Field of fractions, $F$, of $R=k[x,y]/(x^2-y^2+y^3)$ and integral closure of $R$ in $F$

Asked Jul 12 '23 at 10:35

Active Jul 12 '23 at 19:47

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Question:

Let $k$ be a field. Consider the integral domain $R=k[x,y]/(x^2-y^2+y^3)$.

(a) Show that $R$ is not a unique factorization domain.

(b) Let $F$ be the field of fractions of $R$. Find $t\in F$ such that $F=k(t)$.

Answer:

(a) I showed that in $R$, $y^2(y-1)=y^3-y^2\equiv -x^2=x(-x)$ but $y^2$ is not equivalent to one of the factors of $-x^2$ in $R$. Thus, $R$ is not an UFD.

(b),(c) However, for these parts, I have no clue. Any help/hint would be appreciated. Thanks in advance...

edited Jul 12 '23 at 19:47

user26857

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asked Jul 12 '23 at 10:35

confused

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Related – Jyrki Lahtonen Jul 12 '23 at 11:11
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I think the links collected here cover everything. To the extent that this question is a duplicate of that one. As I have earned two applicable dupehammer privileges, I won't vote right away (because my vote would be binding). Please comment. – Jyrki Lahtonen Jul 12 '23 at 11:14
Agreeing with Jyrki Lahtonen's duplicate suggestion: Integral Closure of Ring $k[t,s]/(s^2 -t^2(1+t))$ – KReiser Jul 12 '23 at 19:56

Field of fractions, $F$, of $R=k[x,y]/(x^2-y^2+y^3)$ and integral closure of $R$ in $F$

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