Let $k$ be a division ring, then the ring of upper triangular matrixes over $k$ is hereditary

Question

I'm reading Ring Theory by Louis H. Rowen, and he claimed that The ring of upper triangular matrices over a division ring is hereditary (it's on page 196, Example 2.8.13 of the book).

I think it should be pretty much straight-forward, despite the fact that I'm not clear how it must be true.

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Here's my thought on the problem:

I know that a left (resp, right) hereditary ring is a ring such that all of its left (resp, right) ideals are projective.

Now, let $k$ be a division ring, and the upper triangular matrice ring over $k$, $U_n(k) = \left(\begin{array}{ccccc} k & k & k & \cdots & k \\ 0 & k & k & \cdots & k \\ 0 & 0 & k & \cdots & k \\ \vdots & \vdots &\vdots &\ddots &\vdots \\ 0 & 0 & 0 & \cdots & k \end{array} \right)$, and I'll now try to find the form of its left ideals (i.e, how they look like). Let $I_n(k) = \left(\begin{array}{ccccc} I_{11} & I_{12} & I_{13} & \cdots & I_{1n} \\ 0 & I_{22} & I_{23} & \cdots & I_{2n} \\ 0 & 0 & I_{33} & \cdots & I_{3n} \\ \vdots & \vdots &\vdots &\ddots &\vdots \\ 0 & 0 & 0 & \cdots & I_{nn} \end{array} \right)$ be any $U_n(k)$'s ideal. Since $k$ is a division ring, its ideal are $0$, and itself. Using the fact that $U_n(k) I_n(k) \subset I_n(k)$, I manage to arrive at the following facts:

1. Each $I_{ij}$ must be an ideal of $k$, hence either $0$, or $k$.

2. Consider the column $j$ if $I_n(k)$, $\left(\begin{array}{c} I_{1j}\\ I_{2j}\\ I_{3j} \\ \vdots \\ I_{jj} \\ 0 \\ \vdots \\ 0 \end{array} \right)$, using the fact that $U_n(k) I_n(k) \subset I_n(k)$, we must have the descending chain $I_{1j} \supset I_{2j} \supset \dots \supset I_{jj}$

So, basically, the following form is one possible left ideal of $U_n(k)$: $\left(\begin{array}{ccccccc} k & k & k & 0 & k & \cdots & 0 \\ 0 & k & k & 0 & k & \cdots & 0 \\ 0 & 0 & 0 & 0 & k & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ \vdots &\vdots &\vdots & \vdots &\vdots &\ddots &\vdots \\ 0 &0 &0 & 0 & 0 & \cdots & 0 \end{array} \right)$

Now, I don't think the above should be a $U_n(k)-$projective module at all, since I don't think it is a direct factor of any free $U_n(k)-$ modules.

Where have I gone wrong? :(

Thank you so much guys,

And have a good day,

Answer 1

For each $1\leq i \leq n$, the set $C_i = U_n(k)e_{i,i}$ is a left ideal, which is projective, since $\bigoplus C_i \cong U_n(k)$. Suppose $J$ is any ideal in $U_n(k)$ - let us show that we can write it as $J'\oplus C_i$ for some $i$, hence from the previous argument+induction it will be projective.

Let $i$ be the maximal index such that $e_{i,i}J\neq 0$ and choose some $0\neq v \in e_{i,i}J$. Note that by multiplying $J$ from the right by a matrix $A=I+\alpha e_{i,j}$ for $j>i$ we get an isomorphic left ideal and the index $i$ is still maximal such that $e_{i,i}JA\neq 0$. You can use such multiplications to transform $v$ into $\beta e_{i,j}$ for some $j\geq i$, and wlog assume that $\beta=1$.

The subideal $U_n(k)v$ is now just $Je_{j,j}\cong C_i$, and it is also easy to see that $J'=J(I-e_{j,j})$ is also a subideal so that $J\cong J'\oplus C_i$.