Download Solutions for Problem Set 4 A: Consider the polynomial ring R = Z[x

Solutions for Problem Set 4
A: Consider the polynomial ring R = Z[x]. Let I = (x), the principal ideal of R generated
by x. Show that I is a prime ideal of R, but not a maximal ideal of R.
Solution. Suppose that f (x) and g(x) are in Z[x]. Suppose that their constant terms are
a0 and b0 , respectively. The ring operations in Z[x] were defined in class one day (in a much
more general setting). We need just the following facts. The constant term of f (x) + g(x) is
a0 + b0 . The constant term of f (x)g(x) is a0 b0 . Define a map ϕ : Z[x] → Z by
ϕ f (x) = the constant term of f (x)
for all elements f (x) in the ring Z[x]. The above facts mean that ϕ is a ring homomorphism.
In fact, the map ϕ is just the “evaluation homomorphism” ϕ0 which was defined in class.
The map ϕP: Z[x] → Z is a surjective ring homomorphism. Suppose that f (x) is in Z[x].
Write f (x) = ni=0 ai xi , where a0 , ..., an are in Z. Then
f (x) ∈ Ker(ϕ) ⇐⇒ a0 = 0 ⇐⇒ f (x) = xg(x) f or some g(x) ∈ Z[x] ⇐⇒ f (x) ∈ I .
Thus, we have Ker(ϕ) = I. By the first isomorphism theorem, it follows that
Z[x]/I ∼
= Z .
Since Z is an integral domain, it follows that I is a prime ideal in Z[x]. Since Z is not a
field, it follows that I is not a maximal ideal in Z[x]. We are using theorems 7 and 8 on the
handout listing important theorems about ring homomorphisms and ideals.
B: Consider the polynomial ring R = Q[x]. Let I = (x), the principal ideal of R generated
by x. Show that I is a maximal ideal of R.
Solution The solution is similar to that for problem A. We define ϕ in the same way, except
that now ϕ is a surjective ring homomorphism from Q[x] to Q. The kernel of ϕ is the ideal
I = (x). Hence we have
Q[x]/I ∼
=Q .
Since Q is a field, it follows that I is a maximal ideal of the ring Q[x].
C: This problem is a continuation of problem B from problem set 3. The results from
parts (a), (b), and (c) of that problem will be useful in this problem. Suppose that R is a
commutative ring with identity, that e is an idempotent in R, and that f = 1R − e.
(a) Let S = Re and T = Rf . Thus, S and T are ideals in the ring R. Show that R = S +T .
Solution. Suppose that r ∈ R. Since 1R = e + f , we have
r = r · 1R = r(e + f ) = re + rf
and so, for any r ∈ R, we can write r = s + t, where s = re and t = rf . Note that we have
s ∈ Re = S and t ∈ Rf = T . Thus, r ∈ S + T . This is true for all r ∈ R and so we indeed
have R = S + T .
(b) Suppose that r ∈ R. According to part (a), there exist elements s ∈ S and t ∈ T such
that r = s + t. Prove that s and t are uniquely determined by r.
Solution. In part (a), we have written r = s + t, where s = re and t = rf . Suppose that
we also have r = u + v, where u ∈ S and v ∈ T . We will prove that u = s and v = t. First
note that
re = (u + v)e = ue + ve
and
rf = (u + v)f = uf + vf
.
Furthermore, since u ∈ S and e is the identity element in S, we have u = ue. Since v ∈ T
and f is the identity element in T , we have v = vf . Thus, we obtain the equations
re = ue+ve = ue+(vf )e = ue+v(f e)
and
rf = uf +vf = (ue)f +vf = u(ef )+vf.
Since f e = 0R and ef = 0R , we then obtain the following equations:
s = re = ue + v · 0R = ue = u
and
t = rf = u · 0R + vf = vf = v .
That is, we have u = s and v = t. This proves that s and t are indeed uniquely determined
by r.
(c) Suppose that r ∈ R. Using part (a), one can write r = s + t, where s ∈ S and t ∈ T .
Define a map ϕ : R → S × T by ϕ(r) = (s, t). By part (b), this map ϕ is well-defined.
Prove that ϕ is an isomorphism of the ring R to the ring S × T .
Solution. The map ϕ is surjective because if s is any element of S and t is any element of
T , then r = s + t is an element of R and has the property that ϕ(r) = (s, t).
To see that the map ϕ is a ring homomorphism, suppose that r1 and r2 are in R. We
can write r1 = s1 + t1 and r2 = s2 + t2 , where s1 , s2 ∈ S and t1 , t2 ∈ T . Then we have
r1 + r2 = (s1 + t1 ) + (s2 + t2 ) = (s1 + s2 ) + (t1 + t2 )
and we have s1 + s2 ∈ S, t1 + t2 ∈ T . Therefore, by definition,
ϕ(r1 + r2 ) = (s1 + s2 , t1 + t2 ) = (s1 , t1 ) + (s2 , t2 ) = ϕ(r1 ) + ϕ(r2 ) .
Now we consider multiplication. Let r = r1 r2 . As pointed out in the solution to part
(b), we have r = s + t, where s = re and t = rf . Similarly, we also have the equations
s1 = r1 e, t1 = r1 f, s2 = r2 e, and t2 = r2 f . We obtain
s = re = (r1 r2 )(ee) = (r1 e)(r2 e) = s1 s2
and
t = rf = (r1 r2 )(f f ) = (r1 f )(r2 f ) = t1 t2 .
We have used the associative law and also the assumption that R is a commutative ring in
the above calculation. It follows that
ϕ(r1 r2 ) = ϕ(r) = (s, t) = (s1 s2 , t1 t2 ) = (s1 , t2 )(s2 , t2 ) = ϕ(r1 )ϕ(r2 ) .
We have proved that ϕ(r1 + r2 ) = ϕ(r1 ) + ϕ(r2 ) and that ϕ(r1 r2 ) = ϕ(r1 )ϕ(r2 ) for all
r1 , r2 ∈ R and hence that ϕ is a ring homomorphism from R to S × T .
Finally, we will show that ϕ is injective. It suffices to show that Ker(ϕ) = {0R }. Suppose
that r ∈ Ker(ϕ). Thus, we have ϕ(r) = (0S , 0T ), the additive identity in the ring S × T .
By the definition of ϕ, it follows that r = 0S + 0T . Since S and T are subrings of R, we
know that 0S = 0R and 0T = 0R . Therefore, r = 0R + 0R and so r = 0R . Hence, we have
Ker(ϕ) = {0R }.
We have proved that ϕ is a bijection and that ϕ is a ring homomorphism. Hence ϕ is a
ring isomorphism from R to S × T .
D: This question concerns the following subring of M2 (R).
a b a, b ∈ R
.
R =
b a You may assume that R is a subring of M2 (R). This is a result from problem 4 on the
Suggested Problems handout. Using problem C above, show that the rings R and R × R are
isomorphic. Give an explicit map ϕ : R → R × R which is a ring isomorphism.
Solution. We will use the following two idempotents in the ring R:
1 1
1
− 21
2
2
2
e = 1 1
and
f =
.
− 12 21
2
2
1 0
and that we do actually have f = 1R − e.
Note that 1R is the identity matrix
0 1
a b
in R. Here, a and b are in R. We have
Consider an element r =
b a
a b 12
re =
b a 21
1
2
1
2
=
a+b
2
a+b
2
a+b
2
a+b
2
= (a + b)
1
2
1
2
1
2
1
2
= (a + b)e
and
1
a−b
1
1
a−b
1
−
−
−
a b
2
2
2
2
= (a − b) 2 1 1 2 = (a − b)f .
=
rf =
a−b
−2 2
b a − 12 12
− a−b
2
2
Note that when we write (a + b)e, we mean the product of the real number a + b and the
matrix e, as defined in matrix theory. The meaning of (a − b)f is similar.
The above calculation tells us that S = Re = { ce | c ∈ R } and T = Rf = { cf | c ∈ R }.
We will define a ring isomorphism σ from S to R and a ring isomorphism τ from T to R.
Here is the definition of σ. Suppose that s ∈ S. Thus, we have s = ce, where c ∈ R. We
define σ(s) = c. It is clear that σ is a bijection from S to R. Suppose that s1 and s2 are
elements of S. We have s1 = c1 e and s2 = c2 e, where c1 and c2 are in R. Then
s1 + s2 = c1 e + c2 e = (c1 + c2 )e
and
s1 s2 = (c1 e)(c2 e) = c1 c2 ee = c1 c2 e .
We have used some facts from matrix algebra and the fact that ee = e. It follows that
σ(s1 + s2 ) = c1 + c2 = σ(s1 ) + σ(s2 )
and
σ(s1 s2 ) = c1 c2 = σ(s1 )σ(s2 )
We have proved that σ : S → R is a ring isomorphism.
The definition of τ is similar. Suppose that t ∈ T . Thus, we have t = cf , where c ∈ R.
We define τ (s) = c. Then τ : T → R is a ring isomorphism. The proof is almost identical to
the one given above for τ .
is
We can use σ and τ to define a ring isomorphism χ from S × T to R × R. The definition
χ (s, t) = σ(s), τ (t)
for all elements (s, t) in the ring S × T .
Finally, we have an isomorphism ϕ : R → S × T from problem C which is defined by
ϕ(r) = (re, rf ) for all r ∈ R. Let ψ = χ ◦ ϕ. Then ψ is an isomorphism from R to R × R
which is defined by
a b
a b
a b
= χ ϕ
= (χ ◦ ϕ)
ψ
b a
b a
b a
= χ
= (a + b, a − b)
(a + b)e, (a − b)f
for all a, b ∈ R.
E: Let R = Q[x] (x2 − 4x + 4). Prove that R has infinitely many nilpotent elements.
Solution. Let I = (x2 − 4x + 4). Since x2 − 4x + 4 has degree 2, we know that the elements
in the quotient ring Q[x]/I can be uniquely represented in the form
r(x) + I
where r(x) = a + bx and a, b ∈ Q. The uniqueness will be important in this solution. Now
notice that x2 − 4x + 4 = (x − 2)(x − 2) = (x − 2)2 . Suppose that q ∈ Q. Consider
rq (x) = a + bx, where a = −2q and b = q. Then
rq (x) = − 2q + qx = q(x − 2)
and
rq (x)2 = q 2 (x − 2)2 ∈ I .
It follows that the element rq (x) + I in the quotient ring Q[x]/I has the following property:
2
rq (x) + I
= rq (x)2 + I = I ,
which is the additive identity element in the ring Q[x]/I. Thus, rq (x) + I is a nilpotent
element in the ring Q[x]/I. This is true for each q ∈ Q.
If q1 , q2 ∈ Q and q1 6= q2 , then rq1 (x) 6= rq2 (x). Hence, the cosets rq1 (x) + I and rq2 (x) + I
will be different elements in the ring Q[x]/I. Thus, we have obtained an infinite number of
nilpotent elements in Q[x]/I.
F: Let R = Q[x] (x2 + 1). Show that R is a field.
Solution As explained in class, it suffices to show that (x2 + 1) is a maximal ideal in the
ring Q[x]. For that purpose, it suffices to prove that x2 + 1 is an irreducible element in the
ring Q[x]. Suppose to the contrary that we have a factorization
x2 + 1 = g(x)h(x)
where g(x) and h(x) are in Q[x] and that the degree of g(x) and h(x) are both < 2. This
means that g(x) and h(x) have degree 1. Thus g(x) = a + bx and h(x) = c + dx, where
a, b, c, d ∈ Q, b 6= 0 and d 6= 0. Let J = (a + bx), the principal ideal in Q[x] generated by
a + bx. Then x2 + 1 is in J.
Consider the ring homomorphism ϕθ : Q[x] → Q which
was defined in class. We will
a
take θ = − b , which is an element of Q. Thus, ϕ f (x) = f (θ) for all f (x) ∈ Q[x]. Notice
that
ϕθ (a + bx) = a + bθ = 0
and hence a + bx ∈ Ker(ϕθ ). Therefore, J ⊆ Ker(ϕθ ). (In fact, we have Ker(ϕθ ) = J, but
this fact won’t be needed.)
However, notice that ϕθ (x2 + 1) = θ2 + 1 ≥ 1 since θ is a real number and therefore
θ2 ≥ 0. Hence, ϕθ (x2 + 1) 6= 0. Therefore, x2 + 1 6∈ Ker(ϕθ ). Since J ⊆ Ker(ϕθ ), it follows
that x2 + 1 cannot be in J. This is a contradiction.
follows that x2 + 1 is indeed an
It
irreducible element in the ring Q[x]. Therefore, Q[x] (x2 + 1) is indeed a field.
G: Let R = R[x] (x2 − 2). Show that R is not a field.
Solution We have the following factorization in the ring R[x]:
√
√
x2 − 2 = (x − 2)(x + 2)
which shows that x2 − 2 is a reducible element in the ring R[x]. Therefore, the principal
ideal (x2 − 2) is not a maximal ideal in that ring. Therefore, the quotient ring R[x] (x2 − 2)
is not a field.
H: Let R = Q[x]/I, where I = (x3 + 5x2 + 6x + 7). Let β = x + I. Thus, β is an element
of R. Show that
β 3 + 5β 2 + 6β + 7 · 1R = 0R .
Solution. We will repeatedly use the definition of addition and multiplication in the ring
R. If a(x) and b(x) are in Q[x], then
(a(x)+I) + (b(x)+I) = a(x)+b(x) +I
and
(a(x)+I)(b(x)+I) = a(x)b(x) +I .
Also, recall that 1R = 1 + I. Using this definition, we obtain
β 3 + 5β 2 + 6β + 7 · 1R = (x + I)3 + 5(x + I)2 + 6(x + I) + 7 · (1 + I)
= (x3 + 5x2 + 6x + 7) + I = I .
The last equality holds because the polynomial x3 + 5x2 + 6x + 7 is in the ideal I. Since
I = 0R , we have proved the stated equality.