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Is $R = \mathbb{Q}[x,y]/((x+y)^3-2)$ a UFD? If yes, what are prime ideals of this ring?

UFD is a ring where every element is a product of prime elements. I.e. (hope, I got it right) $(x) \subset p_1 p_2 \cdots p_n $ Note that $p_i$ may be non-principal ideal.

I know that $\mathbb{Q}[x,y]$ is a UFD because it's a polynomial ring over a field (which is PID hence UFD). But what is going on when we factor it over irreducible polynomial?

Here is the statement that $R=K[x,y]/(x^2-y^3+y)$ is not a UFD. I don't know how regularity of the ring relates to UFD'ness.

There $\mathbb{Q}[x,y]/(x^2-y^3)$ is not a UFD as well. There is an answer proposing to factor over a potentially prime ideal and check if the factor is ID.

My attempt. We see that the prime ideals are $(0), (x-a), (y-a), (x-a, y-a)$.

$(f) = (x+y)^3-2$ lies in all ideals of $\mathbb{Q}[x,y]$ where the polynomial has a solution. Alas, this poly has no solutions. Factorization leaves the prime ideals that contain (f), but we see that no such ideal exist. And that means that the result is a field so a UFD. (Here is a mistake I bet)

I'm new in commutative algebra so I have made a bunch of false statements here, I assume.

I would like some geometric interpretation for the question as well, if possible.

user26857
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Lada Dudnikova
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  • Your "no solutions" sounds like the right idea, but it's rather vague of course. I recommend proving that $\mathbb{Q}\left[x,y\right] / \left(\left(x+y\right)^3-2\right) \cong \mathbb{Q}\left[u,v\right] / \left(u^3-2\right)$ as rings first, and then writing the latter ring as $F\left[v\right]$ for $F = \mathbb{Q}\left[u\right] / \left(u^3-2\right)$. – darij grinberg Aug 27 '19 at 09:03
  • @darijgrinberg So, there is $F[v]$ which is a poly ring over field extension $\mathbb{Q}[\omega]$? Is the answer "yes" then? – Lada Dudnikova Aug 27 '19 at 09:13
  • Yes and yes. But you have to find the isomorphism :) – darij grinberg Aug 27 '19 at 09:14
  • @darijgrinberg I thought about this isomorphism but making the ring isomorphsm is a questions itself to me. – Lada Dudnikova Aug 27 '19 at 09:31
  • So you want a $\mathbb{Q}$-algebra isomorphism from $\mathbb{Q}\left[x,y\right]$ to $\mathbb{Q}\left[u,v\right]$ that sends $x+y$ to $u$ (because that will then induce the isomorphism you need on the quotient rings). But a $\mathbb{Q}$-algebra homomorphism out of $\mathbb{Q}\left[x,y\right]$ is uniquely determined by specifying its values at $x$ and $y$. And you want it to be an isomorphism, so there should be an inverse. Can you work from that? – darij grinberg Aug 27 '19 at 09:40
  • Oh, I see, $x = \frac{u+v}{2}$ and $y = \frac{u-v}{2}$ behaves well under multiplication. I think that this is unique Q-algebra isomorphism such that $x+y \to u$. – Lada Dudnikova Aug 27 '19 at 09:53
  • It's not the only one. There is a simpler one. But this one is fine, too. – darij grinberg Aug 27 '19 at 09:54
  • Can someone post an answer so that the question is gone from the unanswered queue? Thanks. – Ayman Hourieh Aug 27 '19 at 12:25

1 Answers1

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If we take an isomorphism, as stated in a comment, we get a polynomial ring over a field which is a UFD and the primes are irreducible polys over $\mathbb{Q}[\omega]$. There are plenty of them, because the field is not algebraically closed.

user26857
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Lada Dudnikova
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