Solutions to MA3417 Homework assignment 3
1. Suppose that < is a monomial ordering on words in {x1 , . . . , xn }. Let us define the recursive path ordering <rp on
words in {x0 , x1 , . . . , xn } as follows:
• If the number of occurrences of x0 in u is less than the number of occurrences of x0 in v, then u <rp v.
• If there is a tie, that is u and v both have m occurrences of x0 , we have factorisations
u = u0 x0 u1 x0 · · · um−1 x0 um
and v = v0 x0 v1 x0 · · · vm−1 x0 vm
with ui , vi being words that do not contain x0 . Let k = min{i : ui 6= vi }; we put u <rp v if uk < vk .
Show that <rp is a monomial ordering.
Solution. Let us demonstrate that <rp is a well ordering. Suppose that there is an infinite decreasing sequence
m1 >rp m2 >rp m3 >rp . . .
The number of occurrences of x0 in these words must stabilise at some point, since it cannot keep decreasing, so
without loss of generality we may assume that they all contain the same number of occurrences of x0 , say l, so that
each of these words gets split into l + 1 parts by the occurrences of x0 . Let us look at the first of those l + 1 parts
in each word. They form a non-increasing sequence which therefore must stabilise at some point, so without loss of
generality we may assume that the first parts are all the same. Then we look at the second parts, they also form a
non-increasing sequence, and therefore must stabilise, etc.
It remains to show that if u <rp v then uw <rp vw and wu <rp wv. If the number of occurrences of x0 in u is less
than the number of occurrences of x0 in v, then the same holds for uw, vw and wu, wv, so the claim follows. Thus,
it is sufficient to look at the case of a tie, so that
u = u0 x0 u1 x0 · · · ul−1 x0 ul
and v = v0 x0 v1 x0 · · · vl−1 x0 vl
with ui , vi being words that do not contain x0 . Suppose that w = w0 x0 w1 x0 · · · wm−1 x0 wm , where wi do not contain
x0 . In this case
uw = u0 x0 u1 x0 · · · ul−1 x0 ul w0 x0 w1 x0 · · · wm−1 x0 wm ,
vw = v0 x0 v1 x0 · · · vl−1 x0 vl w0 x0 w1 x0 · · · wm−1 x0 wm ,
wu = w0 x0 w1 x0 · · · wm−1 x0 wm u0 x0 u1 x0 · · · ul−1 x0 ul ,
wv = w0 x0 w1 x0 · · · wm−1 x0 wm v0 x0 v1 x0 · · · vl−1 x0 vl .
To compare uw and vw, we have to look at the sequences
(u0 , u1 , . . . , ul−1 , ul w0 , w1 , . . . , wm ) and (v0 , v1 , . . . , vl−1 , vl w0 , w1 , . . . , wm ).
It follows that if k = min{i : ui 6= vi } < l, then the same comparison proves that uw <rp vw, and if k = l, then
ul < vl implies that ul w0 < vl w0 since we assume < to be a monomial ordering, therefore uw <rp vw.
To compare wu and wv, we have to look at the sequences
(w0 , w1 , . . . , wm−1 , wm u0 , u1 , . . . , ul ) and (w0 , w1 , . . . , wm−1 , wm u0 , u1 , . . . , ul ).
It follows that if k = min{i : ui 6= vi } > 0, then the same comparison proves that uw <rp vw, and if k = 0, then
u0 < v0 implies that wm u0 < wm v0 since we assume < to be a monomial ordering, therefore uw <rp vw.
2. (a) Show that the monomials {y i xj } form a basis of the ring Chx, yi/(xy − yx − 1) viewed as a C-vector space.
(b) Computing products by hand for small i, j, k, l, guess a formula that expands the product of the basis elements
(y i xj )(y k xl ) as a combination of basis elements,
X
m n
(y i xj )(y k xl ) =
αi,j,k,l
m,n y x ,
and prove the formula you obtained.
Solution.
(a) For the DEGLEX ordering with x > y, the monomial xy is the leading term of xy − yx − 1; it does not form any
critical pairs with itself, so the element xy − yx− 1 forms a reduced Gröbner basis of our ring. Normal monomials are
those not divisible by xy, and they are precisely the monomials y i xj . From class, we know that normal monomials
do indeed form a basis.
(b) Computing for small j, k suggest that
xj y k = y k xj + jky k−1 xj−1 +
j
k k−s j−s
j
k k−l j−l
j(j − 1)k(k − 1) k−2 j−2
y
x
+ · · · + s!
y
x
+ · · · + l!
y x ,
2
s
s
l
l
where l = min(j, k). Let us prove this result; the general formula is obtained from this one by multiplying it by y i
on the left, and by xl on the right. We shall prove this formula in two steps, first for k = 1 and any j by induction
on j, and then for fixed j by induction on k.
For k = 1 and any j, this formula suggests that
xj y = yxj + jxj−1 .
This is true for j = 1 since xy − yx = 1 in the quotient ring. For j > 1 we can use the induction hypothesis, and
write
xj y = x(xj−1 y) = x(yxj−1 + (j − 1)xj−2 ) = xyxj−1 + (j − 1)xj−1 = (yx + 1)xj−1 + (j − 1)xj−1 = yxj + jxj−1 ,
as required.
For fixed j, we use the k = 1 case as the induction hypothesis, and write
xj y k = (xj y k−1 )y =
j
k − 1 k−1−s j−s
j
k − 1 k−1−l j−l
y,
= y k−1 xj + j(k − 1)y k−2 xj−1 + · · · + s!
y
x
+ · · · + l!
y
x
s
s
l
l
after which we can use the formulas xj y = yxj + jxj−1 for various j, transforming
j
k − 1 k−1−s j−s
j
k−1
s!
y
x y = s!
(y k−s xj−s + (j − s)y k−1−s xj−1−s ).
s
s
s
s
The coefficient of y k−s xj−s in the sum becomes
j
k−1
j
k−1
s!
+ (s − 1)!(j − s + 1)
=
s
s
s−1
s−1
(s − 1)!(j − s + 1)j!(k − 1)!
s!j!(k − 1)!
+
=
=
s!(j − s)!s!(k − 1 − s)! (s − 1)!(j − s + 1)!(s − 1)!(k − s)!
(k − 1)!j!
j
k
=
(k − s + s) = s!
,
(j − s)!s!(k − s)!
s
s
as required.
3. For the ideal (y 2 , x2 y + xyx + yx2 ) ⊂ F hx, yi, compute its reduced Gröbner basis for the DEGLEX ordering with x > y.
Solution.
The leading terms y 2 and x2 y of the generators of our ideal form two critical pairs: the overlap
3
2
y = y · y = y · y 2 and the overlap x2 y 2 = x2 y · y = x2 · y 2 . The first of them leads to the S-polynomial yy 2 − y 2 y = 0,
the second one produces the S-polynomial (x2 y + xyx + yx2 )y − x2 (y 2 ) whose reduced form (after eliminating all
monomials divisible by one of our leading terms y 2 and x2 y) can be computed as follows:
(x2 y + xyx + yx2 )y − x2 y 2 = xyxy + yx2 y →
→ xyxy + yx2 y − y(x2 y + xyx + yx2 ) = xyxy − yxyx − y 2 x2 →
→ xyxy − yxyx − y 2 x2 + (y 2 )x2 = xyxy − yxyx.
This new polynomial has the leading term xyxy, and as such creates three critical pairs: xyxy 2 = xyxy · y = xyx · y 2 ,
x2 yxy = x · xyxy = x2 y · xy, and xyxyxy = xyxy · xy = xy · xyxy. Let us compute the corresponding S-polynomials
and their reduced forms:
(xyxy − yxyx)y − xyx(y 2 ) = −yxyxy → −yxyxy + y(xyxy − yxyx) = −y 2 xyx → −y 2 xyx + (y 2 )xyx = 0,
x(xyxy − yxyx) − (x2 y + xyx + yx2 )xy = −xyxyx − xyx2 y − yx3 y →
→ −xyxyx − xyx2 y − yx3 y + xy(x2 y + xyx + yx2 ) = xy 2 x2 − yx3 y → xy 2 x2 − yx3 y − x(y 2 )x2 = −yx3 y →
→ −yx3 y + yx(x2 y + xyx + yx2 ) = yx2 yx + yxyx2 → yx2 yx + yxyx2 − y(x2 y + xyx + yx2 )x = −y 2 x3 →
→ −y 2 x3 + (y 2 )x3 = 0,
(xyxy − yxyx)xy − xy(xyxy − yxyx) = xy 2 xyx − yxyx2 y →
→ xy 2 xyx − yxyx2 y − x(y 2 )xyx = −yxyx2 y → −yxyx2 y + yxy(x2 y + xyx + yx2 ) = yxyxyx + yxy 2 x2 →
→ yxyxyx + yxy 2 x2 − y(xyxy − yxyx)x = yxy 2 x2 + y 2 xyx2 → yxy 2 x2 + y 2 xyx2 − yx(y 2 )x2 = y 2 xyx2 →
→ y 2 xyx2 − (y 2 )xyx2 = 0.
Therefore, all new S-polynomials can be reduced to zero, and
{y 2 , x2 y + xyx + yx2 , xyxy − yxyx}
form a reduced Gröbner basis.
4. Show that there does not exist a monomial ordering for which the ideal (x2 , xy −zx) ⊂ F hx, y, zi has a finite Gröbner
basis.
Solution. Let us consider two cases: xy > zx and xy < zx. In each of those cases, we shall be able to use the
Buchberger-style algorithm to compute the reduced Gröbner basis.
In the case xy > zx, we have x(xy − zx)− x2 ·y = −xzx, xz(xy − zx)− xzx·y = −xz 2 x, . . . , xz k (xy − zx)− xz k x·y =
−xz k+1 x, . . . , and we conclude that the elements xy − zx and xz k x with k ≥ 0 form the reduced Gröbner basis (for
all other critical pairs formed by the leading monomials of these elements, the S-polynomials are equal to zero).
In the case xy < zx, we have (zx−xy)x−z ·x2 = −xyx, (zx−xy)yx−zxyx·y = −xy 2 x, . . . , (zx−xy)y k x−z ·xy k x =
−xy k+1 x, . . . , and we conclude that the elements zx − xy and xy k x with k ≥ 0 form the reduced Gröbner basis (for
all other critical pairs formed by the leading monomials of these elements, the S-polynomials are equal to zero).
Since any Gröbner basis can be made into the reduced one by throwing out elements that have zero remainder after
long division by others, we conclude that no Gröbner basis of this ideal can be finite.