So the question is to prove that if $X = Z(\mathfrak{U})$ is a Zariski-closed subset in $A^n$, then $Y = Z(\mathfrak{P})$ is an irreducible component of $X$ if and only if $\mathfrak{P}$ is a minimal prime ideal containing $\mathfrak{U}$ in $\mathbb{k}[x_1, \ldots, x_n]$

I get that since $A^n$ is Noetherian, there is a finite decomposition into $X = X_1 \cup X_2\ \cup \cdots \ X_r$, with each $X_i = Z(\mathfrak{Q}_i)$ for some prime $\mathfrak{Q}_i$ in $\mathbb{k}[x_1, \ldots, x_n]$, with (I think) $\mathfrak{U} = \mathfrak{Q}_1 \cap \cdots \cap \mathfrak{Q}_r$.

So my job is to show that if $\mathfrak{P}$ is a minimal prime ideal of this kind then $\mathfrak{P} = \mathfrak{Q}_i$ for some $1 \leq I \leq r$, and that if $Y = X_i$ then $\mathfrak{P}$ is a minimal prime ideal &c &c.

And then I'm stuck. I mean, intuitively I can see why it has to be so - if $\mathfrak{P}$ isn't minimal then there's something in $Z(\mathfrak{P})$ which isn't in $Z(\mathfrak{U})$ - or conversely if there's no $\mathfrak{Q}_i$ such that $\mathfrak{P}$ = $\mathfrak{Q}_i$ then you can't have $Y = X_i$, but I can't see a way to do it formally.