Left ideals of the ring $M_2(\Bbb Z_2)$

Question

I am asked to show that there are exactly $5$ left ideals in the ring $M_2(\Bbb Z_2)$ of $2\times 2$ matrices with coefficients in $\Bbb Z_2=\Bbb Z/2\Bbb Z$. Clearly $0$ and $M_2(\Bbb Z_2)$ are left ideals, so it suffices to show that there are $3$ proper nonzero left ideals. I also know that there are only $2^4=16$ elements in $M_2(\Bbb Z_2)$, so maybe I could just try by brutal force, but I'm wondering if there is a clever idea. Any hints?

Answer 1

Theorem 1. Let $\mathbb{K}$ be a field and $n$ a positive integer. Write $R:=\text{Mat}_{n\times n}(\mathbb{K})$. If $\mathcal{L}$ is the set of all left ideals of $R$ and $\mathcal{S}$ is the set of all subspaces of $\mathbb{K}^n$, then define $f:\mathcal{L}\to\mathcal{S}$ and $g:\mathcal{S}\to\mathcal{L}$ via $$f(L):=\bigcap_{\Phi\in L}\,\ker(\Phi)$$ and $$g(S):=\big\{\Phi\in R\,\big|\,S\subseteq\ker(\Phi)\big\}$$ for all $L\in\mathcal{L}$ and $S\in\mathcal{S}$. Then, $f$ and $g$ are inverse functions, thereby, establishing a bijective correspondence between $\mathcal{L}$ and $\mathcal{S}$.

First of all, well-definedness of $f$ is trivial. It is easily seen that $g(S)$ is indeed a left ideal for all $S\in\mathcal{S}$. Therefore, $g$ is also well defined.

Now, let $S\in\mathcal{S}$. The inclusion $S\subseteq (f\circ g)(S)$ is clear. Suppose that $\{s_1,s_2,\ldots,s_k\}$ is a basis of $S$. Extend this set to a basis $\{s_1,s_2,\ldots,s_n\}$ of $\mathbb{K}^n$. There exist linear functionals (considered as row vectors) $\sigma_1,\sigma_2,\ldots,\sigma_n:\mathbb{K}^n\to\mathbb{K}$ such that $\sigma_i(s_j)=\delta_{i,j}$ for $i,j=1,2,\ldots,n$, where $\delta$ is the Kronecker delta. Each row of $\Phi \in g(S)$ is a linear combination of $\sigma_{k+1},\sigma_{k+2},\ldots,\sigma_n$. In particular, if each $\sigma_{k+1},\sigma_{k+2},\ldots,\sigma_n$ appears at least once as a row of $\Phi$, then $$\ker(\Phi)=\bigcap_{i=k+1}^n\,\ker(\sigma_i)=S\,.$$ Ergo, $S\supseteq (f\circ g)(S)$. This means $$(f\circ g)(S)=S$$ for all $S\in\mathcal{S}$.

Let now $L\in\mathcal{L}$. The inclusion $(g\circ f)(L)\supseteq L$ is trivial. For the reversed inclusion, let $\Psi\in L$ be a map with the maximum rank. We claim that every $\Phi\in L$ is equal to $\Xi\Psi$ for some $\Xi\in R$. Define $V$ to be the span of all rows from all elements of $L$. It is easy to show that $f(L)=\bigcap\limits_{\sigma \in V}\,\sigma$, and it follows immediately that the rows of $\Phi$ span $V$, implying that $\ker(\Psi)=f(L)$. Now, each row of $\Phi\in L$ is in $V$, whence each row of $\Phi$ is a span of the rows of $\Psi$. By writing the rows of $\Phi$ as a linear combination of the rows of $\Psi$, we obtain a factorization $\Phi=\Xi\Psi$ for some $\Xi\in R$. Therefore, the ideal $L$ is a principal left ideal generated by $\Psi$. That is, as $f(L)=\ker(\Psi)$, we get $(g\circ f)(L)=\big\{\Phi\in R\,\big|\,\ker(\Psi)\subseteq\ker(\Phi)\big\}$. Hence, each $\Phi\in (g\circ f)(L)$ factors through $\Psi$, namely, $\Phi=\Xi \Psi$ for some $\Xi\in R$. Thus, $(g\circ f)(L)\subseteq L$. This gives $$(g\circ f)(L)=L$$ for all $L\in\mathcal{L}$, as desired.

---

Applying the theorem above to our situation, we conclude that, when $R=\text{Mat}_{2\times 2}(\mathbb{F}_2)$, $R$ has $5$ left ideals: $$0\,,\,\,\left\langle \begin{bmatrix}0&1\\0&1\end{bmatrix}\right\rangle\,,\,\,\left\langle \begin{bmatrix}1&0\\1&0\end{bmatrix}\right\rangle\,,\,\,\left\langle \begin{bmatrix}1&1\\1&1\end{bmatrix}\right\rangle\,,\text{ and }R\,.$$ They correspond, respectively, to the subspaces $$\mathbb{F}_2^2\,,\,\,\text{span}_{\mathbb{F}_2}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}\,,\,\,\text{span}_{\mathbb{F}_2}\left\{\begin{bmatrix}0\\1\end{bmatrix}\right\}\,,\,\,\text{span}_{\mathbb{F}_2}\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}\,,\text{ and }0\,,$$ of $\mathbb{F}_2^2$.

More generally, if $q$ is a perfect power of a prime natural number, then $\text{Mat}_{n\times n}(\mathbb{F}_q)$ has in total $$\sum_{r=0}^n\,\binom{n}{r}_q$$ left ideals. Here, $[x]_q:=\dfrac{q^x-1}{q-1}$ for all $x\in\mathbb{R}$, $[0]_q!:=1$, $[m]_q!:=[1]_q\, [2]_q\, \cdots \, [m]_q$ for all positive integers $m$, and $$\binom{m}{k}_q:=\frac{[m]_q!}{[m-k]_q!\,[r]_q!}$$ for all integers $m$ and $k$ with $0\leq k \leq m$.

---

Here is a generalization. It has a different formulation from Theorem 1. The proof of Theorem 2 is left as an exercise.

Theorem 2. Let $A$ be a unital ring and $n$ a positive integer. Write $R:=\text{Mat}_{n\times n}(A)$. If $\mathcal{L}$ is the set of all left ideals of $R$ and $\mathcal{S}$ is the set of all left $A$-submodules of the left $A$-module $A^n$, then define $f:\mathcal{L}\to\mathcal{S}$ and $g:\mathcal{S}\to\mathcal{L}$ as follows

• for each $L\in\mathcal{L}$, $f(L)$ is the $A$-span of all row vectors of all matrices in $L$, and
• for each $S\in\mathcal{S}$, treat $S$ as a set of row vectors, and set $g(S)$ to be the set of all matrices whose rows belong to $S$.

Then, $f$ and $g$ are inverse functions, thereby, establishing a bijective correspondence between $\mathcal{L}$ and $\mathcal{S}$.