Download Math 403A assignment 7. Due Friday, March 8, 2013. Chapter 12

Math 403A assignment 7. Due Friday, March 8, 2013.
Chapter 12.
Gauss’s lemma.
A nonconstant polynomial f (x) ∈ Q[x] is primitive if it has integer coefficients, the g.c.d. of
its coefficients is 1, and its leading coefficient is positive.
Gauss. The product of primitive polynomials is primitive.
Gauss’s lemma says that the primitive polynomials form a multiplicative subset of Q[x].
We can write each polynomial k(x) ∈ Z[x] as af (x), where f (x) is primitive and a is the
g.c.d. of the coefficients of k(x).
Let J be an ideal in Z[x] such that J ∩ Z = (0) and such that if k(x) lies in J, then so does
the associated primitive polynomial f (x).
We will show in class that such an ideal is principal, generated by a primitive polynomial,
that has the smallest degree among the elements of the ideal.
You can use that result in problem 1.
1.(3.6) Let α be a complex number. Show that the kernel of the homomorphism φ : Z[x] → C,
φ(f (x)) = f (α), is a principal ideal and describe a generator for the ideal.
Solution. Let J be the kernel.
(i). J = (0) if α is not algebraic over Q.
(ii). Suppose that α is algebraic over Q.
Since φ injects Z into C, J ∩ Z = (0). Also, if g(x) is in the kernel, i.e., g(α) = 0 and if we
write g(x) = nf (x), where n = g.c.d. of the coefficients of g(x) and f (x) is primitive, then
f (α) = 0, i.e., f (x) is in the kernel. By the theory above, J is a principal ideal generated by
a smallest degree polynomial in J, and among those polynomials, take a primitive one.
2.(4.3) Decide if the polynomial x4 + 6x3 + 9x + 3 generates a maximal ideal in Q[x].
Solution. Irreducible polynomials generate the maximal ideals since
only if f (x) is irreducible.
Q[x]
(f (x))
is a field if and
By Eisenstein for the prime 3, x4 + 6x3 + 9x + 3 is irreducible in Q[x]. Hence, the ideal
(x4 + 6x3 + 9x + 3) is maximal.
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3.(4.8) How might a polynomial f (x) = x4 + bx2 + c with coefficients in a field F factor in
F [x]?
Explain with reference to the polynomials x4 + 4x2 + 4 and x4 + 3x2 + 4.
Solution. With y = x2 , x4 + bx2 + c = y 2 + by + c. If that quadratic polynomial in y has
its roots α, β in F , then
x4 + bx2 + c = y 2 + by + c = (y − α)(y − β) = (x2 − α)(x2 − β)
is a factoring of x4 + bx2 + c in F [x].
For F = Q, x4 + 4x2 + 4 = (x2 + 2)2 ∈ Q[x], but for x4 + 3x2 + 4, the roots of y 2 + 3y + 4 are
nonreal numbers, so that x4 + 3x2 + 4 does not factor into the form (x2 − α)(x2 − β) in R[x].
4.(4.14) By analyzing the locus x2 +y 2 = 1, show that the polynomial x2 +y 2 −1 is irreducible
in C[x, y].
Solution. If x2 + y 2 − 1 could be factored into a product of linear polynomials, i.e., degree 1
polynomials, then the solutions to x2 + y 2 − 1 = 0 would be a line or a pair of lines, instead
of the unit circle.
Chapter 13.
5.(1.1) Is 21 (1 +
√
5) an algebraic integer?
Solution. Yes, the elements satisfies the monic integer polnomial
√
√
1
1
(x − (1 + 5))(x − (1 − 5)) = x2 − x − 1.
2
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6.(2.1) Show that 2, 3, 1 ±
this ring are ±1.
√
√
−5 are irreducible elements of Z[ −5], and that the units of
√
√
Solution. The norm of m+n −5 is N(m+n −5) = m2 +5n2 . The norm is multiplicative,
and so, if u is a unit, then
1 = N(1) = N(uu−1 ) = N(u)N(u−1 ),
and so, N(u) = 1; hence, units are the elements ±1 of norm 1.
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The smallest norm of a nonunit
is N(2) = 4 and the smallest product of two nonu nits is 16;
√
hence, any
nonunit
a
+
b
−5
of
norm less than 16 is irreducible. In particular, the elements
√
2, 3, 1 ± −5 of norms 4, 9, and 6 are irreducible, as follows.
√
√
√
6 has a nonunique factoring into irreducibles in Z[ −5] since 6 = 2·3 = (1+ −5)(1− −5)
are factorings
that do not differ by unit variation in the factors since the only units are ±1.
√
Thus, Z[ −5] is not a unique factorization domain.
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