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We know that the tensor product of polynomial rings over an algebraically closed field is again a polynomial ring, and hence, a UFD. My question is:

If $R$ and $R'$ are two polynomial rings over an algebraically closed field $\mathbb{K}$, and say $I\subset R$ and $J\subset R'$ are two ideals such that $R/I$ and $R'/J$ are UFDs, then is it true that $R/I \otimes_{\mathbb{K}} R'/J$ is a UFD?

Or, at least, is the above statement true when both $I$ and $J$ are principal ideals in their respective polynomial rings over $\mathbb{K}$?

user26857
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It'sMe
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  • It might happen that $R/I$ or $R/J$ is not even an integral domain. – Bernard Oct 31 '21 at 22:28
  • See: https://math.stackexchange.com/questions/3578801/is-the-tensor-product-of-ufds-again-ufd – morrowmh Oct 31 '21 at 22:35
  • @MichaelMorrow I have seen this, but the discussion seems vague and I don't think it properly answers the question. At least I don't understand, so if you can explain it to me, then that'll be great – It'sMe Oct 31 '21 at 22:36
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    Assume for simplicity that your UFDs are finite type $k$-algebras, with $K$ algebraically closed. The tensor product $R/I \otimes_k R'/J$ is certainly going to be Noetherian and normal. To see normality, recall Serre's condition for normality is that a ring is normal iff it satisfies $R_1$ and $S_2$. By theorem 1.6 of https://arxiv.org/pdf/math/0210359.pdf, $R_n$ and $S_n$ ascend to the tensor product over a perfect field (which is ok since algebraically closed implies perfect). Now recall that a Noetherian normal ring is a UFD iff it has trivial Picard group. – Badam Baplan Nov 01 '21 at 23:12
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    (cont) this doesn't answer your question but it does reduce it the question of when the Picard group of your tensor product will be trivial. e.g. see here https://math.stackexchange.com/questions/493005/picard-group-of-product-of-spaces. The answers there show e.g. that if $K(R/I)$ is a purely transcendental extension of $k$ then you have a positive answer to your question. – Badam Baplan Nov 01 '21 at 23:36
  • @BadamBaplan thanks for your excellent explanation. I was wondering if you can write it in the answer column. Also, I was wondering the result that you mentioned, "a Noetherian normal ring is a UFD iff it has trivial Picard group", isn't the condition on the class group, instead of the Picard group? – It'sMe Nov 02 '21 at 06:20
  • @FreePawn oh geez, yes please replace Picard group by class group in what I said above, it's totally wrong as stated now (e.g. many local normal domains are not UFDs). When writing that I was thinking for some reason that everything was already known to be factorial so the notions would coincide. – Badam Baplan Nov 02 '21 at 06:34

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