Basic Algebra 1 Solutions of Assignment 6

Solution of the problem 1. Starting with prime number 3 and looking for a root, we see that f(x) = x^{2} + 1 has no zero in ℤ_{3}, hence it is irreducible in ℤ_{3}[x]. Next consider 5. This case is actually different. In fact in ℤ_{5}[x] we have the factorization f(x) = (x + 2)(x + 3). Continuing this way, we find that f(x) is irreducible in ℤ_{p}[x] for p = 3, 7, 11, 19, 23 and is reducible for p = 5, 13, 17.

Looking for a general pattern, first note that each of the primes 5, 13 and 17 is of the form 4k + 1, and on the contrary, none of the primes 3, 7, 11, 19 and 23 is in that form. Secondly, observe that

5 = 2^{2} + 1^{2}, 13 = 3^{2} + 2^{2}, 17 = 4^{2} + 1^{2},

while the primes 3, 7, 11, 19 and 23 don’t enjoy such property, namely they cannot be represented as a sum of two squares. In fact one has the following beautiful theorem of Fermat:

An odd prime number p is a sum of two square, i.e., p = a^{2} + b^{2}, if and only if it is of form 4k + 1.

Also look at the solution of the problem 5.

Solution of the problem 2. Here is one example: f(x) = 2x^{2} + 4. Note that f(1) = f(2) = f(4) = f(5) = 0. This does not contradict the theorem proven in class stating that a polynomial in F [x] of degree d has at most d roots and the reason is simple: ℤ_{6} is not a field!

Solution of the problem 3. This can be done by a trial and error search and here is the answer:

[x^{2} + x + 1]^{−1} = [x^{2}].

To verify our answer, notice that since [x^{3} + x + 1] = [0], we have

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

(N.B. 2 = 0 and −1 = 1, because we are working in ℤ_{2}.) For another way of looking at this problem, go to the solution of the next problem.

Solution of the problem 4. Here you are:

[x]^{1} = [x], [x]^{2} = [x^{2}], [x]^{3} = [x + 1], [x]^{4} = [x^{2} + x],
[x]^{5} = [x^{2} + x + 1], [x]^{6} = [x^{2} + 1], [x]^{7} = [1].

So, the smallest j > 0 for which [x]^{j} = [1] is 7, and therefore [x] is a generator for the multiplicative group of nonzero elements of the finite field ℤ_{2}[x]/(x^{3} +x+1).

Back to the solution of the previous problem, note that

[x^{2} + x + 1][x^{2}] = [x]^{5}[x]^{2} = [x]^{7} = [1] !

Solution of the problem 5. Proof by contradiction. Suppose that x^{2} + 1 factors in ℤ_{p}[x]. So, it has a root, a say, in ℤ_{p}, i.e., a^{2} + 1 = 0 in ℤ_{p}. This in turn implies that p | a^{2} + 1 or equivalently a^{2} ≡ −1 (mod p). Now since p is

odd, we can raise both sides of a^{2} ≡ −1 (mod p) to the power \frac{p − 1}{2} to get

a^{p−1} ≡ (−1)^{\frac{p−1}{2}} (mod p).

Comparing with little Fermat, we infer that

1 ≡ (−1)^{\frac{p−1}{2}} (mod p).

Since p > 2, this is impossible unless the last congruence relation becomes equality and that amounts to \frac{p−1}{2} = 2k, hence p = 4k + 1, which is a contradiction!

Solution of the problem 6a. Let R = F [X] where F is a field and let I = F be the set of constant polynomials. Then I isn’t an ideal because i ∈ I ⇔ deg(i) = 0 or i = 0. So for any f ∈ R s.t. deg(f) ≥ 1 and 0 ≠ i ∈ I we have deg(f ∗ i) = deg(i) + deg(f) ≥ 1 ⇒ f ∗ i ∉ I.

Solution of the problem 6b. Let R = ℤ × ℤ and let I = {(m, 0)|m ∈ ℤ}. Then I is an ideal. Let (a, 0), (b, 0) ∈ I. Then (a, 0) + (b, 0) = (a + b, 0) ∈ I, so I is closed under addition and if (m, n) ∈ R is some arbitrary element we have that (m, n)∗(a, 0) = (ma, 0) ∈ I so I is closed under multiplication by arbitrary elements in R. It follows that I is an ideal.

Solution of the problem 6c. Let I be the set of nilpotent elements of a (commutative) ring R i.e.

I = {a ∈ R|∃ m ∈ ℕ s.t. a^{m} = 0}.

Then I is an ideal. Let a, b ∈ I, then ∃m, n ∈ ℕ s.t. a^{m} = b^{n} = 0. If r ∈ R is some arbitrary element then (ra)^{m} = r^{m}a^{m} = r^{m}0 = 0 (the second equality holds because R is commutative), so ra is also nilpotent, hence I is closed under multiplication by arbitrary elements. Now let N = max(n, m) and consider (a + b)^{2N}. First note that if i ≤ N then 2N − i ≥ N. So by the binomial theorem we have:

(a + b)^{2N} = ∑_{i=0}^{2N} \binom{2N}{i}a^{i}b^{2N−i}

= ∑_{i=0}^{N} \binom{2N}{i}a^{i} \underbrace{b^{2N−i}}_{=0} + ∑_{i=N+1}^{2N} \binom{2N}{i} \underbrace{a^{i}}_{=0} b^{2N−i}
= 0

It follows that a + b is also nilpotent, so I is closed under addition, it follows that I is an ideal.

Remark It is important to notice that the statement of this problem is false if we remove the commutativity of R. Here is one example. Let R = M_{2}[ℤ] be the ring of all 2 × 2 integer matrices (with the usual addition and multiplication), and let

A = \binom{0 1}{0 0}, B = \binom{0 0}{1 0}.

One easily sees that

A^{2} = B^{2} = O_{2} (the zero matrix of size 2,)

however, no positive power of A + B = \binom{0 1}{1 0} is zero! (check this). The reason

behind this is that M_{2}[ℤ] is not a commutative ring.

For the next two parts, let R be the ring of functions from ℤ to ℝ with addition +_{R} and multiplication ∗_{R}. For f, g ∈ R, the addition f +_{R} g ∈ R is defined as the mapping:

(f +_{R} g)(x) = f(x) + g(x)

and the product of two functions f and g is defined as the mapping:

(f ∗_{R} g)(x) = f(x)g(x).

From now on, subscripts for the operation signs will be omitted.

Solution of the problem 6d. Let I be the set of functions f s.t. f(0) = f(1). I is not an ideal. Indeed, let f ∈ I be the constant 1 function i.e. for each n ∈ ℤ, f(n) = 1 and let g be the function g(n) = n. Then we have that (f ∗ g)(0) = f(0)g(0) = 1 × 0 = 0 but (f ∗ g)(1) = f(1)g(1) = 1. So f ∗ g ∉ I so I is not closed under multiplication by arbitrary elements of R so it’s not an ideal.

Solution of the problem 6e. Let I = {f ∈ R|f(0) = f(1) = 0}. Let f, g ∈ I then (f+g)(1) = f(1)+g(1) = 0+0 = 0 and similarly (f+g)(0) = 0 so I is closed under addition. Now let h ∈ R be some arbitrary element. Then we find that (h∗f)(1) = h(1)f(1) = h(1)×0 = 0 and (h∗f)(0) = h(0)f(0) = h(0)×0 = 0, so h ∗ f ∈ I. It follows that I is closed under multiplication by arbitrary elements of R. It follows that I is an ideal.

Solution of the problem 7. Let R be the polynomial ring F [X] with coefficients in a field. Then all of its ideals are principal.

Remark. In class the strategy was to show that an ideal in ℤ is generated by its smallest positive element. Recall that in polynomial rings over fields, the notion of size corresponds to the degree of a polynomial.

Proof: Let I ⊂ F [X] = R be an ideal. Then the set S = {n ∈ ℕ : n = deg(f), fo some f ∈ I} ⊂ ℕ is nonempty (why?), so it must have a smallest element. If 0 ∈ S then I contains a constant ⇒ I = R. Otherwise let s > 0 be the smallest element in S. Then there is some f ∈ I s.t. deg(f) = s, i.e. f is the element of minimal degree in I.

Now we claim that I = (f). On one hand f ∈ I ⇒ (f) ⊆ I. On the other hand, suppose I ⊈ (f), then there is some g ∈ I such that f ∤ g. So we can apply the division algorithm to get

g = fq + r

where r ≠ 0 and deg(r) < deg(f) = s. But then we get r = g − fq ∈ I because of closure of ideals under addition and multiplication. And we get that m = deg(r) ∈ S and m < s contradiction minimality of s. It follows that I = (f).