MATH32012: Commutative Algebra
Solutions to Exercises for Section 5
1. (i) Let f = X 6 − X 5 + 3X 2 − 1, g = X 2 + X + 1 in Q[X]; find q, r ∈ Q[X]
such that f = qg + r and deg(r) < deg(g).
Solution: Apply the division algorithm: you should find q = X 4 − 2X 3 + X 2 +
X + 1 and r = −2X − 2.
(ii) Let f = X 5 − 3X 4 + 2X 3 − X 2 + 4, g = X 2 + 2X + 1 in Z5 [X[; find
q, r ∈ Z5 [X] such that f = qg + r and deg(r) < deg(g).
Solution: Apply the division algorithm, working over Z5 : you should find
q = X 3 + X − 3 and r = 2.
2. For each of:
(a) ≺Lex with X ≻ Y ≻ Z;
(b) ≺Lex with Z ≻ Y ≻ X;
(c) ≺Deglex with X ≻ Y ≻ Z
write each of the following polynomials f in standard form and give lm(f ), lc(f ),
lt(f ).
(i) 2X 3 Y 2 Z − XY 4 Z + 4X 2 Y 2 Z 2
Solution: (a) 2X 3 Y 2 Z + 4X 2 Y 2 Z 2 − XY 4 Z, lm(f ) = X 3 Y 2 Z, lc(f ) = 2,
lt(f ) = 2X 3 Y 2 Z; (b) 4X 2 Y 2 Z 2 −XY 4 Z +2X 3 Y 2 Z, lm(f ) = X 2 Y 2 Z 2 , lc(f ) =
4, lt(f ) = 4X 2 Y 2 Z 2 ; (c) 2X 3 Y 2 Z + 4X 2 Y 2 Z 2 − XY 4 Z, lm(f ) = X 3 Y 2 Z,
lc(f ) = 2, lt(f ) = 2X 3 Y 2 Z.
(ii) XY − 3X 2 − XY 2 + Z 3 .
Solution: (a) −3X 2 − XY 2 + XY + Z 3 , lm(f ) = X 2 , lc(f ) = −3, lt(f ) =
−3X 2 ; (b) Z 3 − XY 2 + XY − 3X 2 , lm(f ) = Z 3 , lc(f ) = 1, lt(f ) = Z 3 ; (c)
−XY 2 + Z 3 − 3X 2 + XY , lm(f ) = XY 2 , lc(f ) = −1, lt(f ) = −XY 2 .
3. Prove that the monomial ordering Lex with X1 ≻ X2 ≻ · · · ≻ Xt satisfies
the conditions (i)-(v) of the notes in the definition of a monomial ordering. (Use
the X σ , X τ -type notation and the formal definition of Lex.) [Don’t spend too
long on this.]
Solution: Given two monomials X σ and X τ :
(i) If σ = τ then X σ = X τ ; otherwise there is a least i ≥ 1 such that σi 6= τi - if
σi > τi then (by the definition of Lex) X σ ≻ X τ and if σi < τi then X σ ≺ X τ .
(ii) If we had both X σ ≻ X τ and X σ ≺ X τ then, if i is the least i ≥ 1 such that
σi 6= τi we would have (by the definition of Lex) that both σi > τi and σi < τi ,
which is impossible, as required.
(iii) Also given X µ and supposing that X σ ≺ X τ ≺ X µ let i ≥ 1 be least such
that σi 6= τi - so σi < τi - and let j ≥ 1 be least such that τj 6= µj - so τj < µj .
If i = j then we have σi < µi so X σ ≺ X τ ; if i < j then we have σi < τi = µi
and i is the smallest integer k with σk 6= µk , so X σ ≺ X µ ; if i > j then we have
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σj = τj < µj and j is the smallest integer k with σk 6= µk , so X σ ≺ X µ . So in
each case we have X σ ≺ X µ , as required.
(iv) 1 = X σ where σi = 0 for each i so X τ 6= 1 then 1 ≺ X τ since some τi > 0.
4. Let ≺ = ≺Lex with X ≻ Y ≻ Z.
(i) Reduce f = X 3 + X 2 Y + 2Y 3 completely with respect to g = X 2 − XY .
Write the final remainder in standard form with respect to ≺.
−Xg
Solution: f = X 3 + X 2 Y + 2Y 3 −−−→ X 3 + X 2 Y + 2Y 3 − X 3 + X 2 Y =
−2Y g
2X 2 Y + 2Y 3 −−−−→ 2X 2 Y + 2Y 3 − 2X 2 Y + 2XY 2 = 2XY 2 + 2Y 3 , which is
reduced with respect to g and is in standard form.
(ii) Reduce f = 4X 3 Z − Z 2 completely with respect to f1 = 2X 2 − XY .
−2XZf1
Solution: f = 4X 3 Z − Z 2 −−−−−→ 4X 3 Z − Z 2 − 4X 3 Z + 2X 2 Y Z = 2X 2 Y Z −
−Y Zf1
Z 2 −−−−−→ 2X 2 Y Z − Z 2 − 2X 2 Y Z + XY 2 Z = XY 2 Z − Z 2 , which is reduced
with respect to f1 and is in standard form.
(iii) Reduce f = 4X 3 Z − Z 2 completely with respect to F = {f1 , f2 } where
f1 = 2X 2 − XY , f2 = Y Z − Z + 1.
−XY f2
Solution: Continuing from part (ii), XY 2 Z−Z 2 −−−−−→ XY 2 Z−Z 2 −XY 2 Z+
−Xf2
XY Z − XY = XY Z − XY − Z 2 −−−−→ XY Z − XY − Z 2 − XY Z + XZ − X =
−XY + XZ − X − Z 2 , which is reduced with respect to {f1 , f2 } and in standard
form.
5. (a) Continuing with Exercise 4(iii), use “back-substitution” in the Division
Algorithm to write f there in the form f = q1 f1 + q2 f2 + r for some polynomials
q1 , q2 and where r is the remainder you found.
Solution: Writing the results of successive reductions as h1 = 2X 2 Y Z − Z 2 ,
h2 = XY 2 Z−Z 2 , h3 = XY Z−XY −Z 2 and (h4 ) = r = −XY +XZ−X−Z 2 , we
have the following equations from the reductions: f −2XZf1 = h1 ; h1 −Y Zf1 =
h2 ; h2 − XY f2 = h3 ; h3 − Xf2 = r. Back-substituting, we obtain: r =
h2 −XY f2 −Xf2 = h1 −Y Zf1 −XY f2 −Xf2 = f −2XZf1 −Y Zf1 −XY f2 −Xf2
so, rearranging, we have f = (2XZ + Y Z)f1 + (XY + X)f2 + r.
(b) Take f , f1 and f2 as in Exercise 4(iii) and also set f3 = 2XZ −Z. Continue,
to fully reduce f with respect to {f1 , f2 , f3 } and hence to write f in the form
f = q1 f1 + q2 f2 + q3 f3 + r′ where r′ where r′ is the remainder of f that you
computed with respect to this set.
− 1 f3
Solution: r = −XY + XZ − X − Z 2 −−2−→ −XY + XZ − X − Z 2 − XZ + 12 Z =
−XY − X − Z 2 + 12 Z = r′ , which is reduced with respect to {f1 , f2 , f3 } and
is in standard form. The last reduction gives us the equation r′ = r − 12 f3 so,
combined with the result of the previous exercise, we have f = (2XZ + Y Z)f1 +
(XY + X)f2 + 12 f3 + r′ .
6. Compute S(f1 , f2 ) in each of the following cases with respect to ≺ = ≺Lex
with X ≻ Y ≻ Z, and in each case reduce S(f1 , f2 ) with respect to {f1 , f2 }.
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(i) f1 = 3X 2 Y − XY 2 , f2 = 2XY Z − 1;
Solution: L = lcm(lm(f1 ), lm(f2 )) = 6X 2 Y Z so S(f1 , f2 ) = 2Zf1 − 3Xf2 =
6X 2 Y Z − 2XY 2 Z − 6X 2 Y Z + 3X = −2XY 2 Z + 3X. This reduces (subtract
Y f2 ) to 3X − Y , which is reduced.
(ii) f1 = X + 1, f2 = Y − 1;
Solution: L = XY , so S(f1 , f2 ) = Y f1 − Xf2 = XY + Y − XY + X = Y + X.
−f2
−f1
This reduces: Y + X −−→ Y + X − Y + 1 = X + 1 −−→ X + 1 − X − 1 = 0.
(iii) f1 = X 4 Y 2 Z − 2X 2 Y 3 Z, f2 = 2X 2 Y Z 3 − 4Y 2 Z 3 .
Solution: L = 2X 4 Y 2 Z 3 so S(f1 , f2 ) = sZ 2 f1 − X 2 Y f2 = 2X 4 Y 2 Z 3 −
4X 2 Y 3 Z 3 − 2X 4 Y 2 Z 3 + 4X 2 Y 3 Z 3 = 0.
7. Note that 6(iii) shows (by 5.21) that F = {f1 = X 4 Y 2 Z − 2X 2 Y 3 Z, f2 =
2X 2 Y Z 3 − 4Y 2 Z 3 } is a Gröbner basis. Let I = hf1 , f2 i.
Use 5.10 to answer the following questions: (i) is XY Z ∈
/ I? (ii) is f ′ =
6 6 6
9 6
′′
5 2 3
X Y Z − 8Y Z ∈ I?, (iii) is f = X Y Z ∈ I? In each case where your
answer is positive, write the polynomial as a combination of f1 and f2 .
Solution: (i) Since XY Z is reduced and non-zero it is not in I.
−X 2 Y 4 Z 5 f1
(ii) X 6 Y 6 Z 6 − 8Y 9 Z 6 −−−−−−−−→ X 6 Y 6 Z 6 − 8Y 9 Z 6 − X 6 Y 6 Z 6 + 2X 4 Y 7 Z 6 =
−2Y 5 Z 5 f1
2X 4 Y 7 Z 6 − 8Y 9 Z 6 −−−−−−−→ 2X 4 Y 7 Z 6 − 8Y 9 Z 6 − 2X 4 Y 7 Z 6 + 4X 2 Y 8 Z 6 =
−2Y 7 Z 3 f2
4X 2 Y 8 Z 6 − 8Y 9 Z 6 −−−−−−−→ 4X 2 Y 8 Z 6 − 8Y 9 Z 6 − 4X 2 Y 8 Z 6 + 8Y 9 Z 6 = 0,
so X 6 Y 6 Z 6 − 8Y 9 Z 6 ∈ I. To write X 6 Y 6 Z 6 − 8Y 9 Z 6 in terms of f1 and f2 ,
note that the equations generated by the divisions are: f ′ − X 2 Y 4 Z 5 f1 = h1
(say); h1 − 2Y 5 Z 5 f1 = h2 ; h2 − 2Y 7 Z 3 f2 = 0. These give f ′ = X 2 Y 4 Z 5 f1 +
(2Y 5 Z 5 f1 + (2Y 7 Z 3 f2 )) = (X 2 Y 4 Z 5 + 2Y 5 Z 5 )f1 + 2Y 7 Z 3 f2 .
−XZ 2 f1
−XY 2 f2
(iii) X 5 Y 2 Z 3 −−−−−→ X 5 Y 2 Z 3 − X 5 Y 2 Z 3 + 2X 3 Y 3 Z 3 = 2X 3 Y 3 Z 3 −−−−−−→
2X 3 Y 3 Z 3 − 2X 3 Y 3 Z 3 + 4XY 4 Z 3 = 4XY 4 Z 3 which is reduced and not zero,
hence X 5 Y 2 Z 3 ∈
/ I.
8. Let ≺ = ≺Lex with X ≻ Y ≻ Z. With respect to this ordering find Gröbner
bases for the following ideals:
(i) I1 = hX 2 Y − Y 2 , X 2 Z − Z 2 i;
Is X 2 Y 2 Z 2 − Y Z 4 ∈ I1 ?
Solution: Set f1 = X 2 Y − Y 2 , f2 = X 2 Z − Z 2 . Compute and reduce their Spolynomial: L12 = X 2 Y Z so S(f1 , f2 ) = Zf1 − Y f2 = X 2 Y Z − Y 2 Z − X 2 Y Z +
Y Z 2 = −Y 2 + Y Z 2 which is reduced, so set f3 = −Y 2 + Y Z 2 . Now compute
and reduce the S-polynomials of f3 with the other two.
L13 = X 2 Y 2 Z, S(f1 , f3 ) = Y Zf1 + X 2 f3 = X 2 Y 2 Z − Y 3 Z − X 2 Y 2 Z +
−Z 2 f1
X 2 Y Z 2 = X 2 Y Z 2 − Y 3 Z −−−−→ X 2 Y Z 2 − Y 3 Z − X 2 Y Z 2 − Y 2 Z 2 = −Y 3 Z +
−Y f3
Y 2 Z 2 −−−−→ −Y 3 Z + Y 2 Z 2 + Y 3 Z − Y 2 Z 2 = 0.
L23 = X 2 Y 2 Z so S(f2 , f3 ) = Y 2 f2 + X 2 f3 = X 2 Y 2 Z − Y 2 Z 2 − X 2 Y 2 Z −
2
X Y Z 2 = X 2 Y Z 2 −Y 2 Z 2 and the latter is, you might spot, Z 2 f1 , so will reduce
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to 0.
Therefore {f1 = X 2 Y − Y 2 , f2 = X 2 Z − Z 2 , f3 = −Y 2 + Y Z 2 } is a Gröbner
basis for I.
To test whether or not X 2 Y 2 Z 2 − Y Z 4 ∈ I1 reduce with respect to this
−Y Z 2 f1
Gröbner basis: X 2 Y 2 Z 2 − Y Z 4 −−−−−→ X 2 Y 2 Z 2 − Y Z 4 − X 2 Y 2 Z 2 + Y 3 Z 2 =
Z 2 f3
Y Zf3
Y 3 Z 2 − Y Z 4 −−−→ Y 3 Z 2 − Y Z 4 − Y 3 Z 2 + Y 2 Z 3 = Y 2 Z 3 − Y Z 4 −−−→ Y 2 Z 3 −
Y Z 4 − Y 2 Z 3 + Y Z 4 = 0, so X 2 Y 2 Z 2 − Y Z 4 ∈ I1 .
(ii) I2 = hf1 = X 2 − XZ 2 , f2 = Y 2 − Y Z 3 , f3 = Z 3 − Zi.
Is X 2 Y 2 Z 2 − Y Z 4 ∈ I2 ?
Solution: L12 = X 2 Y 2 , S(f1 , f2 ) = Y 2 f1 = X 2 f2 = −XY 2 Z 2 + X 2 Y Z 3
−Y Z 3 f1
XZ 2 f2
−−−−−→ −XY 2 Z 2 + X 2 Y Z 3 − X 2 Y Z 3 + XY Z 5 = −XY 2 Z 2 + XY Z 5 −−−−→
−XY 2 Z 2 + XY Z 5 + XY 2 Z 2 − XY Z 5 = 0.
−Zf1
L13 = X 2 Z 3 , S(f1 , f3 ) = Z 3 f1 − X 2 f3 = −XZ 5 + X 2 Z −−−→ −XZ 5 +
XZ 2 f3
X 2 Z − X 2 Z + XZ 3 = −XZ 5 + XZ 3 −−−−→ −XZ 5 + XZ 3 + XZ 5 − XZ 3 = 0.
−Zf2
L23 = Y 2 Z 3 , S(f2 , f3 ) = Z 3 f2 −Y 2 f3 = −Y Z 6 +Y 2 Z −−−→ −Y Z 6 +Y 2 Z −
Y Z 3 f3
Y 2 Z + Y Z 4 = −Y Z 6 + Y Z 4 −−−−→ −Y Z 6 + Y Z 4 + Y Z 6 − Y Z 4 = 0.
Since all the S-polynomials reduce to 0, {f1 , f2 , f3 } already is a Gröbner
basis.
−Y 2 Z 2 f1
Now test whether X 2 Y 2 Z 2 −Y Z 4 ∈ I2 : X 2 Y 2 Z 2 −Y Z 4 −−−−−−→ X 2 Y 2 Z 2 −
−XZ 4 f2
Y Z 4 −X 2 Y 2 Z 2 +XY 2 Z 4 = XY 2 Z 4 −Y Z 4 −−−−−→ XY 2 Z 4 −Y Z 4 −XY 2 Z 4 +
−XY Z 4 f3
XY Z 7 = XY Z 7 − Y Z 4 −−−−−−−→ XY Z 7 − Y Z 4 − XY Z 7 + XY Z 5 = XY Z 5 −
Y Z 4 which will not reduce to 0 (you can continue using f3 to reduce it but
not that reduces powers of Z in steps of 2 so it will not lead to the two terms
cancelling). So the remainder of X 2 Y 2 Z 2 − Y Z 4 with respect to this Gröbner
basis is non-zero, hence X 2 Y 2 Z 2 − Y Z 4 ∈
/ I2 .
9. Let ≺ = ≺Lex with X ≻ Y ≻ Z. With respect to this ordering find reduced
Gröbner bases for the following ideals. You may assume, without check (the
first case is covered by the question above), that each generating set is already
a Gröbner basis.
(i) hX 2 − XZ 2 , Y 2 − Y Z 3 , Z 3 − Zi;
Solution: The leading monomials of f1 = X 2 − XZ 2 , f2 = Y 2 − Y Z 3 , f3 =
Z 3 − Z are X 2 , Y 2 , Z 3 respectively. The term −Y Z 3 of f2 is divisible by
Z 3 = lm(f3 ) so this basis is not reduced. Reduce f2 with respect to f3 : Y 2 −
Y f3
Y Z 3 −−→ Y 2 − Y Z 3 Y Z 3 − Y Z = Y 2 − Y Z = f2′ say. Note that each term
polynomial in {f1 , f2′ , f3 } is reduced with respect to the others, so this is a
reduced Gröbner basis.
(ii) hX + Y + Z, Y 2 + Y Z + Z 3 , Z 3 − 1, XY + Y Z + XZ, XY Z − 1i;
Solution: Set f1 = X + Y + Z, f2 = Y 2 + Y Z + Z 3 , f3 = Z 3 − 1, f4 =
XY + Y Z + XZ, f5 = XY Z − 1. The leading monomials are, respectively,
X, Y 2 , Z 3 , XY , XY Z so clearly this is not reduced. Let’s start with f5 :
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−Zf4
f5 = XY Z − 1 −−−→ XY Z − 1 − XY Z − Y Z 2 − XZ 2 = −XZ 2 − Y Z 2 − 1
Z 2 f1
−−−→ −XZ 2 − Y Z 2 − 1 + XZ 2 + Y Z 2 + Z 3 = Z 3 − 1 = f3 , which means that
f5 reduces to 0 with respect to the others, hence can be omitted. We’re not
−Y f1
finished yet since f4 can be reduced using f1 : f4 = XY + Y Z + XZ −−−−→
−Zf1
XY + Y Z + XZ − XY − Y 2 − Y Z = XZ − Y 2 −−−→ XZ − Y 2 − XZ −
f
2
Y Z − Z 2 = −Y 2 − Y Z − Z 2 −→ −Y 2 − Y Z − Z 2 + Y 2 + Y Z + Z 3 = Z 3 − Z 2
−f3
3
2
3
2
−−→ Z − Z − Z + 1 = −Z + 1, which is reduced. This gives a Gröbner basis
f1 = X +Y +Z, f2 = Y 2 +Y Z +Z 3 , f3 = Z 3 −1, f4′ = Z 2 −1. Now we can reduce
−Zf ′
4
Z 3 −1−Z 3 +Z = Z−1 = f3′ say. Which, in turn,
f3 using f4′ : f3 = Z 3 −1 −−−→=
−Zf ′
3
Z 2 − 1 − Z 2 + Z = Z − 1 = f3′ ,
can be used to reduce f4′ : f4′ = Z 2 − 1 −−−→
′
so f4 reduces to 0, hence can be omitted. That leaves f1 = X + Y + Z,
f2 = Y 2 + Y Z + Z 3 , f3′ = Z − 1. And we see we’re not finished yet since f3′ can
be used to reduce both f1 and f2 . Note that the effect of reducing using f3′ is to
replace Z by 1, so we will get f1′ = X +Y +1 and f2′ = Y 2 +Y +1 which, together
with f3′ = Z − 1 finally can be reduced no further. Hence the corresponding
reduced Gröbner basis is {f1′ = X + Y + 1, f2′ = Y 2 + Y + 1, f3′ = Z − 1}.
(iii) hX + Y Z 2 , XZ + Y, Y Z 3 − Y i.
Solution: We can reduce XZ + Y by subtracting off Z(X + Y Z 2 ), to get
Y − Y Z 3 = −(Y Z 3 − Y ), so XZ + Y reduces to 0 using the other two, hence
can be omitted, leaving {X + Y Z 2 , Y Z 3 − Y } which is clearly reduced.
10. Let I = hX 2 − Y, XY 2 − Xi be an ideal in Q[X, Y ]. Decide, by first
computing a Gröbner basis for I, whether or not X 3 − Y ∈ I.
Solution: (in brief) Set f1 = X 2 − Y , f2 = XY 2 − X. Then L12 = X 2 Y 2 so
S12 = Y 2 f1 − Xf2 which reduces to −Y 3 + Y . So set f3 = Y 3 − Y .
Next, L13 = X 2 Y 3 so S13 = Y 3 f1 − X 2 f3 which reduces to 0. Also L23 =
XY 3 , so S23 = Y f2 − Xf3 which also reduces to 0.
Hence {f1 = X 2 − Y, f2 = XY 2 − X, f3 = Y 3 − Y } is a Gröbner basis for I.
−Xf1
Compute the remainder of X 3 − Y with respect to this set: X 3 − Y −−−−→
XY − Y which is reduced but non-zero; hence X 3 − Y ∈
/ I.
11. Let I be as in Exercise 11 above and let J = hX 2 + XY 2 − X − Y i.
Determine whether or not I = J.
Solution: Note (or compute) that X 2 + XY 2 − X − Y is in I (it’s the sum
f1 +f2 ). On the other hand, the ideal, J, generated by X 2 +XY 2 −X−Y consists
of all its multiples so, in particular, f3 = Y 3 − Y , which is clearly not a multiple
of X 2 + XY 2 − X − Y is not in J and we conclude that I 6= J. (Alternatively,
just recall that a generator for a principal ideal always is a Gröbner basis for
that ideal so, by uniqueness of reduced Gröbner bases and noting that the basis
{f1 , f2 , f3 } is reduced, we conclude that I 6= J.)
12. Let I = hX 2 − Y 3 , X − Y 2 i be an ideal in Q[X, Y ]. Find I ∩ Q[Y ]. Also
find I ∩ Q[X].
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Solution: First, set f1 = X 2 − Y 3 , f2 = X − Y 2 , then take the ordering to be
(as usual) ≺Lex with X ≻ Y and compute a Gröbner basis.
We have L12 = X 2 so S12 = f1 − Xf2 = X 2 − Y 3 − X 2 + XY 2 = XY 2 −
−Y 2 f2
Y 3 −−−−→ XY 2 − Y 3 − XY 6 + Y 4 = Y 4 − Y 3 which is reduced so set f3 =
Y 4 − Y 3 . Note that since f1 − Xf2 − Y 2 f2 = f3 , f1 is a linear combination
of f2 and f3 so we can just drop it, leaving just S23 to compute. We have
−Y 3 f2
L23 = XY 4 so S23 = Y 4 f2 −Xf3 = XY 4 −Y 6 −XY 4 +XY 3 = XY 3 −Y 6 −−−−→
Y 2 f3
XY 3 − Y 6 − XY 3 + Y 5 = −Y 6 + Y 5 −−−→ 0. So {f2 = X − Y 2 , f3 = Y 4 − Y 3 }
is a Gröbner basis and hence I ∩ Q[Y ] = hY 4 − Y 3 i.
In order to compute I ∩ Q[X] we use ≺Lex with Y ≻ X. So rewrite f1 =
−Y 3 + X 2 , f− Y 2 + X in standard form with respect to this monomial ordering.
We have L12 = Y 3 so S12 = f1 − Y f2 = −Y 3 + X 2 + Y 3 − Y X = −Y X + X 2
which is reduced, so set f4 = −Y X + X 2 . Again, note that f1 = Y f2 + f4 so
can just be dropped, leaving S24 to be computed. Since L24 = Y 2 X we have
S24 = Xf2 − Y f4 = −Y 2 X + X 2 + Y 2 X − Y X 2 = f4 which reduces to 0. So
{f2 = −Y 2 + X, f4 = −Y X 2 + X 2 } is a Gröbner basis. No polynomial in that
basis is in Q[X] so we deduce that I ∩ Q[X] = {0}.
13. Let I = hX 2 , Y 2 i and J = hXY i be ideals in Q[X, Y ]. Find I ∩ J.
Solution: Introduce a new variable W and find a Gröbner basis for the ideal
A = hf1 = W X 2 , f2 = W Y 2 , f3 = (1 − W )XY = −W XY + XY i with respect
to a suitable ordering, say ≺Lex with W ≻ X ≻ Y .
We have L12 = W X 2 Y 2 so S12 = Y 2 f1 − X 2 f2 = 0.
Also L13 = W Y 2 Y so S13 = Y f1 + Xf3 = W X 2 Y − W X 2 Y + X 2 Y = X 2 Y
which is reduced, so set f4 = X 2 Y .
And L23 = W XY 2 so S23 = Xf2 + Y f3 = W XY 2 − W XY 2 + XY 2 which
is reduced, so set f5 = XY 2 .
Next, L14 = W X 2 Y so S14 = Y f1 − W f4 = 0.
And L24 = W X 2 Y 2 so S24 = X 2 f2 − W Y f4 = 0.
And L34 = W X 2 Y so S34 = Xf3 + W f4 = −W X 2 Y + X 2 Y + W X 2 Y = f4
which reduces to 0.
Similarly we check that all of S15 , S25 , S35 , S45 reduce to 0).
So {f1 , f2 , f3 , f4 , f5 } is a Gröbner basis for A and we intersect with Q[X, Y ],
to get a Gröbner basis {f4 = X 2 Y, f5 = XY 2 } for I ∩ J; that is, I ∩ J =
hX 2 Y, XY 2 i.
[This is probably what you would have guessed anyway and, without much
difficulty, could have proved. But this does illustrate the method. An example
with a less obvious answer would have been more convincing but likely would
have involved a lot more computation - you can try to compute more complicated
examples but you might need a lot of patience or the help of the Gröbner basis
facility in one of the mathematical packages available on the computers.]
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