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Let $K$ be a field and $$R=\begin{pmatrix} K & 0\\ K & K \end{pmatrix}$$ the ring of lower matrix with coefficients in $K$. I want to find the left ideals of $R$ and also prove that $R$ is an hereditary artinian ring.

At some Lam's book and also mentioned here Left and right ideals of $R=\left\{\bigl(\begin{smallmatrix}a&b\\0&c \end{smallmatrix}\bigr) : a\in\mathbb Z, \ b,c\in\mathbb Q\right\}$ he said (if I dualize right) that the left ideals of a lower triangular ring are all of the form $I_{1} \oplus I_{2}$ where $I_{1}$ is a left ideal of $K$ and and $I_{2}$ is a submodule of $K \oplus K$ which contains $K I_{1}$. But still cannot see how this helps, in my case the only left ideals of $K$ are $K$ itself and $\lbrace 0 \rbrace$ so $I_{1}= \lbrace 0 \rbrace$ or $I_{1}= K$ still cannot visualize $I_{2}$ as the only thing I know is that $KI_{1}= \lbrace 0 \rbrace$ or $K I_{1}= K K$.

And for proving $R$ is artinian hereditary I'm suggested to prove all minimal left ideals are isomorphic, but I have three questions here: How do I know proving this solves the problem? How do I compute minimal ideals here? And how do I prove these minimal ideals are isomorphic?

rschwieb
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Cos
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1 Answers1

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Yes, I think you translated the condition correctly. The easiest way for me is to remember this ring is anti-isomorphic (via transposition) to the one in the link you gave.

If $I_1=\{0\}$, then $I_2$ can be any submodule of $K\oplus K$.

If $I_1=K$, then $I_2$ has to contain $\{0\}\oplus K$, so it has to be of the form $K\oplus K$ or $\{0\}\oplus K$.

These last two correspond to

$$\begin{pmatrix} K & 0\\ K & 0 \end{pmatrix}$$

and

$$\begin{pmatrix} K & 0\\ K & K \end{pmatrix}$$

and the ones of the first form look like

$$L_W=\left\{\begin{pmatrix} 0 & 0\\ b & c \end{pmatrix}\middle|\, (b,c)\in W\right\}$$ for a given subspace $W< V$.

Note that if you pick any $1$-dimensional subspace $W$, that is automatically going to make a minimal left ideal. The first two in the list are obviously not minimal because they contain the left ideal of strictly lower triangular matrices.

We can show all minimal left ideals are isomorphic to $L_{K\oplus 0}$. Suppose $(a,b)$ is a nonzero element of $L_W$ where $W$ is one dimensional. If $b=0$ then obviously there is nothing to do. If $b\neq 0$, then

$\begin{pmatrix}0&0 \\ b^{-1}&0\end{pmatrix}$ defines a left $R$ module transformation from $L_W\to L_{K\oplus 0}$ by right multiplication, which is an isomorphism because they're both one dimensional.

So look at what you have:

$$\begin{pmatrix} K & 0\\ K & K \end{pmatrix}\cong \begin{pmatrix} K & 0\\ K & 0 \end{pmatrix}\oplus \begin{pmatrix} 0 & 0\\ 0 & K \end{pmatrix}$$, so both of these pieces are projective, being summands of a free left module.

Then, all of the $L_W$ with one-dimensional $W$ are isomorphic to the second factor, so they are projective as well. If $W$ is two-dimensional, then $L_W=\begin{pmatrix} 0 & 0\\ K & K \end{pmatrix}$ which is obviously a direct sum of two minimal left ideals, and thus projective (since they are.)

So, all left ideals are projective.

rschwieb
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