Solutions to Assignment 11
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Problem 1
Suppose that R and S are unital rings (not necessarily integral domains), then φ : R → S is
a homomorphism, and that φ is onto. Prove that φ(1R ) = 1S .
Solution. Since φ is onto, for all s we can find an element rs ∈ R so that φ(rs ) = s. Therefore,
for all s ∈ S, we have that
s · φ(1R ) = φ(rs )φ(1R ) = φ(rs · 1R ) = φ(rs ) = s
(1)
This is true for all s ∈ S, therefore φ(1R ) = 1S is the unique multiplicative identity in S.
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Problem 2
Prove that U = t · R[t] is a maximal ideal in R[t].
Solution. Suppose that I ⊂ R[t] is an ideal such that t · R[t] ⊂ I and such that P (t) ∈ I
where P (t) ∈
/ t · R[t]. We can write P (t) as
P (t) = a + tQ(t)
(2)
where a ∈ R with a 6= 0 and with Q(t) ∈ R[t].
Since P (t) = a + tQ(t) ∈ I and since tQ(t) ∈ R[t] ⊂ I, we have that a ∈ I. Since
0 6= a ∈ R, a is invertible and we have 1 = a · a−1 ∈ I, so I = R[t].
Therefore there is no ideal containing t · R[t] other than R[t] and itself, so it is maximal.
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Problem 3
Prove that t2 + 1 is a prime in R[t]. Prove that no polynomial of degree 3 is a prime.
Solution. Note that if P, Q ∈ R[t], then deg(P Q) = deg(P )deg(Q), where deg(P ) = deg(P (t))
is the largest power of t appearing in P (t). Therefore, in particular, an element of R[t] is a
unit only if it is a real number, hence the non-zero real numbers are the set of units in R[t].
If t2 + 1 = P (t)Q(t) where neither P nor Q are units, then, since deg(t2 + 1) = 2, it must
be that deg(P )=deg(Q) = 1. Thus t2 + 1 = (at + b)(ct + d), where a and c are nonzero. But
(at + b)(ct + d) has roots at t = −b/a and t = −d/c, whereas t2 + 1 has no real roots. This
is a contradiction, so t2 + 1 is prime.
On the other hand, we can prove that any polynomial of degree 3 has a real root using
the intermediate value theorem. Let
P (t) = a3 t3 + a2 t2 + a1 t + a0
1
(3)
with a3 6= 0. If a3 > 0, then P (t) > 0 for sufficiently large t and P (t) < 0 for sufficiently
small t. If a3 < 0 then P (t), 0 for sufficiently small t and P (t) > 0 for sufficiently small t.
In either case P (t) is a continuous function and therefore by the intermediate value theorem
there exists some t0 so that P (t0 ) = 0.
You may recall from high-school algebra or calculus that a polynomial with a root at t0
is divisible by t − t0 , so P (t) = (t − t0 )Q(t), where deg(Q) = 2. This implies that P is not a
prime. (This fact can be proved using the Remainder Theorem for polynomials – we know
that P (t) = (t − t0 )Q(t) + R(t), where deg(R) < deg(t − t0 ), but that implies deg(R) = 0. If
deg(R) = 0, then R(t) = c for some c ∈ R, but then P (t0 ) = (t0 − t0 )Q(t0 ) + c, so c = 0.)
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Problem 4
p. 149 #2. In a Euclidean ring prove that any two greatest common divisors of a and b are
associates.
Solution. Let d1 and d2 be two greatest common divisors of a and b. Then, by definition,
we have that d1 |d2 and d2 |d1 . In other words, there exists r, s ∈ R so that d2 = rd1 and
d1 = sd2 . Thus, d2 = rsd2 and therefore rs = 1 since we are in an integral domain. Thus r
and s are both units and d1 and d2 are associates.
p. 149 #3. Prove that a necessary and sufficient condition that the element a in the
Euclidean ring be a unit is that d(a) = d(1).
Solution. Suppose that a is a unit. Then we have that a|b for all b ∈ R. That means that
d(a) ≤ d(b) for all b ∈ R. Therefore, in particular d(a) ≤ d(1), and by the same reasoning,
d(1) ≤ d(a), therefore d(a) = d(1).
Now suppose that d(a) = d(1). Then we have that 1 = ta + r for some t ∈ R and some
r ∈ R with either r = 0 or d(r) < d(a). However, since d(r) < d(a) = d(1) is impossible
as mentioned above. It must be that r = 0 and thus 1 = ta for some t ∈ R. Thus a is a
unit.
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