Find all $A$-submodules of $V$, where $A=$ring of diagonal matrices

Question

Question:

Let $V$ be an $n$-dimensional vector space over a field $k$, with a basis $\{v_1,\dots,v_n\}$. Let $A$ be a ring of all $n\times n$ diagonal matrices over $k$. $V$ is an $A$-module under the action: $$\text{diag}(\lambda_1,\dots,\lambda_2)\cdot(a_1v_1+\cdots+a_nv_n)=(\lambda_1a_1v_1+\cdots+\lambda_na_nv_n)$$

Find all $A$-submodules of $V$.

Answer:

It is obvious that for all $i,j,\dots$, $\langle v_{i}\rangle,\langle v_i,v_j\rangle,\dots,\langle v_1,\dots,v_n\rangle$ are $A$-submodules of $V$. However, I don't have a clue how to find the others (if there is any). Any help/hint would be appreciated. Thanks in advance...

Answer 1

You have found all the $A$-submodules. First observe that any $A$-submodule is also a $k$-subspace by considering scalar matrices. Now consider a non-trivial $A$-submodule $W$ and choose $w \in W \setminus \{0 \}$. Expand $w$ in the given basis $$w = \sum_{i=1}^n a_i v_i$$ and let $\mathcal{I}$ denote the set of all indices such that $a_i \neq 0$, which is by assumption non-empty. Hence, $$w = \sum_{i \in \mathcal{I}} a_i v_i.$$ By multiplying by different diagonal matrices, we now may cancel out any of the $a_i$ to conclude that $v_i \in W$ for $i \in \mathcal{I}$. Now if $\text{span}( \{ v_i : i \in \mathcal{I} \}) = W$, we are done. Otherwise, choose $w_1 \in W \setminus \text{span}( \{ v_i : i \in \mathcal{I} \})$ and repeat the same process to obtain a new subset $\mathcal{J}$ (disjoint from $\mathcal{I}$) such that $\text{span}( \{v_i : i \in \mathcal{I} \cup \mathcal{J} \}) \subseteq W$. This process terminates because $W$ is a $k$-subspace and thus has a well-defined dimension over $k$.

Answer 2

The problem could be rephrased as follows:

What are the ideals of the product ring $k^n$?

An easy exercise (with solutions appearing elsewhere on the site) is

What are the ideals of $\prod_{i=1}^n R_i$ for some rings $R_i$ with identity?

The answer is that each one is of the form $\prod_{i=1}^n I_i$ for some collection $I_i\lhd R_i$.

In your case, this is exactly what you described, since each coordinate is a copy of $k$, it only has two possible subspaces: $k$ or $\{0\}$, so it just amounts to selecting a subset of the basis you chose (because this basis is determining the coordinates.)

You can easily re-prove the lemma above if you can't find it on the site. Obviously a product of ideals from each ring is an ideal in the product ring. Conversely, given an ideal of the product ring, you can multiply on the left and/or right with elements that are 'unit vectors' like $(1,0,0,\ldots)$, $(0,1,0,0,\ldots)$ and so on to find a collection of ideals in the $R_i$ that build up the whole ideal.