Solutions to MA3417 Homework assignment 1
1. For each of the following polynomials, rewrite it in the decreasing order of terms using the LEX order, the DEGLEX
order, and the DEGREVLEX order:
(a) 2x + 3y + z + x2 − z 2 + x3 ;
(b) 2x2 y 8 − 3x5 yz 4 + xyz 3 − xy 4 .
Solution.
(a) LEX :x3 + x2 + 2x + 3y − z 2 + z,
DEGLEX: x3 + x2 − z 2 + 2x + 3y + z,
DEGREVLEX: x3 + x2 − z 2 + 2x + 3y + z.
(b) LEX: −3x5 yz 4 + 2x2 y 8 − xy 4 + xyz 3 ,
DEGLEX: −3x5 yz 4 + 2x2 y 8 − xy 4 + xyz 3 ,
DEGREVLEX: 2x2 y 8 − 3x5 yz 4 − xy 4 + xyz 3 ,
2. Do the the previous question for the order of variables z > y > x.
Solution.
(a) LEX :−z 2 + z + 3y + x3 + x2 + 2x,
DEGLEX: x3 − z 2 + x2 + z + 3y + 2x,
DEGREVLEX: x3 − z 2 + x2 + z + 3y + 2x.
(b) LEX: −3x5 yz 4 + xyz 3 + 2x2 y 8 − xy 4 ,
DEGLEX: −3x5 yz 4 + 2x2 y 8 + xyz 3 − xy 4 ,
DEGREVLEX: 2x2 y 8 − 3x5 yz 4 + xyz 3 − xy 4 ,
3. Each of the following polynomials is written in the decreasing order of terms using (exactly) one of the LEX order,
the DEGLEX order, and the DEGREVLEX order. Determine which order was used in each case:
(a) 7x2 y 4 z − 2xy 6 + x2 y 2
(b) xy 3 z + xy 2 z 2 + x2 z 3
(c) x4 y 5 z + 2x3 y 2 z − 4xy 2 z 4
Solution. (a) DEGLEX; (b) DEGREVLEX; (c) LEX.
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4. Let A = (aij ) be a matrix in its reduced row echelon form. Let us consider linear polynomials fi = j aij xj . Find
an order for which fi form a Gröbner basis. (Hint: choose an ordering of variables wisely, and use the LEX order for
that chosen ordering of variables).
Solution. Let us order the variables so that all the principal variables (corresponding to pivots in the rows) are
larger than all the free variables. Let us show that the polynomials fi form a Gröbner basis. We are required to
show that every polynomial in the ideal generated by these polynomials has its leading term divisible by one of the
principal variables. If it is not the case for some f in the ideal, then its leading term is made out of free variables,
and hence because of our chosen ordering all the terms are made out of free variables only. But for any choice of
free variables, we can find a common zero of the polynomials fi with the prescribed values of the free coordinates,
and f , as an element of the ideal, vanishes on all the common zeros of these polynomials. This is a contradiction,
since a nonzero polynomial cannot vanish everywhere.
5. Show that for the ideal I = (y − x2 , z − x3 ) in C[x, y, z], the polynomials y − x2 , z − x3 that generate it do not form
a Gröbner basis for the LEX order with x > y > z.
Solution. Note that the leading terms of these polynomials are x2 and x3 . The polynomial xy−z = x(y−x2 )−(z−x3 )
belongs to the ideal generated by the given polynomials, and its leading term is not divisible by either of the leading
terms. Therefore, these polynomials do not form a Gröbner basis.
6. Show that for the ideal I from the previous question, the polynomials y − x2 , z − x3 that generate it form a Gröbner
basis for the LEX order with z > y > x.
Solution. The leading terms of these polynomials are y and z. Let us show that the leading term of every polynomial
in the ideal generated by these polynomials is divisible by y or z. If it is not the case, there is a polynomial f in the
ideal whose leading term is xm for some m. Because of our choice of ordering, all other terms in that polynomial are
also powers of x, so it is a polynomial in x only. Consider the set S = {(t, t2 , t3 ) : t ∈ C}. Both given polynomials
vanish on that set, so every polynomial in the ideal they generate vanish on S. But a polynomial in x vanishes on
S only if it is identically zero. This contradiction shows that the given polynomials form a Gröbner basis.