#### Q[X,Y]/(Y2−X3) is not a UFD

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User

( 4 months ago )

I'm trying to show that R=Q[X,Y]/(Y2X3)R=Q[X,Y]/(Y2−X3) is not a UFD, but I got stuck.

To prove this, I could try to find two "different" factorisations for one element, but I am not familiar with this, so I tried to use a lemma and one of the previous exercises. If my syllabus is right, in every UFD counts

x is irreduciblex is primex is irreducible⟺x is prime

My syllabus also states that the elements X¯,Y¯RX¯,Y¯∈R are irreducible. So I tried to show that at least one of the elements X¯,Y¯RX¯,Y¯∈R does not generate a prime ideal.

This would mean that I should find two polynomials f,gQ[X,Y]f,g∈Q[X,Y], such that

pQ[X,Y],fgpX(Y2X3

User

( 4 months ago )

As Jared notes, your approach is fine. You also consider directly the relation Y2X3=0Y2−X3=0, which implies that Y2=X3Y2=X3. Does this give you any hints for an element which has two distinct factorizations?

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