MA341D Answers and solutions to homework assignment 1
1. (a) −x21 x23 + x33 + x32 + 3x2 x1 + x1 x3 + 2x1 x2 ;
(b) −x21 x23 + x32 + x33 + 2x1 x2 + x1 x3 + 3x2 x1 ;
(c) −x21 x23 + x32 + 3x2 x1 + 2x1 x2 + x1 x3 + x33 ;
2. (a) Let us first show that this defines a total order.
Irreflexivity: a ≺ a cannot happen since a and a have the same number of occurrences
of x2 , are not proper left divisors of each other, and do not differ in any position.
Asymmetry: if a ≺ b, then either the number of occurrences of x2 in a is less than
the number of occurrences of x2 in b, and hence it is impossible to have b ≺ a, or a
is a proper left divisor of b, and hence b is not a proper left divisor of a, and there
are no positions before the end of a where b and a differ, so it is impossible to have
b ≺ a, or, finally, a and b have the same number of occurrences of x2 , are not proper
divisors of each other, and in the first place where they differ, a has x2 but b has x1 ,
so it is impossible to have b ≺ a.
Transitivity: assume that a ≺ b and b ≺ c. If a has fewer occurrences of x2 than
b, then it follows from our assumption that a has fewer occurrences of x2 than c, so
a ≺ c. Similarly, if b has fewer occurrences of x2 than c, then it follows from our
assumption that a has fewer occurrences of x2 than c, and a ≺ c. So we may assume
that a, b, and c have the same number of occurrences of x2 . If a is a proper left
divisor of b, then since b ≺ c, we we either have b being a proper left divisor of c in
which case a is a proper left divisor of c, or b and c differ in some place, and either
a is a left divisor of c if that place is beyond the last letter of a inside b, or a and
c differ in the very same place; either way, a ≺ c. If a and b differ in some place,
then, since b ≺ c, either b is a proper left divisor of c, in which case a and c differ in
the very same place, or b and c differ in some place, in which case this place might
either occur before the difference of a and b or after; in either case it is easy to see
that a ≺ c.
Consider some set A of monomials. We take the subset A0 ⊂ A which consists
of all monomials with the smallest possible number of occurrences of x2 ; clearly, a
minimal element of A0 will be a minimal element for A too. If monomials in A0 do
not contain x2 at all, then A0 is a subset of the set of all monomials in x1 , and it has
a minimal element since the set of natural numbers is well-ordered. Suppose that
monomials in A0 all contain m > 0 occurrences of x2 . We take the subset A00 ⊂ A0
consisting of monomials in which x2 occurs in the smallest possible position; clearly,
a minimal element of A00 will be a minimal element for A0 too. All these monomials
have the same left divisor ending in x2 , and if we temporarily forget about it, the
remaining part would be a monomial with m − 1 occurrences of x2 . We may assume,
by induction on m, that among those monomials there is a minimal one; restoring
the forgotten left divisor, we obtain a minimal element of A00 .
Let us prove admissibility. Assume that m1 ≺ m2 , and let us show that mm1 ≺ mm2
and m1 m ≺ m2 m. This is true, since by direct inspection for each of the three stages
of the comparison in the definition, if the comparison happens on one of those stages,
then after multiplication by m the same comparison stage applies.
(b) Among x1 , x2 , and x3 , there will be the largest one with respect to the order,
without loss of generality it is x3 . Then x1 < x3 , x2 < x3 , so x1 x2 < x23 , so the answer
is no.
3. (a) The first way: x2 x1 x2 x1 7→ x1 x2 x2 x1 (use the first occurence), then x1 x2 x2 x1 7→
x1 x2 x1 x2 7→ x1 x1 x2 x2 , at each stage we use the only occurrence of x2 x1 . The second way: x2 x1 x2 x1 7→ x2 x1 x1 x2 (use the second occurrence), then x2 x1 x1 x2 7→
x1 x2 x1 x2 7→ x1 x1 x2 x2 , at each stage we use the only occurrence of x2 x1 . The two
results are the same.
(b) The first way: x32 = x22 x2 7→ (x1 x2 −x21 )x2 = x1 x22 −x21 x2 7→ x1 (x1 x2 −x21 )−x21 x2 =
−x31 .
The second way: x32 = x2 x22 7→ x2 (x1 x2 − x21 ) = x2 x1 x2 − x2 x21 . We see that we arrive
at two different results depending on where we apply the reduction.
4. (a) First, note that we have x1 + (x1 − x2 , x1 − x3 ) = x2 + (x1 − x2 , x1 − x3 ) =
x3 + (x1 − x2 , x1 − x3 ), so cosets of powers of x1 form a spanning set in the quotient.
Also, the homomorphism from F hx1 , x2 , x3 i to F [x] sending xi to x for all i = 1, 2, 3
is well defined and surjective, and its kernel I contains (x1 −x2 , x1 −x3 ), which proves
that the linear independence of the corresponding cosets: if they were dependent in
F [x1 , x2 , x3 ]/(x1 − x2 , x1 − x3 ), they would certainly be dependent in F [x1 , x2 , x3 ]/I.
(b) The cosets of powers of x1 are precisely the cosets of monomials whose leading
terms are not divisible by the leading monomials x2 and x3 of these polynomials, so
the criterion from class applies. On the other hand, both leading monomials for the
other order are x1 , so the reduced monomials are all monomials in x2 and x3 , and
those are linearly dependent in the quotient.