Final Exam solutions

Math 116: Finashin, Pamuk, Pierce, Solak June 7, 2011, 13:30–15:30 (120 minutes)

Instructions. Work carefully. Your methods should be clear to the reader. Problem 1 (12 points). Is x^{3} + x + 1 reducible over:

(a) ℚ ?

(b) ℤ_{3} ?

(c) ℤ_{5} ? Solution. Since the polynomial (call it f) has degree 3, it is reducible if and only if it has a root.

(a) If p/q is a root, then p and q divide 1, so p/q = ±1. But f(1) = 3 and f(−1) = −1, so f has no roots and is irreducible over ℚ.

(b) f(1) = 0, so f is reducible over ℤ_{3} (divisible by (x − 1)).

(c) f(0) = 1, f(1) = 3, f(2) = 11, f(3) = 31, f(4) = 69; so f is irreducible over ℤ_{5}.

Problem 2 (8 points). Letting f(x) = 3x^{4} + 5x^{3} + x^{2} + 5x − 2, write f(x) as a product of irreducible polynomials over ℚ. Solution. If p/q is a root of f, then p | 2 and q | 3, so p ∈ {±1, ±2} and q ∈ \{±1, ±3\}, hence p/q ∈ \{±1, ±2, ±\frac{1}{3}, ±\frac{2}{3}\}.

f(1) = 3 + 5 + 1 + 5 − 2 ≠ 0,

f(−1) = 3 − 5 + 1 − 5 − 2 ≠ 0,
f(2) = 48 + 40 + 4 + 10 − 2 ≠ 0,
f(−2) = 48 − 40 + 4 − 10 − 2 = 0.

Since −2 is a root, f(x) is divisible by (x + 2). Using the division algorithm,

f(x) = (x + 2)(3x^{3} − x^{2} + 3x − 1).

Also 1/3 is a root of 3x^{3} − x^{2} + 3x − 1, and

3x^{3} − x^{2} + 3x − 1 = (3x − 1)(x^{2} + 1).

Therefore f(x) = (x + 2)(3x − 1)(x^{2} + 1) . Since (x + 2) and (3x − 1) have degree 1, they are irreducible. Since x^{2} + 1 > 0 for all x in ℚ, it has no rational roots and is therefore irreducible. Problem 3 (5 points). Is every integral domain a field? Explain. Solution. No, consider the ring of integers (ℤ, +, ·) which is an integral domain, since it has no zero divisors. But it is not a field, since the non-zero elements except ±1 do not have multiplicative inverses. Problem 4 (15 points). Find a polynomial f(x) of least positive degree with the given properties. (Your answer should show the coefficients of f(x).)

(a) f(x) is over ℂ, and f(2i) = 0 = f(1 + i).

(b) f(x) is over ℝ, and 2i and 1 + i are zeros of it.

(c) f(x) is over ℤ_{2}, and 1 (that is, [1]) is a zero of multiplicity 4. Answers.

(a) f(x) = (x − 2i)(x − 1 − i) = x^{2} − (1 + 3i)x + 2i − 2 over ℂ.

(b) f(x) = (x − 2i)(x + 2i)(x − 1 − i)(x − 1 + i) = (x^{2} + 4)(x^{2} − 2x + 2) = x^{4} − 2x^{3} + 6x^{2} − 8x + 8 over ℝ.

(c) f(x) = (x − 1)^{4} = x^{4} − 4x^{3} + 6x^{2} − 4x + 1 = x^{4} + 1 over ℤ_{2}. Problem 5 (5 points). Over a field K, suppose f(x) is a polynomial with no zeros in K. Must f(x) be irreducible over K? Explain. Solution. No. For example, the polynomial x^{4} +2x^{2} +1 over ℝ has no zeros in ℝ but it is reducible over ℝ.

Problem 6 (15 points). Working over ℤ_{3}, letting

f(x) = x^{5} + x + 1,  g(x) = x^{2} + 1,

find s(x) and t(x) such that f(x) · s(x) + g(x) · t(x) = 1. Solution.

f(x) = x^{5} + x + 1 = g(x) · (x^{2} + 2x) + 2x + 1
g(x) = x^{2} + 1 = (2x + 1) · (2x + 2) + 2.

Then

2 = g(x) − (2x + 1)(2x + 2)
= g(x) − [f(x) − g(x)(x^{3} + 2x)](2x + 2)
= f(x)(x + 1) + g(x)[1 + (x^{3} + 2x)(2x + 2)]
= f(x)(x + 1) + g(x)(2x^{4} + 2x^{3} + x^{2} + x + 1)

Hence,

1 = f(x)(2x + 2) + g(x)(x^{4} + x^{3} + 2x^{2} + 2x + 2).

So,

s(x) = 2x + 2,  t(x) = x^{4} + x^{3} + 2x^{2} + 2x + 2.

Problem 7 (20 points). Let R be the subring {x + yi: x, y ∈ ℤ} of ℂ, and let I be the ideal {x + yi: x, y ∈ 2ℤ} of R.

(a) How many additive cosets has I in R? List them clearly.

(b) Is the quotient R/I cyclic as an additive group? Explain.

(c) Show that the function ϕ from R to ℤ_{2} given by

ϕ(x + yi) = [x + y]  (∗)

is a ring homomorphism.

(d) Does the same formula (∗) define a ring homomorphism from R to ℤ_{3}? Explain.

Solution.

(a) Four cosets: I, 1 + I, i + I, and 1 + i + I.

(b) No: R/I has order 4, but each element has order 1 or 2.

(c)  \begin{aligned}ϕ((a + bi) + (x + yi)) & = ϕ(a + x + (b + y)i) \\ & = [a + x + b + y] \\ & = [a + b] + [x + y] \\ & = ϕ(a + bi) + ϕ(x + yi), \end{aligned}

ϕ((a + bi) · (x + yi)) = ϕ(ax − by + (ay + bx)i)

= [ax − by + ay + bx] = [ax + by + ay + bx] = [a + b] · [x + y] = ϕ(a + bi) · ϕ(x + yi).

(d) No, since ϕ(i^{2}) = ϕ(−1) = [−1] = [2], while ϕ(i)^{2} = [1]^{2} = [1].

Remark. For part (d), many people observed that the computations of (c) for multiplication were not justified modulo 3. This is correct, but one must show that the system of equations fails in at least one case. (In fact the system is correct when 3 | by, but fails in all other cases.)