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It is well-known that $R=k[x^2,x^3]$ is not a UFD, since $x^2x^2x^2=x^3x^3$ are two different decompositions of $x^6$ to irreducibles.

Question 1: Are $A=k[x^2,x^3+x]$ and $B=k[x^3,x^2+x]$ UFD's?

It seems to me that they are, but I have not found a proof (or a counterexample).

Question 2: If $A$ and $B$ are UFD's, is it because one of the generators ($x^3+x$ or $x^2+x$) is separable (=have different roots)? In $R$ both generators are not separable.

Thank you very much!

user237522
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1 Answers1

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In $A$, write $f = x^2, g = x^3 + x$. Then we have

$$g^2 = (x^3 + x)^2 = x^2 (x^2 + 1)^2 = f(f + 1)^2.$$

In $B$, similarly write $f = x^3, g = x^2 + x$. Then we have

$$g^3 = (x^2 + x)^3 = x^3 (x^3 + 3x^2 + 3x + 1) = f(f + 3g + 1).$$

Probably a more geometric approach is possible that would give some conditions under which $k[f, g]$ isn't a UFD.

Qiaochu Yuan
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  • Very nice, thank you! I wonder if it is possible to 'find' all $u=u(x), v=v(x) \in k[x]$ such that $k[u,v]$ is a UFD and $\dim(k[u,v])=1$ (namely, $k \subsetneq k[u,v]$). – user237522 Sep 03 '20 at 21:45
  • Of course, if $u=v$ with $\deg(u) \geq 1$, then trivially $k[u,v]=k[u,u]=k[u]$ is isomorphic to $k[x]$, hence a UFD. – user237522 Sep 03 '20 at 21:50
  • By A-M-S (Abhyankar-Moh-Suzuki) theorem: If $\deg(u),\deg(v) \geq 2$ and $\gcd(\deg(u),\deg(v))=1$, then $k[u,v] \subsetneq k[x]$. I wonder if in such a situation (two relatively prime degrees, each degree is strictly bigger than $1$), $k[u,v]$ is necessarily not a UFD? (perhaps the above idea of not being integrally closed can be applied?). Thank you. – user237522 Sep 03 '20 at 22:00
  • Is $k[x^2+x,x^3+x]$ a UFD? What is its field of fractions? If it is a UFD then my claim in the previous comment is wrong. – user237522 Sep 03 '20 at 23:42
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    @user237522 Let's assume $k$ is algebraically closed to make our lives simple. If $A\subseteq k[x]$ is a UFD of dimension $1$ then one gets a dominant map $\mathbb{A}^1\to U$ where $U:=\mathrm{Spec}(A)$ is a smooth integral curve. This then extends to a dominant map $\mathbb{P}^1\to X$ where $X$ the smooth compactification of $A$. This implies that $X\cong\mathbb{P}^1$. So then, one has that $U$ looks like the complement of a closed subscheme of $\mathbb{P}^1_k$. Thus, $U$ should be isomorphic to $k[t,\frac{1}{t-a_1},\ldots,\frac{1}{t-a_n}]$ for some $a_i$. etc. – Alex Youcis Sep 03 '20 at 23:50
  • @AlexYoucis, thank you very much! You can write your comment as another answer, if you like. Unfortunately, I still know very little about algebraic geometry, so could not understand your comment. Please, what is the relation between the original ring $A$ and $U$? – user237522 Sep 04 '20 at 00:01
  • @AlexYoucis, please another question: Is it possible to adjust the ideas in your comment to $S \subseteq k[x,y]$ a UFD of dimension $2$? Thank you. – user237522 Sep 04 '20 at 00:05
  • @user237522 Not that I can see. Curve was leveraged at several points during what I said. In particular, there's no reason that you can extend your map to a map $\mathbb{P}^2$ to a compactification of $\mathrm{Spec}(S)$. – Alex Youcis Sep 04 '20 at 00:11
  • @user237522 In fact, to be clear I think that my comment shows above that $A\cong k[t]$. So, $A$ must be generated by a single element in $k[t]$. – Alex Youcis Sep 04 '20 at 00:25