Download immerse 2010 - UNL Math Department

IMMERSE 2010
Algebra Course
Problem Set 11 Solutions
1. Prove that Z[x] is not a PID.
Proof by (Ron, Caitlyn, Nathan S, Derrek). We need to show that there exists an ideal, I of Z[x] such
that I 6= hf (x)i, where f (x) ∈ Z[x]. Let I be the ideal of all polynomials with even constant term. It
is easy to see that this is an ideal, as multiplication with any other polynomial will have a constant
term equal to the product of the constant terms, which must be even, and subtraction between two
polynomials in I will still have an even constant term. If I is principle, then we must have I = hf (x)i,
for some f (x) ∈ I. So, suppose that I is generated by f (x). We know that 2x + 2 ∈ I, and so
2x + 2 = f (x)g(x). The only way to express this polynomial as a product of two polynomials, with
one of them having even constant term, is if f (x) = 2 and g(x) = x + 1. Then we must have I = h2i.
But x + 2 ∈ I, and x + 2 6= 2h(x) for any h(x) ∈ Z[x]. Thus, our assumption that I is principle is false.
That is, there exist ideals of Z[x] that are not principle, and so Z[x] is not a principle ideal domain.
2. Show that if f (x) ∈ Z[x] is a nonconstant polynomial that is irreducible over Z, then f (x) is primitive.
Proof by (Lilith, Lisa, Nathan M, Rob). We will prove the contrapositive version of this statement. “If
f (x) ∈ Z[x] is a nonconstant polynomial f (x) that is not primitive, then f (x) is reducible over Z.”
Let f (x) ∈ Z[x], f = an xn + an−1 xn−1 · · · a0 be a nonconstant polynomial that is not primitive. Then
n−1
there exists d > 1 such that d divides ai for all 0 ≤ i ≤ n. Let g(x) = adn xn + an−1
· · · ad0 . Then
d x
f (x) = dg(x) however neither d nor g(x) can be a unit in Z[x] since the only units are 1 and -1, so
f (x) is reducible.
3. Determine whether each of the following polynomials is irreducible over Q[x]:
(a) x5 + 9x4 + 12x2 + 6
(b) x3 − 11x2 + 45
(c) x4 − x2 + 3
Proof by (Chris, Tim, Kirsten, Annette). (a) We will show that the polynomial x5 + 9x4 + 12x2 + 6
is irreducible over Q[x] by using Eisenstein’s Criterion. Let p = 3, a prime number. We see that
3 - an = 1, but since 3|9, 3|12, 3|0 and 3|6, 3 divides all the other coefficients of x5 +9x4 +12x2 +6.
We also see that 32 = 9 - 6 = a0 . Thus we have met Eisenstein’s Crierion, and so x5 +9x4 +12x2 +6
is irreducible over Q[x].
(b) We will show that x3 − 11x2 + 45 is irreducible over Q[x] using the mod p test. We will use
p = 2, a prime, and consider x3 − 11x2 + 45 over Z2 . When we reduce x3 − 11x2 + 45 over Z2 we
get the polynomial x3 + x2 + 1. We also see the the degrees of x3 − 11x2 + 45 and x3 + x2 + 1 are
equal. Since x3 + x2 + 1 has degree 3, showing that it does not have any zeros in Z2 is sufficient for
showing it is irreducible. If we evaluate the polynomial at 0 we get 03 +02 +1 = 1 6= 0. Evaluating
at 1 gives us 13 + 12 + 1 = 1 6= 0. Since {0, 1} = Z2 , we see that x3 + x2 + 1 is irreducible over
Z2 . Therefore by the mod p test we see that x3 − 11x2 + 45 is irreducible over Q[x].
(c) Consider the polynomial p(x) = x4 − x2 + 3. Let q(x) = x2 + x + 3 be another polynomial over
Q[x]. Notice that q(x2 ) = x4 − x2 + 3 = p(x). Thus if we can show that q(x) is irreducible over Q
we will have that q(x2 ) = p(x) is irreducible over Q[x]. Consider q(x) over Z2 . When we reduce
q(x) over Z2 we get x2 + x + 1. Notice that the degree of q(x) and the degree of q(x) reduced
over Z2 are the same. Since q(x) over Z2 has degree 2, showing it has no zeros in Z2 is sufficient
to show it is irreducible over Z2 . Evaluating q(0) = 1 mod 2 and q(1) = 1 mod 2 we see that
q(x) has no zeros over Z2 . Thus q(x) is irreducible over Z2 . Therefore by the mod p test, q(x)
is irreducible over Q[x]. Thus q(x2 ) = x4 − x2 + 3 is irreducible over Q.
4. Let f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 ∈ Z[x], where an 6= 0. Prove that if r and s are relatively
prime integers and f (r/s) = 0, then r|a0 and s|an .
Proof by (Zach, Anne, Linda, Jayna). We know that
f
r
s
= an
r n
s
+ an−1
r n−1
s
+ . . . + a1
r
s
+ a0 = 0.
Multiplication through by sn gives
an rn + an−1 rn−1 s + . . . + a1 rsn−1 + a0 sn = 0.
We can rearrange terms
an rn + an−1 rn−1 s + . . . + a1 rsn−1 = −a0 sn
in order to note an interesting fact: Since r divides every term on the left hand side of the equation
(LHS), r must also divide the right hand side (RHS). Thus r|a0 sn . Since r and s are relatively prime,
r 6 |sn . Therefore r|a0 . Recall the equation
an rn + an−1 rn−1 s + . . . + a1 rsn−1 + a0 sn = 0.
Subtract an rn from both sides to get
an−1 rn−1 s + . . . + a1 rsn−1 + a0 sn = −an rn .
Note that s divides everything on the LHS of the equation. Therefore it must divide the RHS. So
s|an rn . Since s and r are relatively prime, s|an .
5. Let F be a field, and p(x) ∈ F [x] be irreducible over F . If E is some field containing F , and there
exists a ∈ E such that p(a) = 0, show that the map
φ : F [x] → E
f (x) 7→ f (a)
is a ring homomorphism with kernel (p(x)). [Note: (p(x)) = {f (x)p(x)|f (x) ∈ F [x]} is the ideal
generated by p(x).]
Proof by (Robert, Julia, Sarah, Matthew). In order to show φ is a ring homomorphism, we want to show
that φ is well-defined, preserves addition, and preserves multiplication. Let f (x), g(x) ∈ F and h(x) =
g(x). Observe h(x) = hn xn + hn−1 xn−1 + . . . + h1 x1 + h0 and g(x) = gn xn + gn−1 xn−1 + . . . + g1 x1 + g0 .
Thus hi = gi for 0 ≤ i ≤ n and φ(h(x)) = h(a) = g(a) = φ(g(x)). Therefore φ is well-defined.
Now let f (x), g(x) ∈ F [x]. Then we have
φ((f · g)(x)) = φ(f (x)g(x)) = f (a)g(a) = φ(f (x))φ(g(x))
φ((f + g)(x) = φ(f (x) + g(x)) = f (a) + g(a) = φ(f (x)) + φ(g(x))
Hence φ preserves the operations and is a homomorphism.
Now we want to show that ker φ = (p(x)). Let g(x) ∈ (p(x)) then g(x) = f (x)p(x) for some f (x) ∈ F [x].
Hence
φ(g(x)) = φ(f (x)p(x)) = f (a)p(a) = f (a) · 0 = 0.
Thus g(x) ∈ ker φ and we have (p(x)) ⊆ ker φ ⊆ F [x]. Since (p(x)) is a maximal ideal and ker φ
is an ideal, we know either ker φ = F [x] or (p(x)) = ker φ. Suppose ker φ = F [x]. Then for all
f (x) ∈ F [x], φ(f (x)) = f (a) = 0. Let g(x) = c where c is a nonzero constant in F [x]. Then φ(g(x)) =
g(a) = c 6= 0 which contradicts ker φ = F [x]. Therefore (p(x)) = ker φ as desired.