
Lemma 15.47.3. Let $R$ be a regular ring. Let $f \in R$. Assume there exists a derivation $D : R \to R$ such that $D(f)$ is a unit of $R$. Then $R[z]/(z^ n - f)$ is regular for any integer $n \geq 1$. More generally, $R[z]/(p(z) - f)$ is regular for any $p \in \mathbf{Z}[z]$.

Proof. By Algebra, Lemma 10.157.10 we see that $R[z]$ is a regular ring. Apply Lemma 15.47.2 to the extension of $D$ to $R[z]$ which maps $z$ to zero. This works because $D$ annihilates any polynomial with integer coefficients and sends $f$ to a unit. $\square$

Comment #3525 by Dario Weißmann on

Doesn't this work for any regular ring and any polynomial of the form $p+f$ where $p\in \mathbf{Z}[z]$ is regarded as a polynomial in $R[z]$?

A reference to the fact that a polynomial algebra over a regular ring is regular would be nice. I can't find it in the project so here is what I came up with (compiles fine for me but not as a comment?): Let $R$ be a regular ring. To show that $R[z]$ is a regular ring it suffices to show that the localization at all maximal ideals of $R[z]$ are regular local rings. Let $\mathfrak{m}$ be a maximal ideal of $R[z]$. Write $\mathfrak{p}=R\cap \mathfrak{m}$. Then $R[z]_{\mathfrak{m}}=R_{\mathfrak{p}}[z]_{\mathfrak{m}}$ and we can reduce to $R$ is a regular local ring, $\mathfrak{m}$ is a maximal ideal of $R[z]$ lying over $\mathfrak{m}_R$. Then $R[z]/\mathfrak{m}=\kappa[z]/\overline{\mathfrak{m}}$ is a field where $\overline{\mathfrak{m}}$ denotes the image of $\mathfrak{m}$ in $\kappa[z]$. Let $\overline{f}\in \kappa[z]$ be a generator of $\overline{\mathfrak{m}}$ and $f\in R[z]$ be a lift. Further let $f_1,\dots,f_r$ be a regular sequence in $R$ generating the maximal ideal. Then $f_1,\dots,f_r,f$ defines a regular sequence in $R[z]_{\mathfrak{m}}$ generating $\mathfrak{m}$.

Comment #3526 by Dario Weißmann on

OK, so it only didn't compile in the preview?

Comment #3529 by on

The preview and the final render are dealt with by different pieces of the software, but of course they should be as close together as possible. I'll see what I can do to improve this in the future, thanks for telling me about it.

Comment #3530 by on

Any smooth algebra over a regular ring is regular, see 10.157.10. The proof is essentially what you said in the case of polynomial rings.

Unfortunately, I don't understand your first comment. Can you clarify?

Comment #3531 by Dario Weißmann on

So the proof works by lemma 07PF for any polynomial mapping to a unit in R[z] under some derivation. We have $D(z^k)=kz^{k-1}D(z)=0$ by definition of $D$. By linearity the same is true for any polynomial $p$ with integer coefficients regarded as a polynomial in $R[z]$ via $\mathbf{Z}[z]\to R[z]$. So $D(p+f)=D(f)$ is a unit in $R[z]$.

We didn't need the $\mathbf{F}_p$-algebra structure and it holds for a larger class of polynomials.

Thank you for the reference!

Comment #3534 by on

OK, I understand now. Thank you. I will change the lemma when I next go through the comments.

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