# $X \times \mathbf{A}^1$ is regular in codimension 1

This is Proposition II.6.6 in Hartshorne. We assume that $X$ is a Noetherian, integral, and separated scheme which is regular in codimension 1, i.e. every local ring of dimension one is regular.

For this question, I am solely interested in the Type I points, namely, points $x \in X \times \mathbf{A}^1$ whose closures are of codimension one and under the image of the projection map to $X$, is another point $y$ of codimension one. Hartshorne claims

$x$ is the generic point of $\pi ^{-1} (y)$. Its local ring $\mathcal{O}_x \cong \mathcal{O}_Y [t] _{m_y}$ is clearly a DVR because $\mathcal{O}_Y$ is a DVR. The corresponding prime divisor $\overline{\{ x \}}$ is just $\pi ^{-1} ( \overline{\{ y \} })$.

I truly have no idea what Hartshorne is trying to say here. If someone is able to explain rigorously what is quoted above, I would appreciate it.

In addition, I have attempted my own alternative proof and I hope if someone can proofread it and also compare it to the quoted proof by Hartshorne.

Reduce to the affine case so $X = \textrm{Spec } A$ and $x$ corresponds to some $p \subset A[t]$. Then being of type I, $\textrm{ht} (p) = 1$, $p' = p \cap A \ne 0$, $p = p' A[t]$. Note that $$A[t]_p = (A_{p'}[t])_p$$ Moreover, since $X$ is regular in codimension one, then $p'A_{p'} = (g)$ and thus $pA[t]_p = g A[t]_p$. I have thus shown that every point of codimension has a regular ring for its stalk.