Showing that $X \times \mathbb{A}^1$ is regular in codimension one.

Question

I am reading the proof of Proposition 6.6 in Hartshorne which states that $\operatorname{Cl} X \cong \operatorname{Cl} (X \times \mathbb{A}^1)$.

Let $X$ be a noetherian, integral, separated scheme which is regular in codimension one. We want to show $X \times \mathbb{A}^1$ is regular in codimension one.

Hartshorne starts out be describing points in $X \times \mathbb{A}^1$ which are of codimension one. I am already confused at this point.

First of all, we talk about codimension for closed, irreducible subsets of $X$ but I am not sure how we know that a point in $X \times \mathbb{A}^1$ is necessarily closed and irreducible.

More importantly, how does knowing what the points of codimension one are help us show that the stalks of dimension one are regular?

I don't understand the motivation here at all.

I notice the next claim in the proof was asked about but never answered: $X \times \mathbf{A}^1$ is regular in codimension 1

(I am also very lost as to what Hartshorne is trying to say in that claim however I didn't add it to this question as it has already been asked but was never answered).

Answer 1

1. The dimension of the stalk at a point is equal to the codimension of the closure of that point.

2. There are two types of points whose closure will have codimension 1. It is helpful to think of points of codimension 1 as hypersurfaces of $X \times \mathbb{A}^1$.

In order for the closure of a point in $X \times \mathbb{A}^1$ to have codimension one, its closure needs to be either $X$, or letting $H$ denote a hypersurface of $X$, $H \times \mathbb{A}^1$. The latter are the points of type 1 and the former are those of type 2.