0
$\begingroup$

My question is about tensor product and Noetherian ring. We know that that tensor product of two general comutative ring is not Noetherian in general and even tensor product of two Noetherian ring also not Noetheria in general. However if we have a ring $R$ Notherian and $S$ is finitely generated (both over the field $k=\mathbb{Q}_p$ or its finite extension), then the tensor product $R \otimes_{k} S$ is also Noetherian. I think in this case if we consider scalars from p-adic ring $\mathbb{Z}_p$ then also the same result holds i.e., $R \otimes_{\mathbb{Z}_p} S$ is also Noetherian with the above two conditions.

Now if we consider the ring of witt vectors $W(\kappa)$ over the residue field $\kappa$ of characteristic $p$ of a finite extension of the $p$-adic field $\mathbb{Q}_p$ and if I am not wrong then this ring $W(\kappa)$ is not a Noetherian ring. Next, consider a finite flat $\mathbb{Z}_p$ algebra $R$ and take the action of it on $W(\kappa)$ by $R \otimes_{\mathbb{Z}_p} W(\kappa) $. I don't know whether the ring $W(\kappa)$ or $W(\kappa)[[x]]$ is finitely generated but probably no and in fact $W(\kappa) \cong \mathbb{Z}_p$ (due to $\textbf{Witt}$). Hence the above can not be applied possibily, though I am not sure. So the question:

$(i)$ What conditions on $R$ make the ring $R \otimes_{\mathbb{Z}_p} W(\kappa) $ Noetherian ?

$(ii)$ What conditions on $R$ make the ring $(R \otimes_{\mathbb{Z}_p} W(\kappa))[[x]] $ Noetherian ?

$(iii)$ If we rather consider the ring $W_n(\kappa)$ of $n$-truncated $p$-typical witt vectors, then can we say whether $R \otimes_{\mathbb{Z}_p}W_n(\kappa)$ is Noetherian ?

Because here $W_n(\kappa)$ is probably finitely generated ring.

Any help with questions $(i), (ii), (iii)$ please

$\endgroup$
  • $\begingroup$ Obviously your question (which doesn't make sense as it is) is how to construct $W(\kappa)$ and which kind of $\Bbb{Z}_p$-module $R$ is supposed to be. If you assume that $W(\kappa)\cong \Bbb{Z}_p$ as $\Bbb{Z}_p$-module then $R \otimes_{\mathbb{Z}_p} W(\kappa)\cong R \otimes_{\mathbb{Z}_p} \Bbb{Z}_p\cong R$. Finitely generated as what ? $\endgroup$ – reuns Jan 18 '20 at 23:33
  • $\begingroup$ @reuns, my question is- If $R$ is a finite flat $\mathbb{Z}_p$-algebra and $W(\kappa)$ is the ring of witt vectors, then can we say $R \otimes_{\mathbb{Z}_p} W(\kappa)$ is Noetherian under some conditions on $R$ or on $W(\kappa)$? $\endgroup$ – Why Jan 19 '20 at 5:16
2
$\begingroup$

If $k$ is a finite extension of $\Bbb F_p$ then $W(k)$ is a principal ideal domain and so it is noetherian so you don't need any more condition on $R$.

In general by your condition $R$ would be a faithfully flat extension of $\Bbb Z_p$: any finite flat extension of a local ring is faithfully flat and you can easily see that if $S$ is not noetherian then $S\otimes R$ is not noetherian (just take an ideal if after base change it become finitely generated by faithfull flatness it must be finitely generated before base change.)

$\endgroup$
  • $\begingroup$ Is the ring of witt vectors $W(k)$ is a PID ? $\endgroup$ – Why Jan 18 '20 at 21:23
  • 1
    $\begingroup$ Yes, it's a finite unramified extension of $\Bbb Z_p$. Thus every ideal is principal and generated by $p^n$ for some $n$. You can find this in most introductory treatments of the Witt vectors. $\endgroup$ – KReiser Jan 19 '20 at 0:05
  • $\begingroup$ @KReiser, so you mean $W(k)$ is always a PID for the residue field $k$ of characteristic $p$. Is it? $\endgroup$ – Why Jan 19 '20 at 5:17
  • 1
    $\begingroup$ The answerer and I made this claim for $k$ a finite extension of $\Bbb F_p$. I don't remember the general case (when $k$ may not be finite) off hand. Perhaps you could go look it up. $\endgroup$ – KReiser Jan 19 '20 at 5:19
  • 1
    $\begingroup$ @M.A.SARKAR w(k) is a p.i.d for any prefect field of charactetistic p. $\endgroup$ – ali Jan 19 '20 at 6:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.