# What does abstract algebra have to say about the determinant?

The determinant is a homomorphism from the multiplicative monoid of matrices to the multiplicative monoid of a field (right?). I find this to be the most intuitive way to interpret some of the determinant's properties (notably the invertibility condition: obviously a matrix is only invertible if a homomorphism maps it to an invertible element of the field).

So from an algebraic point of view, can anything interesting be said about the determinant?

Some fairly specific technical questions:

1. How does it fit into the big picture of all homomorphisms $M_n(\Bbb K)\to\Bbb K^\star$? Is it the only one? How are the others related to it? The invertibility condition is one of the most useful things about the determinant, but any homomorphism would give the same condition, so why use the determinant?
2. Are there any ring homomorphisms $M_n(\Bbb K)\to\Bbb K^\star$? How does the determinant relate to them?

Some vaguer, softer questions:

1. Are there any intuitive proofs for the formulae for calculating determinants based on the fact that it's a homomorphism?
2. This one's pretty out there. The characteristic polynomial is a polynomial with coefficients in a field which is the homomorphic image of a polynomial with coefficients in the ring of matrices. Is there any way of explaining the various relationships between a matrix and its characteristic polynomial based on the structure-preserving properties of the determinant?
• math.stackexchange.com/a/21617/1650 Nov 2, 2013 at 23:00
• @MartinBrandenburg Good god that's abstract. A lot (all) of that answer is beyond me, but thanks for the link, I'll try and read it some time. Nov 2, 2013 at 23:03
• @JackM What's interesting is that the determinant's existence is predicted by algebraic geometry. Namely, "elimination of quantifiers", which in algebraic geometry language says that the structure map $\mathbb{P}^n_A\to \text{Spec }A$ is closed (where $A=k[a,b,c,d,e,f,g,h,i]$ for example), tells us that there should be some polynomial $D(a,b,c,d,e,f,g,h,i)$ such that the system $\{ax+by+cz,dx+ey+fz,gx+hy+iz\}$ has a non-trivial solution if and only if $D(a,b,c,d,e,f,g,h,i)= 0$. Of course, this polynomial is just $\det\begin{pmatrix}a & b & c\\ d & e & f\\ g & h & i\end{pmatrix}$. Nov 2, 2013 at 23:17

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 $\mathbb K^*$.

1. 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.