Midterm 1 Suggested Review Problems
9.4.1. Determine whether the following polynomials are irreducible over the given rings.
(a) Claim: x2 + x + 1 is irreducible in F2 [x].
Proof: The polynomial x2 + x + 1 is quadratic, and has no roots in F2 . Hence by
Proposition 9.10, it is irreducible.
(b) Claim: x3 + x + 1 is reducible F3 [x].
Proof: The polynomial x3 + x + 1 has 1 as a root, and factors as (x + 2)(x2 + x + 2)
in F3 [x].
(c) Claim: x4 + 1 is reducible in F5 [x].
Proof: The polynomial x4 + 1 factors as (x2 + 2)(x2 + 3) in F5 [x].
(d) Claim: x4 + 10x2 + 1 is irreducible in Z[x].
Proof: The polynomial x4 +10x2 +1 has no integer roots, since for all values of x ∈ Z the
polynomial returns values greater than or equal to 1. Therefore it has no linear roots.
Further, the polynomial is monic, and hence has no constant factors. Therefore, it could
only factor as a product of irreducible quadratics x4 +10x2 +1 = (x2 +ax+b)(x2 +cx+d).
That is x4 + 10x2 + 1 = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd. Since bd = 1,
either b = d = 1 or b = d = −1. Further, a + c = 0. But then ac + b + d = 10 cannot
hold, since ac ≤ 0 and b + d ≤ 2. Therefore x4 + 10x2 + 1 is irreducible.
9.4.2. Prove that the following polynomials are irreducible in Z[x].
(a) Claim: x4 − 4x3 + 6 is irreducible.
Proof: The polynomial is irreducible by Eisenstein’s Criterion, with p = 2.
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(b) Claim: x6 + 30x5 − 15x3 + 6x − 120 is irreducible.
Proof: The polynomial is irreducible by Eisenstein’s Criterion with p = 3.
(c) Claim: x4 + 4x3 + 6x2 + 2x + 1 is irreducible.
Proof: Consider the reduction of this polynomial to the ring (Z/3Z)[x]. There, it is
x4 + x3 + 2x + 1. This has no roots, so it can have no linear factors, Hence, if it is
reducible, it must factor as two irreducible quadratic polynomials. There
are 6 monic
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quadratic polynomials in (Z/3Z)[x] and 3 possible factors, hence 2 = 3 reducible
quadratics. Checking by hand, we see the 3 irreducible quadratics are x2 + 1, x2 + x + 2
and x2 + 2x + 2. No pair of these multiplies to x4 + x3 + 2x + 1. Hence by Proposition
9.12, the polynomial is irreducible in Z[x].
(d) Claim:
(x+2)p −2p
x
is irreducible when p is an odd prime.
2p−i−1 p!
p
Proof: The coefficient of xi in this polynomial is given by 2p−i−1 i+1
= (i+1)!(p−i−1)!
.
Since p is prime, this is divisible by p for i < p − 1. For i = p − 1 this is 1, so the
polynomial is monic. However, the constant term is 2p−1 p which is not divisible by p2
when p is odd. Hence by Eistenstein’s Criterion the polynomial is irreducible.
9.6.10. Claim: If F is a field, and I is a monomial ideal of F [x1 , . . . , xn ] generated by
m1 , . . . , mk , then f ∈ I if and only if every monomial of f is a multiple of an mi . Further,
x2 yz + 3xy 2 ∈ I = (xyz, y 2 ) but x2 yz + 3xy 2 6∈ I 0 = (xz 2 , y 2 ).
Proof: If f ∈ F [x1 , . . . , xn ] is a sum of multiples of the monomials mi , then by definition
f ∈ I. Now suppose f ∈ I. By definition, f = f1 m1 +· · ·+fk mk for some fi ∈ F [x1 , . . . , xn ].
Plj
P Plj
mij mi is a
Write each fi = j=1
mij as the sum of its monomials. Then f = ki=1 j=1
sum of multiples of mi .
Following this result, since x2 yz + 2xy 2 = x(xyz) + 3x(y 2 ) it is in I. But since the
monomial x2 yz is not a multiple of xz 2 or y 2 , it is not in I 0 .
9.6.11. Claim: If F is a field, and I is a monomial ideal of F [x1 , . . . , xn ] with Gröbner basis
{g1 , . . . , gm }, then h ∈ LT (I) if and only if h is a sum of monomial terms each divisible by
some LT (gi ).
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Proof: The ideal LT (I) is generated by LT (g1 ), . . . , LT (gm ). By the previous exercise,
h ∈ LT (I) if and only if all of its monomial terms are multiples of LT (gi ).
9.6.12. Claim: If I is a monomial ideal with generators g1 , . . . , gm then {g1 , . . . , gm } is a
Gröbner basis for I.
Proof: Suppose I is a monomial ideal generated by g1 , . . . , gm . Suppose f ∈ I. Then by
9.6.10, the leading term of f is divisible by some gi . Therefore the ideal LT (I) generated
by the leading terms of polynomials f ∈ I is generated by gi as well. Therefore g1 , . . . , gm
area a Gröbner basis for I.
9.6.40. Claim: If M and N are monomial ideals of F [x1 , . . . , xn ], then M ∩N is a monomial
ideal as well.
Proof: Suppose M is generated by monomials m1 , . . . , mr and N is generated by monomials
n1 , . . . , ns Let I be the monomial ideal generated by monomials ei,j where ei,j is the least
common multiple of mi and nj . We claim that I = M ∩N . First, we show that I ⊆ M ∩N .
If f ∈ I, then by 9.6.10, every monomial in f is a multiple of some ei,j . But each ei,j is a
multiple of mi , so f ∈ M . Also, ei,j is a multiple of nj , so f ∈ N . Therefore f ∈ M ∩ N .
Now suppose f ∈ M ∩ N . Then each monomial of f is divisible by some mi and some
nj . Hence it is divisible by the least common multiple ei,j . Therefore f ∈ I. This proves
I = M ∩ N is a monomial ideal.
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