Solutions to MA3417 Homework assignment 4
1. Using the Gröbner basis for the ideal I = (y 2 , x2 y + xyx + yx2 ) ⊂ F hx, yi that you computed in the previous
homework, compute the dimension of the space of elements of degree n in F hx, yi/I.
Solution. The Gröbner basis in question is
{y 2 , x2 y + xyx + yx2 , xyxy − yxyx},
so the leading monomials are y 2 , x2 y, and xyxy. Let us attempt to describe normal words. In principle, a word may
begin with one of the two-letter combinations
xx, xy, yx, yy.
The last one cannot occur in a normal word, since y 2 is a leading monomial of our Gröbner basis. The first one
cannot be followed by y (since x2 y is not allowed), so it has to be followed by x, which in turn has to be followed by
x, etc., giving a normal monomial xn of the given length n. The second one cannot be followed by y (since y 2 is not
allowed), so it must be followed by x, which then must be followed by x again, since xyxy is not allowed, which must
be followed by x since x2 y is not allowed, etc., giving a normal monomial xyxn−2 of the given length n. Therefore,
for each length starting from 2, there are two monomials starting from x of that length. If a monomial starts from
y, the second letter must be x, and there are no further restrictions, so there are two monomials starting from y of
each length l ≥ 3. We conclude that the dimension dn in question is given by

1, n = 0,



2, n = 1,

3, n = 2,



4, n ≥ 3.
2. Choose some ordering of monomials and compute the reduced Gröbner basis for the ideal
I = (x2 + z 2 , yz − zy, yx − xy, zx + xz) ⊂ F hx, y, zi.
Describe the normal monomials with respect to the Gröbner basis you computed, and use them to compute the
dimension of the space of elements of degree n in F hx, y, zi/I.
Solution. Let us take the DEGLEX ordering with x > y > z. Normalising the given elements so that the leading
coefficients are 1, we get
x2 + z 2 , yz − zy, xy − yx, xz + zx,
so that there are three overlaps: x3 = x2 ·x = x·x2 , x2 y = x2 ·y = x·xy, and x2 z = x2 ·z = x·xz. The corresponding
S-polynomials and their remainders are
(x2 + z 2 )x − x(x2 + z 2 ) = z 2 x − xz 2 → z 2 x − xz 2 + (xz + zx)z = z 2 x + zxz → z 2 x + zxz − z(xz + zx) = 0,
(x2 + z 2 )y − x(xy − yx) = z 2 y + xyx → z 2 y + xyx − (xy − yx)x = z 2 y + yx2 →
→ z 2 y + yx2 − y(x2 + z 2 ) = z 2 y − yz 2 → z 2 y − yz 2 + (yz − zy)z = z 2 y − zyz → z 2 y − zyz + z(yz − zy) = 0,
(x2 + z 2 )z − x(xz + zx) = z 3 − xzx → z 3 − xzx + (xz + zx)x = z 3 + zx2 → z 3 + zx2 − z(x2 + z 2 ) = 0.
It follows that the generators of this ideal form a Gröbner basis. The leading monomials are xy, yz, xz, x2 , so normal
monomials are
z k y l and z k y l x,
n monomials of the first type for the given n, and n − 1 ones of the second type. We conclude that the dimension
dn in question is given by
dn = 2n + 1.
3. Pick some ordering of monomials and compute the reduced Gröbner basis for the ideal
I = (x2 + yz, x2 + 3zy) ⊂ F hx, y, zi.
Describe the normal monomials with respect to the Gröbner basis you computed, and use them to compute the
dimension of the space of elements of degree n in F hx, y, zi/I.
Solution. Let us take the DEGLEX ordering with z > y > x. The leading monomials of the generators of our ideal
are yz and zy; they form the overlaps yzy and zyz.The corresponding S-polynomials are
1
1
(yz + x2 )y − y(zy + x2 ) = x2 y − yx2
3
3
and
1
1
(zy + x2 )z − z(yz + x2 ) = x2 z − zx2 .
3
3
The leading monomials here are yx2 and zx2 , leading to two new overlaps, zyx2 and yzx2 . The corresponding
S-polynomials and their remainders are
1
1
1
1
(zy + x2 )x2 − z(yx2 − 3x2 y) = x4 + 3zx2 y → x4 + 3zx2 y + 3( x2 z − zx2 )y =
3
3
3
3
1 4
1
1
= x + x2 zy → x4 + x2 zy − x2 (zy + x2 ) = 0
3
3
3
and
1
1
1
1
(yz + x2 )x2 + y( x2 z − zx2 ) = x4 + yx2 z → x4 + yx2 z + (x2 y − yx2 )z = x4 + x2 yz → x4 + x2 yz − x2 (x2 + yz) = 0,
3
3
3
3
so the elements
1
x2 + yz, x2 + 3zy, yx2 − 3x2 y, zx2 − x2 z
3
form a reduced Gröbner basis of our ideal. Their leading terms are yz, zy, yx2, zx2 . Let us describe the respective
normal words. Let Xn , Yn , Zn be the numbers of normal words of length n whose first letters are x, y, and z
respectively. Let us show that
Xn+1 = Xn + Yn + Zn ,
Yn+1 = Yn + Yn−1 + Zn−1 ,
Zn+1 = Zn + Yn−1 + Zn−1 .
Indeed, if a normal word of length n + 1 starts with x, the remaining part can be any normal word of length n. If a
normal word starts with y, the remaining word cannot start with z, if it starts with y, there are no restrictions, and
if it starts with x, then the remaining part must start with y or z. If a normal word starts with z, the reasoning is
the same.
These recurrence relations immediately imply Yn = Zn for all n, so we have
Xn+1 = Xn + 2Yn ,
Yn+1 = Yn + 2Yn−1 .
The second relation can be solved directly using one of many methods of solving recurrence relations; turns out that
Yn = 2n−1 . In that case, it follows that Xn = 2n −1, so the dimension dn in question is 2n −1+2n−1 +2n−1 = 2n+1 −1.
4. Pick some ordering of monomials and compute the reduced Gröbner basis for the ideal
I = (x2 + yz + zy, y 2 + zx + xz, z 2 + xy + yx) ⊂ F hx, y, zi.
Describe the normal monomials with respect to the Gröbner basis you computed, and use them to compute the
dimension of the space of elements of degree n in F hx, y, zi/I.
Solution. Let us take the DEGLEX ordering with x > y > z. The leading monomials of these relations then are x2 ,
xy, and xz. There are two overlaps, x2 y = x2 · y = x · xy and x2 z = x2 · z = x · xz. The corresponding S-polynomials
and their remainders are
(x2 + yz + zy)y − x(xy + yx + z 2 ) = yzy + zy 2 − xyx − xz 2 → yzy + zy 2 − xyx − xz 2 + (xy + yx + z 2 )x =
= yzy + zy 2 − xz 2 + yx2 + z 2 x → yzy + zy 2 − xz 2 + yx2 + z 2 x + (xz + zx + y 2 )z = yzy + zy 2 + yx2 + z 2 x + zxz + y 2 z →
→ yzy + zy 2 + yx2 + z 2 x + zxz + y 2 z − y(x2 + yz + zy) = zy 2 + z 2 x + zxz → zy 2 + z 2 x + zxz − z(xz + zx + y 2 ) = 0,
and
(x2 + yz + zy)z − x(xz + zx + y 2 ) = zyz + yz 2 − xzx − xy 2 → zyz + yz 2 − xzx − xy 2 + (xy + yx + z 2 )y =
= zyz + yz 2 − xzx+ yxy + z 2 y → zyz + yz 2 − xzx+ yxy + z 2 y + (xz + zx+ y 2 )x = zyz + yz 2 + yxy + z 2 y + zx2 + y 2 x →
→ zyz + yz 2 + yxy + z 2 y + zx2 + y 2 x − y(xy + yx + z 2 ) = zyz + z 2 y + zx2 → zyz + z 2 y + zx2 − z(x2 + yz + zy) = 0,
so the generators of our ideal do form a Gröbner basis. The normal monomials are those not containing x2 , xy, and
xz. In other words, a normal monomial of length n has all letters except for the last one equal to y or z, and the
last letter may be x, y, or z. Altogether, the dimension dn in question is 3 · 2n−1 .