) $\endgroup$ – rschwieb Dec 8 '17 at 21:36

This is just matrix multiplication in disguise. Specifically, if you identify $(a_1,a_2,a_3)$ with the matrix $\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}$, these operations are the usual matrix operations: $$\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}+\begin{pmatrix}b_1 & b_3 \\ 0 & b_2\end{pmatrix}=\begin{pmatrix}a_1+b_1 & a_3+b_3 \\ 0 & a_2+b_2\end{pmatrix}$$ $$\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}\begin{pmatrix}b_1 & b_3 \\ 0 & b_2\end{pmatrix}=\begin{pmatrix}a_1b_1 & a_1b_3+a_3b_2 \\ 0 & a_2b_2\end{pmatrix}$$

  • $\begingroup$ Oh of course! Nice and simple, thanks, the ring is much more motivated now! $\endgroup$ – Kezer Dec 8 '17 at 21:32
  • $\begingroup$ Unfortunately I realized that's exactly what I was describing a minute too late... $\endgroup$ – rschwieb Dec 8 '17 at 21:32

It is isomorphic to the ring of matrices

$$ \left\{\begin{bmatrix}a_1&a_3\\0&a_2\end{bmatrix}\,\middle|\,a_1, a_2,a_3\in \mathbb R\right\} $$

It's a semiprimary ring whose Jacobson radical is the subset with $a_1=a_2=0$. The Jacobson radical is nilpotent, and $R/J(R)\cong\mathbb R\times\mathbb R$. Here is a list of more properties of such a ring.

This sort of ring is fairly famous, and has nice interpretations. One of them is that if you select a chain of subspaces $\{0\}<V<W<\mathbb R\times \mathbb R$ ($W$ of dimension $1$, $V$ of dimension $2$) then the linear transformations of $\mathbb R\times\mathbb R$ which stabilize this chain is isomorphic to this triangular matrix ring. That is, $\phi$ stabiliezes the chain if $\phi(V)\subseteq\phi(W)$.

Incidentally, you are always going to be able to extract some sort of matrix presentation for a multiplication like you are describing, because you can rely on it being a finite dimensional algebra. If it really is a valid ring multiplication, it's bilinear, and so you can work on figuring out what a logical 'basis' is and then deduce what it looks like with matrices.

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    $\begingroup$ This should be the accepted and most upvoted answer. $\endgroup$ – Xam Dec 9 '17 at 5:54
  • $\begingroup$ In your last paragraph, what exactly do you mean by "a multiplication like you are describing"? $\endgroup$ – Jack M Dec 9 '17 at 11:07
  • $\begingroup$ @JackM Multiplication of $n$-tuples over a field, at least. Because that defines a representation with square matrices. $\endgroup$ – rschwieb Dec 9 '17 at 13:18
  • $\begingroup$ @rschwieb Surely not any ring structure on $\mathbb F^n$ is isomorphic to a ring of $\mathbb F$-matrices. Maybe a ring structure in which addition is component wise, and multiplication is given by quadratic polynomials? $\endgroup$ – Jack M Dec 9 '17 at 13:29
  • $\begingroup$ @JackM You're right, I should emphasize the pointwise addition. Remember I'm stipulating that the multiplication is actually a valid ring multiplication, not any rule with tuples. To rephrase, "if multiplication on elements of $F^n$ defines a valid ring structure, then the ring can be represented with $n\times n$ matrices." This is easy to see, because you just identify each tuple with the linear transformation it produces with the multiplication. $\endgroup$ – rschwieb Dec 9 '17 at 14:22

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