What's the proper way to take a Groebner basis with respect to a quotient polynomial ring?

Question

Suppose I have the quotient ring $R=\mathbb{Q}[x,y]/I$ for some ideal $I$, and I want to find a Groebner basis for another ideal $J\subseteq R$. When computing the basis, does it make a difference if I consider $J$ as a subset of $\mathbb{Q}[x,y]$, find a basis, then quotient by $I$ vs. finding the basis in $R$ and 'quotienting' as I go along, for lack of a better word.

For reference, I'm trying to do this computation in Sage, and they seem to be giving incompatible answers:

R.<x,y> = PolynomialRing(QQ)
Rq = R.quotient([x*y-1])

ideal1 = Rq.ideal([2*x^2 - 4*y + 1, 3*x^3-2*x^2+y^2-1])
ideal2 = R.ideal([2*x^2 - 4*y + 1, 3*x^3-2*x^2+y^2-1])
print(ideal1)
print(ideal2)
print(ideal1.groebner_basis())
print(ideal2.groebner_basis())
print(x*y in ideal2.groebner_basis())

gives the output:

Ideal (2*xbar^2 - 4*ybar + 1, 3*xbar^3 - 2*xbar^2 + ybar^2 - 1) of Quotient of Multivariate Polynomial Ring in x, y over Rational Field by the ideal (x*y - 1)
Ideal (2*x^2 - 4*y + 1, 3*x^3 - 2*x^2 + y^2 - 1) of Multivariate Polynomial Ring in x, y over Rational Field
[1]
[y^3 - 319/4*y^2 + 45/8*x + 51*y - 9/2, x^2 - 2*y + 1/2, x*y + 1/6*y^2 - 1/4*x - 2/3*y]
False

so the ideal in the quotient ring is $\langle 1\rangle$, but in the original ring $\mathbb{Q}[x,y]$, it's that big mess of three polynomials. Further, $xy$ isn't in the ideal in the original ring, which suggests (to me) that I wouldn't get the same ideal if I took the quotient of the basis. Which of these is the 'right' basis? Or is my understanding of something a bit off?

I've also tried using a few different monomial orderings in the quotient ring, but that hasn't led to any progress either.

Answer 1

I am trying to give an answer, for this, i need to clarify some points during the answer. In the OP, the ideal $J$ appears once as an ideal in a quotient ring of the polynomial ring (anneau) $A=\Bbb Q[x,y]$, then as an ideal in $A$. This is maybe the point causing problems, so i will restate, hoping to hit the point of the question.

---

Let $R$ be the ring $R=\Bbb Q[x,y]$. Let $I,J$ be two ideals of $R$. Then we have the chains of inclusions (injections): $$ \begin{aligned} &0\to I\to I+J\ ,\text{ and}\\ &0\to J\to I+J\ . \end{aligned} $$ Passing to quotients of $R$ we obtain morphisms (surjections): $$ \begin{aligned} &R\to R/I\to R/(I+J)\ ,\text{ and}\\ &R\to R/J\to R/(I+J)\ . \end{aligned} $$ The $R/I$-module $(I+J)/I$ may be considered an ideal.

(The $R/J$-module $(I+J)/J$ may be considered an ideal.)

Let $Q=$Rq be the quotient mentioned in the OP.

Then we have isomorphisms: $$ \underbrace{\frac{R/I}{(I+J)/I}}_{\displaystyle=\frac Q{(I+J)/I}}\cong \frac R{I+J}\cong \frac{R/J}{(I+J)/J}\ . $$

---

Now i am trying to understand the question inside of the last quotient ring(s), based on the given example. We start explicitly with: $$ \begin{aligned} R &= \Bbb Q[x,y]\ ,\\ I &= (\ xy-1\ )\ ,\\ Q &= A/I\ ,\\ J &= (\ 2x^2 - 4y + 1\ , \ 3x^3 - 2x^2 + y^2 - 1\ )\ . \\ &\qquad\text{ Then:} \\ I+J &= (\ xy-1\ ,\ 2x^2 - 4y + 1\ , \ 3x^3 - 2x^2 + y^2 - 1\ )\ . \end{aligned} $$ (I am using I instead of ideal1 and J instead of ideal2.)

We consider the ideal "generated by $J$" inside the quotient $Q=A/I$. Strictly speaking we map $J$ to $Q=A/I$, and consider the image as an ideal. This ideal is $(J+I)/I$.

The code from the OP is building a Groebner basis now for this ideal of $Q$. Well, i am modest here, any system of generators is enough for this discussion now. One such basis may be the union of the given generators for $I$ and $J$. And taking them modulo $I$ makes the $I$-generators zero in the quotient, so the $Q=A/I$-ideal $(J+I)/I$ is generated by $(xy-1)$ modulo $I$. OK, now we can try to "do more", and ask for a Groebner basis. Let us type some code for the story so far.

R.<x,y> = PolynomialRing(QQ, 2)
I = R.ideal(x*y - 1)
J = R.ideal([2*x^2 - 4*y + 1, 3*x^3 - 2*x^2 + y^2 - 1])
Q.<X,Y> = R.quotient(I)
print(f'Q is the ring:\n{Q}')
print(f'Q(x), Q(y) are now {Q(x)}, {Q(y)}\n')
# so X, Y are the images of x, respectively y in Q = R/I
JQ = Q.ideal(J.gens())
print(f'The ideal JQ is:\n{JQ}\n')
# so already we have other generators for JQ = (J+I)/I 
print(f'... the generators of JQ are:\n')
for gen in JQ.gens():
    print(f'    {gen}')
print(f'\nJQ has the groebner pasis:\n{JQ.groebner_basis()}')

We copy+paste into the sage interpreter, obtain...

Q is the ring:
Quotient of Multivariate Polynomial Ring in x, y
    over Rational Field by the ideal (x*y - 1)
Q(x), Q(y) are now X, Y

The ideal JQ is:
Ideal (2*X^2 - 4*Y + 1, 3*X^3 - 2*X^2 + Y^2 - 1)
    of Quotient of Multivariate Polynomial Ring in x, y
    over Rational Field by the ideal (x*y - 1)

... the generators of JQ are:

    2*X^2 - 4*Y + 1
    3*X^3 - 2*X^2 + Y^2 - 1

JQ has the groebner pasis:
[1]

The "other way" from the OP is not so clear for me. We start with $J$, ideal in $R$. Then pick the Groebner basis of $J$. Then we have to pass somehow to the quotient modulo $I$. When done explicitly as above, we obtain the generator $1$.

We can also do "this" at $R$-level, considering the ideal $K=I+J$ with generators the union of the generators for $I,J$, then asking for a Groebner basis of $K$. Then passing to the quotient. We obtain a new ideal of the quotient, let us denote it by KQ in the code.

In a dialog with the sage interpreter, this is in the given example...

sage: K = R.ideal(I.gens() + J.gens())
sage: K
Ideal (x*y - 1, 2*x^2 - 4*y + 1, 3*x^3 - 2*x^2 + y^2 - 1)
    of Multivariate Polynomial Ring in x, y over Rational Field
sage: K.groebner_basis()
[1]

So there is no difference. Both ways are the same. We can consider also the ideal KQ:

sage: KQ = Q.ideal(K.gens())
sage: KQ
Ideal (0, 2*X^2 - 4*Y + 1, 3*X^3 - 2*X^2 + Y^2 - 1)
    of Quotient of Multivariate Polynomial Ring in x, y
    over Rational Field by the ideal (x*y - 1)

(As seen, the $I$-generator goes to zero when passing to the generators of KQ.)

---

So what does in an "unexpected" way the code in the OP? (In this new notation, that is more structural, seen from where i am staying now.)

I am really not expecting to see $xy$ in $J$, so also not in the small list of elements of a Groebner basis of $J$. Even $XY$ is only "by chance" in the Groebner basis of $K$, since $XY=1$ and $1$ is by chance computed as the only generator in the Groebner basis of $K$.

sage: x*y in J
False
sage: x*y in K
True
sage: X*Y in KQ
True
sage: x*y in J.groebner_basis()    # there is no chance for this, since xy not in J
False
sage: X*Y in KQ.groebner_basis()
True
sage: KQ.groebner_basis()
[1]
sage: X*Y
1