Math 593 Assignment # 4 Solutions

Bryden R. Cais

October 9, 2002

1. (a) We differentiate two cases:

• For all m ∈ ℤ \ {0}, we have ϕ(m) ≠ 0 ∈ R. In this case, ϕ is an injective map (to its image) and hence ℤ is isomorphic to ϕ(ℤ) (since ϕ is obviously surjective onto its image).

• There exists m ∈ ℤ \ {0} such that ϕ(m) = 0 ∈ R. Observe that if such an m is negative then we also necessarily have ϕ(−m) := −ϕ(m) = 0 so that we can take m to be the least positive integer for which ϕ(m) = 0. Suppose that m is not prime, say m = ab for positive integers 1 < a, b < m. Then we have

0 = ϕ(m) = ϕ(ab) = ϕ(a)ϕ(b).

Since R is an integral domain, and ϕ(a), ϕ(b) ∈ R have product zero, we must have ϕ(a) = 0 or ϕ(b) = 0. In either case, we have contradicted the minimality of m and hence m = p is prime.

(b) Take, for example, R_{p} = ℤ/(p)[X], the ring of polynomials in X with coefficients in ℤ/(p). Clearly this is infinite since it is a (countably) infinite dimensional vector space over ℤ/(p). Moreover, let r, s ∈ R_{p} have degree d, e respectively. Then we can write

r = aX^{d} + ⋯
s = bX^{e} + ⋯
rs = abX^{d+e} + ⋯,

so that if rs = 0 then in particular, ab = 0 ∈ ℤ/(p). Since ℤ/(p) is an I.D., we must have a or b equal to 0, which is absurd. Hence R_{p} is an integral domain.

(c) Let a ∈ R. Then we have

p · a = p · 1_{R} · a = ϕ(p) · a.

Since R has characteristic p, the image of ℤ under ϕ must be ℤ/(p), that is, ϕ(p) = 0 ∈ R so we have p · a = 0 · a = 0, as required.

(d) It is obvious that F preserves the multiplicative structure of R and that F (1) = 1, F (0) = F (0). So it suffices to check that F is a homomorphism of abelian groups (under the addition of R). Since the binomial theorem holds in any commutative ring, we have

F (x + y) = (x + y)^{p} = ∑_{i=0}^{p} \binom{p}{i}x^{p−i}y^{i}.

It is not difficult to see that \binom{p}{i} is divisible by p for 1 ≤ i ≤ p − 1 and therefore, since R has characteristic p, we find that \binom{p}{i} = 0 ∈ R for 1 ≤ i ≤ p − 1 and hence that

F (x + y) = x^{p} + y^{p} = F (x) + F (y).

2. Let

f(x) = x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{0} ∈ R[x]

and suppose that a_{i} ∈ P, where P is a prime ideal of R, but that a_{0} ∉ P^{2}. Suppose that f is not irreducible, that is, we have a factorization

f = gh.

Reducing modulo P gives a factorization

f̄ = ḡh̄

where f̄, ḡ, h̄ are the images of f, g, h in R/P [x] under the natural quotient map. By hypothesis, we have f̄ = x^{n}. Since R/P is an integral domain (P being afterall a prime ideal) it follows that ḡ = x^{i}, h̄ = x^{j} with i + j = n and 0 < i, j. Writing

g = α_{i}x^{i} + ⋯ + α_{0}
h = β_{j}x^{j} + ⋯ + β_{0},

it follows that α_{0}, β_{0} ∈ P. By our factorization of f, we must have a_{0} = α_{0}β_{0} ∈ P^{2}, a contradiction. Hence f is irreducible.

3. (a) Let p be prime and set

Φ_{p}(x) = x^{p−1} + x^{p−2} + ⋯ + x + 1 ∈ ℤ[x].

Observe that

Φ_{p}(x) = \frac{x^{p} − 1}{x − 1}.

Now Φ_{p}(x) is irreducible if and only if Φ_{p}(x + 1) is irreducible. But we have

Φ_{p}(x + 1) = \frac{(x + 1)^{p} − 1}{x} ≡ x^{p−1} mod p

by part d) of question 1. On the other hand, the constant term in Φ_{p}(x + 1) is clearly

\binom{p}{p − 1} = p

so that Φ_{p}(x + 1) is irreducible since it is p-Eisenstein.

(b) It is false that for any natural number the polynomial

x^{m−1} + ⋯ x^{m−2} + ⋯ + x + 1

is irreducible. A simple counterexample is m = 4 when we have

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

More generally, define

Φ_{m}(x) = \begin{cases}x − 1 & if m = 1 \\ \frac{x^{m} − 1}{∏\begin{matrix}d|m \\ d<m \end{matrix} Φ_{d}(x)} & otherwise \end{cases} .

Then it can be shown that Φ_{m}(x) ∈ ℤ[x] is irreducible for every m > 0. The polynomial Φ_{m}(x) is called the m^{th} cyclotomic polynomial.

4. This is a fairly simple counting argument. If a degree two monic polynomial p(x) is reducible, then it splits as the product of two monic linear factors. Moreover, these monic linear factors are uniquely determined by p(x). To see this, observe that a monic linear polynomial is completely determined by its constant term, and that a, b are uniquely determined by the requirement

(x + a)(x + b) = x^{2} + cx + d

in F [x], since the system of equations a + b = c, ab = d has a unique (up to transposition of a, b) solution (if any) modulo p. It follows that the number of degree 2 reducible polynomials is the same as the number of pairs (where order doesn’t matter) of monic linear polynomials. There are

\frac{p(p − 1)}{2}

such pairs if the factors are different, and

p

pairs if they are the same, for a total of

\frac{p(p + 1)}{2}

reducible degree two monic polynomials. Therefore, there are

p^{2} − \frac{p(p + 1)}{2} = \frac{p(p − 1)}{2}

monic irreducible degree 2 polynomials in F [x].

5. (a) Suppose that d < 0 is squarefree and that a + b\sqrt{d} is a unit. Then there exist c, d such that

(a + b\sqrt{d})(c + e\sqrt{d}) = 1.

Observe that if one of b, e is 0, so is the other, giving the units ±1. Otherwise, both b, e ≠ 0. There is a unique, nontrivial automorphism of ℚ[\sqrt{d}] fixing ℚ pointwise—namely, complex conjugation, which sends \sqrt{d} to −\sqrt{d}. Applying this automorphism gives

(a + b\sqrt{d})(a − b\sqrt{d})(c + e\sqrt{d})(c − e\sqrt{d}) = (a^{2} − db^{2})(c^{2} − de^{2}) = 1.

Now since a^{2} − db^{2}, c^{2} − de^{2} ∈ ℤ we have a^{2} − db^{2} = ±1. If d < −1, this requires b = 0, a = ±1. If d = −1 we can have a = ±1, b = 0 or b = ±1, a = 0. The same goes for c, e. In any case, we have determined that the group of units in ℤ[\sqrt{d}] is precisely

{±1}  if d ≠ −1
{±1, ±i}  if d = −1

(b) Observe that (\sqrt{2} + 1)(\sqrt{2} − 1) = 1 so that \sqrt{2} − 1 is a unit in ℤ[\sqrt{2}]. It follows that (\sqrt{2} + 1)^{n} is a unit for every n ∈ ℤ. Moreover, this gives an infinitude of units as |\sqrt{2} + 1| > 1 so that |(\sqrt{2} + 1)^{n}| ≠ |(\sqrt{2} + 1)^{m}| unless n = m. In fact, every unit in ℤ\sqrt{2} has the form ±(1 + \sqrt{2})^{n}, for some n ∈ ℤ (in other words, the unit group is isomorphic to ℤ ⊕ ℤ/(2)).