Artin Algebra Chapter 11
This has been answered here. My questions are about the solution of Brian Bi:
- By stronger, does he mean that $C \ne L$ and $f$ is irreducible $\implies l \nmid f?$ If so, here is my proof. Is it correct?
Since $f$ is irreducible, $l$ divides $f$ if and only if $\frac f l =: k$ is a nonzero constant. Then
$$C=\{f=0\}=\{kl=0\}=\{l=0\}=L$$
- Is $a=0=b$ ruled out by definition of linear?
If so, is this in book, or is it understood because $a=0=b$ gives either $\emptyset$ or $\mathbb C^2$? If not, how do we prove $|C \cap L| \le d$ in the case where $c=0$? I was able to show just $C \cap L = C$.
- Why is it that $g$ has degree $\le d$?
I think this has something to do with:
- irreducibility of $f$ and
- what it means for $d$ to be the degree of $f$.
I think of writing $f=e_0(y)+e_1(y)x^1 + ... + e_m(y)x^m$. If each monomial has degree $\le d$, then because $y=px+q$ has degree 1, $g$ has degree $\le d$, being made up of monomials in one variable that have degree $\le d$. Therefore,
- it's irrelevant that $f$ is irreducible, and
- the definition that $d$ is the degree of $f$ is that all its monomials have degree $\le d$, and one of them has degree $d$?