Ideal $(y^2-x^3-x^2)$ is a prime ideal in $k[x,y]$

Question

I have been able to show that $(y^2-x^3-x^2)$ is a prime ideal in $k[x,y]$. However I am not so sure if the quotient ring is integrally closed in its field of fractions. I have done a proof using the fact that $k[x,y]$ is a UFD. Is my result about it being integrally closed right?

Answer 1

The quotient ring is not integrally closed. This is is because we have the identity $$ (y/x)^2=\frac{y^2}{x^2}=\frac{x^3+x^2}{x^2}=x+1 $$ showing that $y/x$ belongs to the integral closure.

Geometrically this shows as a double point of the curve $y^2=x^3+x^2$ at the origin. When $k=\Bbb{R}$ that curve is the $\alpha$-curve you may have seen in some calculus course.

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In this case we can also describe the integral closure using the above. Let $t=y/x$. We just saw that $x+1\in k[t]$. Hence $x\in k[t]$ and also $y=xt\in k[t]$. Because the integral closure is a ring, we can deduce that $k[t]$ is contained in the integral closure. OTOH, as a ring of univariate polynomials over a field, $k[t]$ is integrally closed. The fields of fractions of $k[t]$ and $k[x,y]$ coincide (leaving that to you), so we can conclude that the integral closure of $k[x,y]$ is equal to $k[t]$.