Mathematics 351 – Abstract Algebra I Solutions for Midterm Exam 1 – October 1, 2007

I. Terminology. A) (5) Give an example of a ring R that is not a commutative ring.

Solution: The ring of 2 × 2 matrices with real entries, M_{2×2}(R), is one example. B) (5) What does it mean to say that R is a ring with identity?

Solution: R is a ring with identity if there is a multiplicative identity in R: an element 1_{R} satisfying 1_{R} · a = a · 1_{R} = a for all a ∈ R.

C) (5) What does it mean to say that u ∈ R is a unit?

Solution: u is a unit if u has a multiplicative inverse in R (an element x such that u · x = 1_{R} = x · u. Note that this only makes sense if R is a ring with identity(!)

D) (5) Is every commutative ring with identity an integral domain? If so, say why. If not, give a counterexample. Solution: The answer is no. The ring Z_{6} is a commutative ring with identity, but it is not an integral domain, since for instance, [2][3] = 0, but [2] ≠ [0] and [3] ≠ [0].

II. Let R be a ring. Let b be any one particular element of R, and let

C_{b} = {a ∈ R : a · b = b · a}.

A) (10) Show that C_{b} is a subring of R.

Solution: We use the criterion from Theorem 3.6 in Hungerford. First notice that C_{b} is not empty, since 0_{R} · b = b · 0_{R} = 0_{R}. Hence 0_{R} ∈ C_{b}. Let a, a′ ∈ C_{b}. Then by the distributive law

(a − a′) · b = a · b − a′ · b = b · a − b · a′ = b · (a − a′).

This shows a − a′ ∈ C_{b} so C_{b} is closed under differences. Similarly by associativity of multiplication,

(a · a′) · b = a · (a′ · b) = a · (b · a′) = (a · b) · a′ = (b · a) · a′ = b · (a · a′). Hence C_{b} is also closed under products, and hence is a subring of R.

B) (10) Let S be a second ring, and let φ : R → S be a ring homomorphism. Show that if b ∈ R, then φ(C_{b}) ⊆ C_{φ(b)} ⊆ S.

Solution: Let a ∈ C_{b}, so a · b = b · a. By the definition of a ring homomorphism,

φ(a) · φ(b) = φ(a · b) = φ(b · a) = φ(b) · φ(a).

This shows φ(a) ∈ C_{φ(b)}. Hence φ(C_{b}) ⊆ C_{φ(b)}. (φ(C_{b}) ⊆ S by definition, so it is not necessary to prove that part).

III. Let F [x] be the ring of polynomials with coefficients in a field F. A) (10) Show that the quotient and remainder under the division of f(x) by g(x) in F [x]

are unique.

Solution: Suppose f(x) = q(x)g(x) + r(x) and f(x) = q_{1}(x)g(x) + r_{1}(x), where both r(x) and r_{1}(x) are either zero or have degree less than the degree of g(x). Then subtracting these equations shows 0 = (q(x) − q_{1}(x))g(x) + r(x) − r_{1}(x), so r(x) − r_{1}(x) = (q_{1}(x) − q(x))g(x). If r(x) − r_{1}(x) ≠ 0, then it is a polynomial of degree ≤ deg(g). However, deg((q_{1}(x)−q(x))g(x)) = deg(q_{1}(x)−q(x))+deg(g(x)) ≥ deg(g(x)). This gives a contradiction, so r(x) = r_{1}(x) and q(x) = q_{1}(x).

B) (5) Define: f(x) is irreducible in F [x].

Solution: f(x) is irreducible if in every factorization f(x) = g(x)h(x) in F [x], one of g(x), h(x) is a unit (a nonzero constant polynomial), and the other is an associate of f(x).

C) (10) In what sense are factorizations of polynomials into irreducibles unique? (State the theorem that answers this question; you do not need to prove it.)

Solution: The sense in which irreducible factorizations are unique in F [x] is that if

f(x) = p_{1}(x) ⋯ p_{s}(x)  and  q_{1}(x) ⋯ q_{s}(x),

where all the p_{i}(x) and q_{j}(x) are irreducible in F [x], then r = s, and for each 1 ≤ i ≤ s, there exists some j such that p_{i}(x) and q_{j}(x) are associates.

IV. A) (10) Let F be a field. Show that a ∈ F is a root of f(x) ∈ F [x] if and only if

(x − a)|f(x) in F [x].

Solution: Use the division algorithm in F [x], dividing x − a into f(x) to write f(x) = q(x)(x − a) + r(x), where either r is zero, or else deg(r) < deg(x − a) = 1. If a ∈ F is a root of f(x), then 0 = f(a) = q(a)(a − a) + r, so r = 0. This shows (x − a)|f(x). Conversely, if (x − a)|f(x), then r = 0, so f(a) = q(a)(a − a) = 0, so a is a root of f(x).

B) (10) Let F be a field. If a_{1}, …, a_{n} are distinct roots of f(x) ∈ F [x] in F, show that ((x − a_{1})(x − a_{2}) ⋯ (x − a_{n})) |f(x). (Hint: induction on n).

Solution: Following the hint, part A is the base case for the induction. So assume that whenever a polynomial g(x) in F [x] has k distinct roots a_{1}, …, a_{k}, then ((x − a_{1}) · · · · · (x − a_{k}))|g(x). Assume now that f(x) is a polynomial with k + 1 distinct roots in F, say a_{1}, …, a_{k+1}. By part A, we know that (x − a_{k+1})|f(x) so write

(1)  f(x) = (x − a_{k+1})q(x)

for some q(x) ∈ F [x]. If we substitute x = a_{i}, for any i, 1 ≤ i ≤ k, then 0 = f(a_{i}) = (a_{i} − a_{k+1})q(a_{i}). The first factor cannot be zero since the a_{j}’s are distinct. Hence q(x) must have a_{1}, …, a_{k} as roots, and by induction ((x − a_{1}) · · · · · (x − a_{k}))|q(x). This means

(2)  q(x) = ((x − a_{1}) · · · · · (x − a_{k}))s(x)

for some s(x) ∈ F [x]. Hence combining (1) and (2),

f(x) = (x − a_{k+1})((x − a_{1}) · · · · · (x − a_{k}))s(x)

which shows what we wanted to prove.

C) (5) What are the roots of x^{2}+2x ∈ Z_{8}[x] in Z_{8}? Why is your result not a contradiction to part B?

Solution: x = 0, 2, 4, 6 all satisfy x^{2} + 2x = 0 in Z_{8}. This is not a contradiction to part B since Z_{8} is not a field. The statement in part B does not apply.

V. All parts of this question refer to

f(x) = 5x^{5} − 12x^{3} + 36x^{2} + 18 ∈ Q[x].

A) (5) List all rational numbers that could be roots of f according to the Rational Root Test.

Solution: The possible rational roots of f(x) are all rational numbers in lowest terms of the form p/q, where p|18 and q|5, so:

x = ±1, ±2, ±3, ±6, ±9, ±18, ±\frac{1}{5}, ±\frac{2}{5}, ±\frac{3}{5}, ±\frac{6}{5}, ±\frac{12}{5}, ±\frac{18}{5}.

B) (5) Use any applicable method to decide whether f is irreducible in Q[x], and state which criterion or criteria you are using.

Solution: By the Eisenstein Criterion with p = 2, f(x) is irreducible in Q[x]: 2 |̸ 5, 2|(−12), 2|36, 2|18, but 2^{2} = 4 |̸ 18.