I am looking for help proving the following statement, and if that is possible to mention me a book containing the proof, and thanks in advanced.
Let $V$ denote an affine variety in $\mathbb{A}_{K}^{n}$ (over an algebraically closed field $K$). Prove that an affine variety $W \subset V$ is an irreducible component of $V$ if and only if $I(W)$ is a minimal prime ideal over $I(V)$.
I tried the following
: $\Rightarrow $: If $W=W_{1},W_{2},...,W_{k}$ are the irreducibles components of $V$, then we will prove that $I(W_{1}),...,I(W_{k})$ are the minimal prime ideals lying over $I(V)$. Let $P$ be a prime ideal in lying over $I(V)$. Then: $\prod_{i=1}^{k}I(W_{i})\subseteq \bigcap_{i=1}^{k}I(W_{i})=I(V)\subseteq P$. Since $P$ is prime, this implies $I(W_{i})\subseteq P$ for some $i$. This proves that every minimal prime ideal lying over $I(V)$ belongs to the set $\{I(W_{1}),...,I(W_{k})\}$. It remains to check that there are no inclusion relation between the $I(W_{i})$: If $I(W_{i})\subseteq I(W_{j})$ for some $i \neq j$ then $W_{i}=\mathbb{V}(I(W_{i}))\supseteq \mathbb{V}(I(W_{j}))=W_{j}$, a contradiction. (Where $\mathbb{V}$ is the variety or the zero set).
$\Leftarrow $: Since $I(V)$ is a radical ideal in $K[x_{1},...,x_{n}]$ and $K[x_{1},...,x_{n}]$ is Noetherian, that we know $I(V)=\sqrt{I(V)}=\bigcap_{i=1}^{k}Q_{i}$, the intersection of all distinct prime ideals $Q_{i}$ containing $I(V)$ (and they are only finitely many). Or we can describe these components in another way, if we go to the coordinate ring $A(V)$ of $V$ which is reduced and Noetherian. We know then that there are only finitely many minimal prime ideals in $A(V)$, say $P_{i}$, where $1\leq i\leq k$. Now we have $(0)=\sqrt{(0)}=\bigcap_{i=1}^{k}P_{i}$ in $A(V)$ and hence $V=\bigcup_{i=1}^{k}\mathbb{V}(P_{i})$. (We have $P_{i}=Q_{i} \ mod \ I(V)$).