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.
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.