user2154420

Does an algebraic geometer have an opinion on this:
https://math.stackexchange.com/questions/4053116/prime-ideal-after-modding-out-by-homogenizing-variable-implies-original-ideal-pr

@TedShifrin

One last question. In this post: https://math.stackexchange.com/a/70265/94106

Matt E's answer says:
"We may write $f(x)=g(x^{p^d}),f(x)=g(x^{p^d})$, where $g$ is irreducible and separable and $d\geq 0$.

Why is this true?

@TedShifrin I guess I ask this because of this special case:

Suppose $f$ and $g$ are two monic irreducible polynomials over $\mathbb{F}_q$ of degrees $16$ and $20$ respectively. Let $F$ be the splitting field of $fg$ over $\mathbb{F}_q$. Then I want to say that $F$ is really just the composite of $\mathbb{F}_{q^{16}}$ and $\mathbb{F}_{q^{20}}$, which I guess is $\mathbb{F}_{q^{80}}$? So this is a degree $80$ extension?

I keep confusing myself with the wording of "algebra" and "module"

The question at hand is "Is $\mathbb{M}_n(\mathbb{C})$ a simple $\mathbb{C}$-algebra?" So I thought it was sufficient to prove that for any nonzero $A\in\mathbb{M}_n(\mathbb{C})$, $(A)=\mathbb{M}_n(\mathbb{C})$. Then since the only ideals are the trivial ones, $\mathbb{M}_n(\mathbb{C})$ is simple