I wrote this up, because i don't want to forget it, I made a pdf files about variety, but hasn't finished it yet... It is hard to keep things alive in your mind, if you don't get used to it a lot, after I manage to bear algebraic geometry in my faculty through a student skripsi while ago I managed to write some stuffs about variety in an undergraduate level (of course for those who have familiar with commutative algebra).. Now it's going to fade again, maybe because they are are not used with "many definitions" math (that is modern math :D)

Let's recall an important thing I wrote up in the previous post. For a closed set $X$ on the affine space $\mathbb{A}^n$, define the ideal

\[\mathfrak{A}_X :=\{f\in k[T_1,\cdots,T_n] | f(X)=0\}\]

The above is indeed an ideal, we can see it by construction of the map $\tau: k[T] \rightarrow k[X]$ where $k[X]$ is the ring of regular function.

Theorem

Let $X$ be a closed set of affine space $\mathbb{A}^n$, then the following are equivalent:
i)   $X$ be an irreducible closed set
ii)  $\mathfrak{A}_X$ is prime ideal
iii)  $k[X]$ is integral domain.

Recall that for the ring $R$ and Ideal $I$, $R/I$ is integral domain iff $I$ is prime ideal (it's standard exercise in undergraduate algebra). By the isomorphism applied on $\tau:k[T]\rightarrow k[X]$, we have $k[X]\cong k[T]/\mathfrak{A}_X$, so we have the equivalence of ii) and iii).
Now we want to prove the equivalence of iii) and i) by negating the statement first and obtain $X$ is reducible if and only if $k[X]$ is not integral domain, or equivalently $k[X]$ has zero divisors.

Suppose that $X$ is defined by polynomials $F_i$.  Let $r$ and $s$ be a zero divisor of $k[X]$, the set $A=\{y\in X | r(y)=0\}$ and $B=\{y\in X | s(y)=0\}$ define closed sets (be cautious that $r$ and $s$ are not necessary polynomials, but still $A$ and $B$ are closed sets), which are propers subset of $X$, hence $A\cup B \subset X$, furthermore for any $x\in X$ we have $s(x)r(x)=0$ therefore $x\in A$ or $x\in B$, that is $x\in A\cup B$, and we also have $X\subset A\cup B$. Therefore $X=A\cup B$, and $X$ is reducible.

Suppose that $X$ is reducible, then we can find the closed set $A$ and $B$ such that $X=A\cup B$,  consider $\mathfrak{A}_A$ and $\mathfrak{A}_B$, since $A\subset  X$ and $B \subset X$ we have by Theorem on previous post  $\mathfrak{A}_X \subset \mathfrak{A}_A$ and $\mathfrak{A}_X \subset \mathfrak{A}_B$. Since the subset is proper, we can take $F\in \mathfrak{A}_A$ and $G\in \mathfrak{A}_B$ such that $F,G\not\in \mathfrak{A}_X$, by this condition the regular function $f$ and $g$ in $k[X]$ whose representation is  $F$ and $G$ respectively are both nonzero elements of $k[X]$ (if it is zero then $F,G\in \mathfrak{A}_X$ contradicting our choice of $F$ and $G$).  Therefore $F(x)G(x)=0$ for all $x\in X$, thus $FG\in \mathfrak{A}_X$ and we conclude that  in $k[X]$ the regular function $fg$ associate to a zero function, and since $f$ and $g$ are both nonzero in $k[X]$ with $fg=0$  in $k[X]$, thus $f$ and $g$ are zero divisor.

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