Solutions To Ma3417 Homework Assignment 2

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Solutions to MA3417 Homework assignment 2
1. For each of the following systems of polynomials f
, compute a Gr¨ o bner basis for the ideal (f
). Use
, f
, f
, f
, f
1
2
3
1
2
3
the LEX order with x > y > z.
3
2
2
2
2
(a) f
= x
yz
xz
, f
= xy
z
xyz, f
= x
y
z;
1
2
3
2
2
2
2
(b) f
= x
y + xz + y
= xz
= xyz
;
z, f
yz, f
y
1
2
3
2
2
2
2
2
(c) f
= xy
, f
= x
= y
.
z
z
y
y, f
z
1
2
3
Solution. In this question, we just give answers for the reduced Gr¨ o bner bases; these should serve you for checking
your computations.
2
2
2
2
2
2
2
2
2
3
(a) xy
, yz
, x
.
z
xyz, x
y
z, x
yz
z
z
z
z
2
2
2
2
3
2
3
2
2
2
2
(b) x
y + xz + y
, y
+ y
+ yz
, xy
+ y
+ yz.
z, xz
yz, xyz
y
z
z
2
2
1
2
1
(c) x
y
y, y
+
z, z
+
z, xz + z.
2
2
3
2
5
3
3
2
2. (a) Does polynomial xy
z
+ y
z
belong to the ideal ( x
+ y, x
y
z) ⊂ C[x, y, z]? Explain your answer.
3
2
2
(b) Does polynomial x
z
2y
belong to the ideal (xz
y, xy + 2z
, y
z) ⊂ C[x, y, z]? Explain your answer.
Solution. (a) Let us use the LEX ordering with z > x > y. Then the leading monomials of the generators of the ideal
3
3
2
5
3
are z and x
, which have no common divisors, so they form a Gr¨ o bner basis. The normal form of xy
z
+ y
z
can be computed by following steps of long division:
3
2
5
3
3
2
5
3
2
2
xy
z
+ y
z
→ xy
z
+ y
z
z
(x
y
z) =
3
2
5
2
2
3
2
5
2
2
2
2
3
2
5
4
2
= xy
z
+ y
z
x
y → xy
z
+ y
z
x
y
x
yz(x
y
z) = xy
z
+ y
x
y
z →
3
2
5
4
2
4
2
2
3
2
5
6
3
3
2
5
6
3
2
→ xy
z
+ y
x
y
z
x
y
(x
y
z) = xy
z
+ y
x
y
→ xy
z
+ y
x
y
z(x
y
z) =
3
5
6
3
2
3
5
6
3
2
2
2
3
5
6
3
4
2
= xy
+ y
yz → xy
+ y
z) = xy
+ y
x
y
x
x
y
x
yz
x
y(x
y
x
y
x
y
3
5
6
3
4
2
3
3
3
3
5
4
2
3
4
→ xy
+ y
( x
+ y) = xy
+ y
x
y
x
y
x
y
x
y
x
y
3
5
4
2
3
4
4
3
3
4
2
3
4
2
2
3
→ xy
+ y
x
y
x
y
y
( x
+ y) = xy
x
y
→ xy
x
y
xy
( x
+ y) = 0.
We see that the normal form is zero, so the polynomial is in the ideal.
(b) Let us use the LEX ordering with y > x > z. Computing a Gr¨ o bner basis of the ideal in question, we get
2
3
2
z, 2z
+ z, xz
z. The normal form of x
2y
can be computed by following steps of long division:
y
z
3
2
3
2
2
2
2
2
2
2y
→ x
2y
(xz
z) = x
2y
→ x
2y
z) =
x
z
z
x
z
z
x(xz
2
2
2
2
= xz
2y
→ xz
2y
(xz
z) = z
2y
→ z
2y
+ 2y(y
z) =
2
2
2
= z
2yz → z
2yz + 2z(y
z) = z
2z
→ z
2z
+ (2z
+ z) = 2z.
We see that the normal form is nonzero, so the polynomial is not in the ideal.
3. Use Gr¨ o bner bases to show that the first of the two following systems of equations has infinitely many solutions, and
the second one has only finitely many solutions (over complex numbers), and find all solutions to the second system.
2
2
xy = z
+ z,
xy = z
+ z,
 
 
2
2
(a)
yz = x
+ x,
(b)
yz = x
+ x,
2
2
 
 
xz = y
+ y.
xz = y
y.
Solution.
(a) Computing the reduced Gr¨ o bner basis for the corresponding ideal with respect to the ordering LEX with x > y > z,
we get the Gr¨ o bner basis
2
2
2
2
2
+ yz + y + z
+ z, xz + yz + z
+ z, xy
+ x
y
z
z, x
yz.
This means that the elimination ideal I
is {0}. Using the Extension Theorem, we conclude that since there is
2
2
a polynomial in our Gr¨ o bner basis with the leading term y
, every z can be extended to a solution (y, z) to the
2
2
elimination ideal I
= (y
+ yz + y + z
+ z). Using the Extension Theorem again, we conclude that because of the
1
2
leading term x
, every solution to I
can be extended to a solution (x, y, z) of I.
1

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