This is exercice 5.4 from Miles Reids Undergraduate Commutative Algebra.
Let $k$ be an algebraically closed field. For $f \in k[x_1,...,x_n]$, write $V(f) \subset k^n$ for the hypersurface defined by $f=0$.
Using Hilberts Nullstellensatz, one can show that:
Claim: if $f$ is irreducible and $f$ does not divide $g$, then $V(f) \not\subset V(g)$ (see this post of mine).
From this result, I want to deduce the following.
If $g= $ cst $\cdot \, \,\Pi \, \,f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$.
If I understand correctly, this follows from the counterpositive statement of the claim above.
Indeed, for any $f_i$, we have that $V(f_i) \subset V(g)$ and hence $f_i$ is either irreducible or $f_i$ divides $g$. If $f_i$ is irreducible, we are done. If not, we can reiterate the process until we get irreducibility.
Does this make sense?
Thank you for all your comments and remarks.
You can't just use the contrapositive of the quoted result, you need the if and only if form. I.e., that $f\mid g$ if and only if $V(f)\subseteq V(g)$.
Your proof is phrased rather weirdly, and I can't quite tell what you're saying. I thought that the $f_i$s were the irreducible factors of $g$, but then you conclude that $f_i$ is either irreducible or $f_i$ divides $g$, which makes no sense as a conclusion if the $f_i$s are irreducible factors, since we know both of those things about $f_i$ from the start.
The correct proof goes like this.
Let $g=c\prod_i f_i^{n_i}$ with the $f_i$s irreducible polynomials with $f_i$ not an associate of $f_j$ for $i\ne j$, and $c\in k$. Then $f_i\mid g$, so $V(f_i)\subseteq V(g)$. Moreover, if $a\in k^n$ with $g(a)=0$, then $\prod_i f_i(a)=0$, so for some $i$, $f_i(a)=0$. Thus $a\in V(f_i)$ for some $i$. Thus $V(g)\subseteq \bigcup_i V(f_i)$. Thus we have that $$ V(g) = \bigcup_i V(f_i).$$
It just remains to check that $V(f_i)$ is irreducible. Suppose not, then there are $g$ and $h$ such that $V(f)\subseteq V(gh)$, but $V(f)\not\subseteq V(g)$ and $V(f)\not\subseteq V(h)$. Using that $f\mid g$ if and only if $V(f)\subseteq V(g)$, this translates to the existence of $g$ and $h$ such that $f_i\mid gh$, but $f_i\nmid g$ and $f_i\nmid h$, which contradicts the irreducibility of $f_i$, since irreducibles are prime in $k[x_1,\ldots,x_n]$.