What does abstract algebra have to say about the determinant?

Solution 1:

Let's consider $M_n(\mathbb K)$ and $\mathbb K$ as monoids under multiplication. Here are fairly complete answers to your technical questions:

1. Every monoid homomorphism $f$ from $M_n(\mathbb{K})$ to $\mathbb{K}$ is of the form $f=\phi \circ \text{det}$ for some monoid endomorphism $\phi:\mathbb{K}\to\mathbb{K}$. In other words, the determinant has the universal property that every monoid homomorphism from $M_n(\mathbb{K})$ to $\mathbb{K}$ factors through it. To make things more concrete, take, for example, $\mathbb{K}=\mathbb{R}$: if we restrict to continuous homomorphisms, the result implies that the only such homomorphisms are $f=(\text{sgn}(\text{det} A))^{\varepsilon}\cdot|\text{det} A|^r$ for $\varepsilon\in\{0,1\}$ and $r\in\mathbb{R}$.

Here is how we can prove this: Let $f : M_n(\mathbb{K}) \to \mathbb{K}$ be a monoid homomorphism. First, we claim that either $f(A)=1$ for all $A$, or $f(A)=0$ for all non-invertible $A$: We have $f(0)=f(00)=f(0)^2$, so either $f(0)=0$ or $f(0)=1$. If $f(0)=1$, then for all $A\in M_n(\mathbb{K})$, $f(A)=f(A)f(0)=f(A0)=f(0)=1$. So assume $f(0)=0$. Let $J$ be a nilpotent matrix of rank $n-1$, e.g., take $J$ to be a single Jordan block with zeros on the diagonal. Then $J^n=0$, so $f(J)^n=f(J^n)=f(0)=0$, hence $f(J)=0$. Now, any matrix $A$ of rank $n-1$ may be written $A=SJT$ for suitable matrices $S$ and $T$, and it follows that $f(A)=f(S)f(J)f(T)=0$. Finally, since any non-invertible matrix $A$ may be written as a product of several rank $n-1$ matrices, we obtain $f(A)=0$ for all non-invertible $A$, which proves the claim.

Thus, if we ignore the trivial case $f=1$ and restrict attention to the case where $f(A)=0$ for all non-invertible $A$, then $f$ is completely determined by its restriction to $GL_n(\mathbb{K})$, which is a group homorphism into the multiplicative group $K^*$, whose kernel is a normal subgroup $N \unlhd GL_n(\mathbb{K})$. If we exclude the exceptional situation where $n=2$ and $\mathbb{K}=\mathbb{F}_2$ or $\mathbb{F}_3$, then either $N$ is a subgroup of scalar matrices or $N$ contains $SL_n(\mathbb{K})$. In the former case, $GL_n(\mathbb{K})/N$ has the group $PGL_n(\mathbb{K})$ as a homomorphic image, which is non-abelian (for $n\geq 2$), in contradiction to the fact that the First Isomorphism Theorem applied to $f$ provides an embedding of $GL_n(\mathbb{K})/N$ into the abelian group $\mathbb{K}^*$. Therefore, $N$ must contain $SL_n(\mathbb{K})$, which is the kernel of the determinant homomorphism $\text{det} : GL_n(\mathbb{K}) \to \mathbb{K}^*$. It follows that $f$ factors through the determinant homomorphism, i.e., for $A\in GL_n(\mathbb{K})$ we have $f(A) = \phi(\text{det} A)$ for some group endomorphism $\phi$ of $\mathbb{K}^*$. If we extend $\phi$ to a monoid endomorphism on $\mathbb{K}$ by setting $\phi(0)=0$, then the equation $f(A)=\phi(\text{det} A)$ also holds for non-invertible matrices $A$ since in this case both $f(A)=0$ and $\phi(\text{det}(A))=\phi(0)=0$. Thus $f = \phi \circ \text{det}$.

In the exceptional situation that $n=2$ and $\mathbb K=\mathbb F_2$ or $\mathbb F_3$, it is straightforward to check that although there are some additional normal subgroups of $GL_2(\mathbb K)$, none of them give rise to any new group homomorphisms from $GL_2(\mathbb K)$ into $\mathbb K^*$.

2. There are no ring homomorphisms $M_n(\mathbb K) \to \mathbb K$ except for $n=1$, in which case the ring homomorphisms are just the field automorphisms of $\mathbb K$.

This is just a consequence of the fact that $M_n(\mathbb K)$ is a simple ring.