Irreducible polynomials and irreducible varieties I need some help with an exercise (proved half of it but now I'm stuck on the other half). This is an exercise from Smith's book about algebraic geometry:
Show that a hypersurface in $\mathbb A^n$ is irreducible if and only if the defining equation $F$ is a power of an irreducible polynomial $G$. 
Here is what I have so far:
Proof: $\implies$: Let $V(F)$ be a hypersurface in $\mathbb A^n$. Assume $F$ is not a power of an irreducible polynomial. Then $F$ can be factored as $F = GH$ for some non-constant polynomials $G,H$. But then $V(F) = V(G)\cup V(H)$ is a union of two non-empty proper subvarieties hence reducible. 

Is this correct so far?

And this is the part that I'm stuck with:
$\Longleftarrow$: Now assume $F=G^n$ where $G$ is irreducible. Then $V(F) = \bigcup_{i=1}^n V(G) = V(G)$. 
I want to say something like "Then $V(G)$ does not contain two non-empty proper subvarieties because $G$ is irreducible" but it's not so clear to me how to prove this.

Please could someone help me finish this proof?

Note/Edit
This exercise is on page 12 and so far we only have the definition of variety, subvariety, morphism and dimension. Ideals and radicals have not been mentioned. 
 A: This works without the notion of ideals. Affine varieties are considered to be the common zero set of a collection of polynomials.
Let $G$ be an irreducible polynomial. Let $$V(G)=V(F_i, i \in I) \cup V(H_j, j \in J) = V(F_iH_j, i \in I, j \in J)$$
be the union of two closed subsets. We have to show, that one of those is not proper.
Fix $i \in I$ and assume $G \not\mid F_i$. We have $V(G) \subset V(F_iH_j)$ for all $j \in J$, hence  by Study's Lemma (**), we obtain $G|F_iH_j$. $G$ is irreducible, the polynomial ring is factorial, thus we get $G|H_j$ for any $j$. This shows that we can assume $G|F_i$ for any $i$ (If not, we have $G|H_j$ for any $j$). This yields 
$$V(G) \subset \bigcap_i V(F_i) = V(F_i, i \in I),$$
thus we have shown that $V(G)$ is irreducible, since it is not the union of two proper closed subsets.

(**) I assume the field to be algebraically closed, since the statement is wrong without that assumption anway, as the irreducible one-point set $V(x(x^2+1)) \subset \mathbb A^1_{\mathbb R}$ shows. As far as I can remember, your book does only treat the complex case, at least in the first chapter.
A: You can show that a closed set $X$ in $\mathbb{A}^n$ is irreducible if and only if the ideal $I(X)$ corresonding to $X$ is a prime ideal in $k[X_1, ... , X_n]$.  When does a single element $G \in k[X_1, ... , X_n]$ generate a prime ideal?
