Algebra 1 Assignment 7 Solutions

Problem 1 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.

Problem 2 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) = (m ∗ a, 0) ∈ I so I is closed under multiplication by arbitrary elements in R. It follows that I is an ideal.

Problem 3 Let I be the set of nilpotent elements of a 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 (r ∗ a)^{m} = r^{m} ∗ a^{m} = r^{m} ∗ 0 = 0 (the second equality holds because R is commutative) so r ∗ a is also nilpotent so 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.

For the next two problems, let R be the ring of functions from ℤ to ℝ with addition +_{R} and multiplication ∗_{R}. For f, g ∈ R, addition the sum 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.

Problem 4 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.

Problem 5 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.

Problem 6 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|f ∈ I, n = deg(f)} ⊂ ℕ is nonempty 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, 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). □