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.

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    $\begingroup$ They are 'considered the same' in the sense that they are isomorphic. Without any information on your background, it is extremely hard to answer this at a level that is appropriate for you. For instance, do you see there is an obvious surjective map from $R$ to $K[x^4,x^3y,xy^3,y^4]$? $\endgroup$ Apr 17, 2014 at 15:03
  • $\begingroup$ yes there is an obvious surjective map. I'm reading bruns herzog book (it's hard) $\endgroup$
    – user142666
    Apr 17, 2014 at 15:09
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    $\begingroup$ Ok, what I would do is be lazy and use Grobner basis in some computer algebra tool to show that $(ad - bc, \dots)$ is the relation ideal between $x^4, x^3y, xy^3, y^4$. Would that help you? Doubtlessly there are elementary ways to prove that $(ad - bc, \dots)$ is exactly the kernel, but I don't believe they would be very insightfull. $\endgroup$ Apr 17, 2014 at 15:11
  • $\begingroup$ what is definition of "the relation ideal" $\endgroup$
    – user142666
    Apr 17, 2014 at 15:14
  • $\begingroup$ It's the ideal of all $f(a,b,c,d)$ of $K[a,b,c,d]$ such that $f(x^3,x^3y,xy^3,y^4) = 0$, i.e., the ideal of all the relations between the four monomials. $\endgroup$ Apr 17, 2014 at 15:15

1 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);


Ideal of Polynomial ring of rank 4 over Rational Field
Order: Lexicographical
Variables: a, b, c, d
Homogeneous, Dimension >0
    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.)

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    $\begingroup$ Thanks for your answer and your fantastic link :+1 $\endgroup$ Apr 17, 2014 at 17:01
  • $\begingroup$ Magdiragdag: Thanks for your answer in details. now i have another question:" here we have two sides and want to find isomorphism. what if we had one side and wan to find the other side?" is this a simple question or i should ask it as a separate question $\endgroup$
    – user142666
    Apr 18, 2014 at 14:20
  • $\begingroup$ Well, if you only had $k[x^4,\dots]$, the computation I gave gives you exactly the other side. Now the other way around, from $k[a,b,c,d]/(ad-bc,\dots)$ compute polynomials $f_1,\dots,f_4$ such that it is isomorphic to $k[f_1,\dots,f_4]$ is different. It won't be possible in general - I mean, not even if the ideal you're modding out is prime. I'll have to think about it to see if there is an algorithm for this. $\endgroup$ Apr 18, 2014 at 14:35

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