How to prove determinant is a group homomorphism and onto?.

Question

I posted this question I am struggling with previously but it was put on hold for lack of context, I hope this is now clearer.

Consider the determinant function

Det: Mn($\mathcal{F}$) $\to$ $\mathcal{F}$, where $\mathcal{F}$ is a field

i) Explain how to restrict the domain and range of Det to obtain a group homomorphism. State any important properties of the determinant function that are used to prove that the resulting map is a homomorphism.

(Do not need to prove these properties).

ii) Is Det: M3($\mathcal{F}$)$\to$ $\mathcal{F}$ one-to-one, or onto , or both, or neither? Explain your answer with examples.

My thinking For part i) of this question could the answer be that the domain could be the complex numbers but the range could be the complex numbers excluding 0, as if the output of the determinant was zero the matrix would not have an inverse so it would not be a group?.

For part ii) proving that the function is not one-to-one is routine but I do not know how we could prove or disprove that the determinant is onto? .

Answer 1

It isn't one to one (what is the determinant of the matrix with diagonal -1,-1,1? 1,1,1?). For onto, consider the diagonal matrix consisting of a,1,1 where a is some arbitrary field element.

As for (i), the domain is matrices, not complex numbers, so try again.

Answer 2

The "property" of $\det$ you are meant to use in (i) is:

$\det(AB) = \det(A)\det(B)$.

How could you turn this into a group homomorphism (hint: a field is a ring, what is its group of units)?

What is another name for the pre-image (under $\det$) of the group of units of $\mathcal{F}$?

Alternatively: what do we call matrices that map to $\mathcal{F}\setminus U(\mathcal{F})$ (that is which matrices map to element(s?) of $\mathcal{F}$ that are not units)?

Does the multiplicative monoid of $\text{Mat}_{n \times n}(\mathcal{F})$ contain a subgroup? What is its usual name?

It is a general theorem of ring theory, that for any two rings, $R,S$ with a ring-homomorphism $\phi:R \to S$, that the homomorphism $\phi$ induces a group homomorphism:

$\phi^{\ast}:U(R) \to U(S)$.

Is $\det$ that homomorphism here?