$\newcommand{\scrI}{\mathcal{I}}$ $\newcommand{\scrV}{\mathcal{V}}$ $\newcommand{\scrO}{\mathscr{O}}$ $\newcommand{\A}{\mathbb{A}}$ Context: Here we're considering the Zariski topology, we denote $\scrO(\A^n) = k[x_1,\dots,x_n]$; i.e the collection of all regular maps defined on $\A^n \rightarrow k$. The field $k$ is infinite and algebraically closed. The maps $\scrI$ and $\scrV$ denote the maps from the Nullstellensatz, i.e $\scrI$ carries algebraic sets to the ideal of all polynomials vanishing on the entire set, and $\scrV$ carries ideals to the subset of affine space which vanish every polynomial in the ideal. An algebraic set $X$ is said to be reducible if $X = Y \cup Z$ where $Y,Z$ are proper algebraic subsets of $X$.
Problem: Suppose that $\scrI(X) = (f)$ for some $f \in \scrO(\A^n)$. Show that $X$ is irreducible if and only if $f$ is irreducible in $\scrO(\A^n)$.
I think I have one direction down, but it's the other that's giving me issue. For refernce, here is the attempt.
Attempt: To prove the forward direction we prove the contrapositive, i.e $f$ reducible implies that $X$ is reducible. If $f$ is reducible then $f = gh$ for some $g,h \notin \scrO(\A^n)^\times = k\setminus \{0\}$. Because neither $g$ or $h$ is a unit, we know that $(g)$ and $(h)$ are not the entire ring $\scrO(\A^n)$. We know that $(gh) = (g)(h)$, i.e the product ideal of the principal ideals, so that $\scrI(X) = (f) = (gh) = (g)(h)$. By the Nullstellensatz, since $X$ is algebraic $\scrV(\scrI(X)) = X$ and so $X = \scrV((g)(h))$, which by the properties of vanishing sets is just $\scrV(g) \cup \scrV(h)$. We also know that $X = \scrV(f)$, moreover because $f = gh$ we know that $(f) \subset (g)$ and $(f) \subset (h)$. The contravariance of $\scrV$ gives that $\scrV(g) \subset \scrV(f)$ and $\scrV(h) \subset \scrV(f)$ and we conclude that $X = \scrV(g) \cup \scrV(h)$ is a reducible decomposition of $X$.
I'm having trouble showing the reverse direction. If $f$ is irreducible, what does that tell me about $X$? I know $\scrI(X) = (f)$ so if $f = gh$ where one has to be a unit, say wlog $h$, it follows that $\scrI(X) = (g)(h) = (g)\scrO(\A^n) = (g)$. But what does this give? Then $X = \scrV(g)$, but how can I then show this is irreducible? Thanks in advance for the hint, and if you could let me know if the first direction could be done faster I'd appreciate it. This was a 2 liner example in a book, so I might be missing an easy solution.