Are $R=K[a,b,c,d]/(ad-bc, a^2c-b^3, bd^2-c^3, ac^2-b^2d)$ and $K[x^4,x^3y,xy^3,y^4]$ isomorphic?

Question

I saw somewhere that $R=K[a,b,c,d]/(ad-bc, a^2c-b^3, bd^2-c^3, ac^2-b^2d)$ and $K[x^4,x^3y,xy^3,y^4]$ are considered the same. Is it true? Why?

I'm a beginner so please answer in detail.

Answer 1

They're isomorphic. Consider the map $K[a,b,c,d] \to K[x^4,x^3y,xy^3,y^4]$ given by $a \mapsto x^4$, $b \mapsto x^3 y$, $c \mapsto xy^3$, $d \mapsto y^4$. What we need to see is that the kernel of this map is the ideal $(ad - bc, a^2c - b^3, bd^2 - c^3, ac^2 - b^2d)$.

One way to do this is look at the ideal $I := (a - x^4, b - x^3y, c - xy^3, d - y^4)$ of $K[a,b,c,d,x,y]$; the relation ideal is then $J := I \cap K[a,b,c,d]$. (Try verifying yourself that this is indeed the kernel of the map above.)

The ideal $J$ is called an elimination ideal as it eliminates the variables $x, y$. Now choose as term ordering an elimination order (one in which every monomial containing an $x$ or $y$ is larger than every monomial in just $a, b, c, d$) and compute a Grobner basis $G$ of $I$ with respect to that ordering. Then the elements of $G$ that are purely polynomials in $a, b, c, d$, i.e., those in $G \cap K[a,b,c,d]$, generate the relation ideal $J$.

For instance, Magma (online calculator) can do this all by itself.

Q := RationalField();
R<x,y> := PolynomialRing(Q,2);
P<a,b,c,d> := PolynomialRing(Q,4);
RelationIdeal([x^4,x^3*y,x*y^3,y^4],P);

gives

Ideal of Polynomial ring of rank 4 over Rational Field
Order: Lexicographical
Variables: a, b, c, d
Homogeneous, Dimension >0
Basis:
[
    a^2*c - b^3,
    a*c^2 - b^2*d,
    b*d^2 - c^3,
    a*d - b*c
]

(And we're lucky, because it gives exactly the relations you want.)