Let $k$ be a field and is algebraically closed. Let $f\in k[x_{1},\cdots, x_{n}]$ be a polynomial (not necessarily irreducible). As $k[x_{1},\cdots, x_{n}]$ is a UFD, we can write $f=ug_{1}\cdots g_{n}$ where $u$ is a unit, and the $g_{i}$ are irreducible. I want to prove the following:
Every irreducible component of $Z(f)$ is of the form $Z(g_{i})$ for some $i$.
As a set $W\subseteq Y$ is an irreducible component if and only if $W$ is a maximal irreducible susbset in $Y$, we only need to show that
- each $Z(g_{i})$ is a maximal irreducible subset of $Z(f)$.
- $Z(f)=\bigcup_{k=1}^{n}Z(g_{k})$.
The second claim is easy to prove: note that since $k$ is a field, the unit $u$ must be a nonzero constant, and thus $Z(u)=\varnothing$. Hence, $$Z(f)=Z(ug_{1}\cdots g_{n})=Z(u)\cup Z(g_{1})\cup\cdots\cup Z(g_{n})=\bigcup_{k=1}^{n}Z(g_{k}).$$
However, I got stuck in how to show $Z(g_{i})$ is a maximal irreducible subset of $Z(f)$.
On one hand, it is clear that $Z(g_{i})$ is irreducible. Indeed, we take $Z(g_{1})$ as an example. As $g_{1}$ is irreducible and $k[x_{1},\cdots, x_{n}]$ is a UFD, $\langle g_{1}\rangle$ is prime and thus $I(Z(g_{1}))=\sqrt{\langle g_{1}\rangle}=\langle g_{1}\rangle$ is prime, which means that $Z(g_{1})$ is irreducible.
On the other hand, let $W$ be another irreducible subset of $Z(f)$ such that $W\supseteq Z(g_{1})$. How could I prove the inverse inclusion? The first attempt I had is to prove that if $W\neq Z(g_{1})$, then $W$ cannot be irreducible.
To this end, suppose that $Z(g_{1})\subsetneq W$, then $\langle g_{1}\rangle=I(Z(g_{1}))\supsetneq I(W).$ So, let $h\in I(W)$, then $h\in\langle g_{1}\rangle$, and thus we can write $h=rg_{1}$ for some $r\in k[x_{1},\cdots, x_{n}]$. Therefore, $rg_{1}\in I(W)$. As $W$ is irreducible, $I(W)$ is a prime ideal, and thus either $r\in I(W)$ or $g_{1}\in I(W)$.
But $g_{1}$ cannot belong to $I(W)$, otherwise $\langle g_{1}\rangle \subseteq I(W)$, and thus we can equality $\langle g_{1}\rangle=I(W)$, contradicting the properness.
Hence, $r\in I(W)$. But then $r\in I(W)\subsetneq \langle g_{1}\rangle$, and thus $r=r_{2}g_{1}$ for some $r_{2}\in k[x_{1},\cdots, x_{n}]$. By the same argument, $r_{2}\in I(W)$. Also, we can write $h=rg_{1}=r_{2}g_{1}^{2}.$
As $k[x_{1},\cdots, x_{n}]$ is an integral domain and $g_{1}$ is irreducible (so it is not constant and thus cannot be zero), we have $$rg_{1}=r_{2}g_{1}^{2}\implies g_{1}(r-r_{2}g_{1})=0\implies r=r_{2}g_{1}.$$
By the same argument, we will have $$h=rg_{1}=r_{2}g_{1}^{2}=r_{3}g_{1}^{3}\implies r_{2}=r_{3}g_{1}.$$
Well, I can go like this forwever: $$h=rg_{1}=r_{2}g_{1}^{2}=r_{3}g_{1}^{3}=\cdots,$$ this is already weird enough but I cannot get a contradiction..
Best Answer
What you've written at the end of your post is indeed a contradiction: since $g_1$ is irreducible, it shows up to some maximum power in the factorization of $h$ because we are working in a UFD. But you've shown $g_1^n$ divides $h$ for all $n$, contradiction.