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IMMERSE 2010
Algebra Course
Problem Set 10 Solutions
1. Let D be an integral domain and let F be the field of quotients of D. Show that if E is any field that
contains D, then E contains a subfield that is ring isomorphic to F .
Proof by (Robert, Julia, Sarah, Matthew). Let a, b ∈ D with b 6= 0. Since E is a field, b−1 ∈ E. Thus
define the function φ : F → E as φ(a, b) = ab−1 . We want to show that φ is well defined, so suppose
(a, b), (a0 b0 ) ∈ F such that (a, b) = (a0 b0 ). (Note that b, b0 6= 0.) Then by the definition of equivalence
classes in the field of quotients, we have ab0 = ba0 ⇒ ab−1 = a0 b0−1 ⇒ φ((a, b)) = φ((a0 , b0 )). Hence φ
is well-defined.
Now we want to show that φ preserves addition and multiplication. Let (a, b), (c, d) ∈ F (b, d 6= 0).
Then
φ((a, b)+(c, d)) = φ((ad+bc, bd)) = (ad+bc)(d−1 b−1 ) = add−1 b−1 +bcd−1 b−1 = ab−1 +cd−1 = φ((a, b))+φ((c, d))
and
φ((a, b)(c, d)) = φ((ac, bd)) = acd−1 b−1 = ab−1 cd−1 = φ((a, b))φ((c, d))
Thus φ preserves the operations and is a homomorphism.
Note that φ : F → φ(F ) is onto by definition. Next we want to show that φ is one-to-one. Suppose that
φ((a, b)) = φ((c, d)). Then ab−1 = cd−1 which implies ad = bc. Hence (a, b) = (c, d) by the definition
of equivalence classes in the field of quotients and therefore φ is one-to one. Since φ is a bijection, φ is
an isomorphism.
We know that φ(F ) is a field by its construction. φ(F ) is an integral domain (D) plus the inverses of
the elements of D.
Therefore we have a subfield of E, specifically φ(F ) that is isomorphic to F .
2. (a) Let D be an integral domain and f (x), g(x) ∈ D[x]. Prove that
deg f (x)g(x) = deg f (x) + deg g(x).
(b) Show that the above result does not necessarily hold in R[x] if R is not an integral domain.
Proof by (Lilith, Lisa, Nathan M, Rob). (a) Let D be an integral domain. Then a product in D is
only 0 if one of the factors of that product is 0. Then, let f (x), g(x) ∈ D such that
2
f (x) = an xn + an−1
n−1 + · · · + a2 x + a1 x + a0 , an 6= 0, ai ∈ D
2
g(x) = bm xm + bm−1
m−1 + · · · + b2 x + b1 x + b0 , bm 6= 0, ai ∈ D
Note that the degree of f (x) is n and the degree of g(x) is m. Now, it is easy to see that
m−1
2
m
2
n m
f (x)g(x) = (an xn +an−1
n−1 +· · ·+a2 x +a1 x+a0 )(bm x +bm−1 +· · ·+b2 x +b1 x+b0 ) = an bm x x +...+a0 b0
Then since an and bm are not equal to zero, an bm 6= 0, since D is an integral domain. Thus, the
degree of f (x)g(x) comes from an bm xn xm = an bm xn+m and is n + m. Note that the degree of
f (x) + g(x) is also n + m. Therefore,
deg f (x)g(x) = deg f (x) + deg g(x).
(b) We can show that the above result does not necessarily hold in R[x] if R is not an integral domain
through a counterexample. Consider Z6 [x]. This is not an integral domain, because while neither
2 nor 3 are equal to 0, 2 × 3 = 0. Define f (x) and g(x) as follows:
f (x)
g(x)
= 3x3 + 4x + 5
= 2x2 + x + 1
Note that the degree of f (x) = 3 and the degree of g(x) = 2, so adding gives us deg f (x) +
deg g(x) = 5. However, note that
f (x)g(x) = 3x4 + 5x3 + 2x2 + 4x + 1,
which has degree 4. Since 4 6= 5, the above result does not necessarily hold in R[x] if R is not an
integral domain.
3. Let I be an ideal of a ring R.
(a) Show that I[x] is a ideal of R[x].
(b) Show that R[x]/I[x] ∼
= (R/I)[x].
Proof by (Rob, Caitlyn, Nathan S, Derrek). (a) First, we note that since I is an ideal of a ring R,
then for all a, b ∈ I and r ∈ R, then a − b ∈ I and ra, ar ∈ I. So, we let a(x), b(x) ∈ I[x], and
r(x) ∈ R[x] where deg (a(x)) = n, deg (b(x)) = m and deg (r(x)) = k. Then,
n
X
(ai − bi )xi
a(x) − b(x) =
i=0
where we arbitrarily let n ≥ m and bi = 0 where i > m. Since ai −bi ∈ I then the sum of (ai −bi )xi
over the possible values of i must be in I[x]. Next, by the commutativity of the multiplication of
functions, we have that a(x)r(x) = r(x)a(x). Then,
! n
!
k
X
X
h
i
r(x)a(x) =
rh x
ai x .
h=0
i=0
When we break this down, all of the terms will be of the form rs xs at xt for some s ≤ k and
t ≤ n. Then, by the commutativity of real multiplication, this equals rs at xs+t . Since rs at ∈ I
then the sum of terms like these will be in I[x]. Since we have shown a(x) − b(x) ∈ I[x] and
r(x)a(x), a(x)r(x) ∈ I[x], then I[x] is also an ideal in R[x] when I is an ideal in R.
(b) We will show this using the First Isomorphism Theorem for Rings, but in order to do this, we must
first show that φ : R[x] → (R/I)[x] is a homomorphism. We will define φ for any element in R[x]
like a(x) = an xn + an−1 xn−1 + · · · + a0 which is operated on by φ(a(x)) = (an + I)xn + (an−1 +
I)xn−1 +· · ·+(a0 +I). Thus, if we let b(x) = bm xm +bm−1 xm−1 +· · ·+b0 , then we want to show that
φ is operation preserving if φ(a(x)b(x)) = φ(a(x))φ(b(x)) and φ(a(x) + b(x)) = φ(a(x)) + φ(b(x)).
Subsequently, as long as we assume without loss of generality that n ≥ m and bi = 0 where i > m,
then we see that

φ(a(x)b(x)) = φ 
n+m
X

ai bj xi+j 
i+j=0
=
n+m
X
(ai bj + I)xi+j
i+j=0
=
n+m
X
(ai + I)(bj + I)xi+j
i+j=0
=
n
X

m
X
j
 (bj + I)x 
!
(ai + I)x
i
i=0
j=0
= φ(a(x))φ(b(x))
so that φ preserves the multiplicative operation. Then, to show that it preserves the additive
operation, we again assume without loss of generality that n ≥ m and bi = 0 where i > m.
Therefore, we see that
!
n
X
i
i
φ(a(x) + b(x)) = φ
ai x + bi x
=φ
i=0
n
X
!
(ai + bi )x
i
i=0
=
n
X
(ai + bi + I)xi
i=0
n
X
((ai + I) + (bi + I))xi
=
i=0
n
X
=
(ai + I)xi + (bi + I)xi
i=0
n
n
X
X
=
(ai + I)xi +
(bi + I)xi
i=0
i=0
= φ(a(x)) + φ(b(x)).
Thus, φ preserves both operations which means that it is a ring homomorphism. Next, we easily
see that I[x] = ker φ since φ(an xn + · · · + a0 ) = (an + I)xn + (an−1 + I)xn−1 + · · · + (a0 + I)
which will be the identity polynomial only if ai ∈ I for all i. Thus, ker φ = I[x]. Then, we need
(R/I)[x] to be the image of R[x]/I[x]. If we pick some element r(x) + I[x] ∈ R[x]/I[x], then
φ(r(x) + I[x]) =
k
X
i=0
k
X
(ri + I + I)x =
(ri + I)xi ∈ (R/I)[x].
i
i=0
Given that we have shown the homomorphism, kernel, and image, then we can utilize the First
Isomorphism Theorem for Rings to say that R[x]/I[x] ∼
= (R/I)[x].
4. Let F be a field, a ∈ F and f (x) ∈ F [x]. Show that f (a) is the remainder in the division of f (x) by
x − a. Conclude that a is zero of f (x) iff x − a is a factor of f (x).
Proof by (Chris, tim, Kirsten, Annette). Let F be a field, a ∈ F , and f (x) ∈ F [x]. To show the first
statement, note that by the Division Algorithm, we can write f (x) = q(x)(x − a) + r(x) for some
q(x), r(x) ∈ F [x], where deg r(x) < 1. Then deg r(x) = 0, and r(x) = c for some constant c ∈ F .
Evaluating f (x) at a yields f (a) = q(a) · 0 + c. This implies f (a) = c. Thus, r(x) = f (a), and we have
proved the first statement.
To prove the second statement, first assume a is a zero of f (x). By the first statement of (4), f (x) =
q(x)(x − a) + f (a) for some q(x) ∈ F [x]. Since f (a) = 0 by our assumption, we have f (x) = q(x)(x − a).
Thus, x − a is a factor of f (x). To prove the other direction, suppose x − a is a factor of f (x). Then
f (x) = q(x)(x − a) for some q(x) ∈ F [x]. This implies f (a) = q(a)(a − a) = 0. Thus, a is a zero of
f (x). This completes the proof of the second statement.
5. Let f (x) = 4x3 + 2x2 + x + 3 and g(x) = 3x4 + 3x3 + 3x2 + x + 4.
(a) With f (x), g(x) ∈ Z5 [x], find f (x) + g(x) and f (x)g(x).
(b) With f (x), g(x) ∈ Z3 [x], find f (x) + g(x) and f (x)g(x).
Proof by (Anne, Zach, Linda, Jayna). (a) In Z5 [x], we see that
f (x) + g(x)
=
3x4 + 7x3 + 5x2 + 2x + 7
=
3x4 + 2x3 + 2x + 2
and that
f (x)g(x)
=
12x7 + 18x6 + 21x5 + 22x4 + 30x3 + 18x2 + 7x + 12
=
2x7 + 3x6 + x5 + 2x4 + 3x2 + 2x + 2
(b) In Z3 [x], we see that
f (x) + g(x)
=
3x4 + 7x3 + 5x2 + 2x + 7
= x3 + 2x2 + 2x + 1
and that
f (x)g(x)
=
12x7 + 18x6 + 21x5 + 22x4 + 30x3 + 18x2 + 7x + 12
= x4 + x