Show that the ideal A= {xf(x)+2g(x)|f(x), g(x)€Z[x]} of Z[x] is a maximal ideal in Z[x]
Answers
Answer:
Step-by-step explanation:
Maximal ideals of Z[x].
The maximal ideals of Z[x] are of the form (p, f(x)) where p is a prime number and
f(x) is a polynomial in Z[x] which is irreducible modulo p. To prove this let M be a
maximal ideal of Z[x]. Assume first that M ∩ Z 6= (0). As Z/(M ∩ Z) injects into Z[x]/M,
we conclude that Z/(M ∩ Z) is a domain, so M ∩ Z is a prime ideal of Z so M ∩ Z = (p),
where p is a prime number. Let M0
be the image of M in Z/(p)[x] and you may check
that Z[x]/M is isomorphic to Z/(p)[x]/M0
. So M0 = (f0(x)) where f0(x) ∈ Z/(p)[x] is
irreducible. Take now f(x) ∈ Z[x] reducing to f0(x) modulo p and it should be clear that
M = (p, f(x)) as described above.
We still need to show that M ∩ Z 6= (0). Let’s assume the contrary and get a contradiction.
Let M1 be the ideal of Q[x] generated by M. It is a proper ideal of Q[x] so
M1 = (f(x)), deg f(x) > 0 and we may assume without loss of generality that f(x) is a
polynomial in Z[x] of content 1. We show that M = (f(x)). Indeed, if h ∈ M, then h is
a multiple of f by an element of Q[x] and using Gauss lemma we get that this element of
Q[x] is actually in Z[x].
To finish the proof we show that Z[x]/(f(x)) is not a field if deg f(x) > 0. For this
purpose, choose a ∈ Z, f(a) 6= 0, ±1 and a prime p dividing f(a). Let φ : Z[x] → Z/(p) be
the unique homomorphism with φ(x) = a mod p. Then φ factors through Z[x]/(f(x)) since
φ(f(x)) = 0. Now, Z[x]/(f(x)) is infinite, so φ¯ : Z[x]/(f(x)) → Z/(p) is not bijective. If
we show that φ¯ is not the zero map, then ker φ¯ will be a non-trivial ideal of Z[x]/(f(x))
and it won’t be a field. If φ¯ is the zero map, then φ¯(1) = 0, i.e., there exists polynomials
u, v ∈ Z[x] with 1 = u(x)f(x) + pv(x). Putting x = a we get a contradiction since
u(a)f(a) + pv(a) is divisible by p as well as being equal to 1