Exercises on ideals generated by a subset

Exercise 1. Prove that if R is a (commutative ring) with unit, then

Exercise 2. Prove that the ideal of ℚ[x,y] made up by polynomials with null constant term is not a principal ideal.

Exercise 3. Prove the following relations given a and b ∈ ℤ

Generalize the results for multiple elements.

Exercise 4. Prove that

Exercise 5. Prove the result of exercise2 is a generalization of the Chinese remainder theorem.

Exercise 6. Determine the polynomial generator of the ideal I = (x⁴ + x −1, x³ − 2) in ℝ[x] nad give the polynomials of I.

Exercise 7. Let a₁, a₂, ..., a_n elements of a ring R. Give the elements of (a₁, a₂, ..., a_n) in case of R non-commutative and without unit.

Exercise 8. Determine the maximal and prime ideals of ℤ₂₄ and ℤ₃₀.

Exercise 9. Consider the ring (with respect to the operations of addition and multiplication of real functions)

Let M = {f ∈ R | f(2) = 0}. Prove that M is a maximal ideal of R.

Exercise 10. Let R be any commutative ring with 1, and let I a proper ideal of R. Prove that I is maximal iff I + aR = R ∀a ∈ R \ I.

Exercise 11. Let R be any commutative ring and let I₁ and I₂ two of its ideals such that

prove that I₁ ∩ I₂ is not a prime ideal of R.

Exercise 12. Prove that ℝ[x] / (x² − 1) ≃ ℝ ⊕ ℝ

Solutions

1. If 1 ∈ R, among the elements r_ia_i there are the a_i. Indeed the ideal generated by the subset, S = {a₁, a₂, ..., a_n} is defined as

   (a₁, a₂, ..., a_n) = {set of linear combinations of a_ir_i terms} = {∑^t_i=1r_ia_i | r_i ∈ R}

   The obtain a_i, it suffices to take the linear combination wtih r_i = 1 and a_{j ≠i} = 0. ■

2. (x,y) consists of those polynomials whose constant term is zero and is an ideal of ℚ[x,y]. (x,y) = {fx + gy |f,g ∈ ℚ[x,y]}.

   If (x,y) were a principal ideal (p), it would contain along with other elements the polynomials x and y

   x = fp,   y = gp

   for suitable choice of g and f in ℚ[x,y]. Since the polynomials of degree 0 have inverses, and any element with an inverse generates the entire ring, p cannot be of degree 0. Since (x,y) ≠ (0), p cannot be the zero polynomial and therefore must have positive degree. Moreover, we have that for the ring ℚ[x,y] the degree of a product of nonzero polynomials is the sum of the degree of the factors. Hence the relation above for x and y would show that g and f are of degree 0 while p is of the first degree. It is easily verified that this cannot happen for any such p, and hence that (x,y) cannot be a principal ideal.

3. Let (m) = (a ) ∩ (b). m ∈ (m) implies m ∈ (a) ∩ (b), which implies m ∈ (a) and m ∈ (b). Hence m = ah and m = bk for some h,k ∈ ℤ i.e. a|m and b|m. Thus m is a common multiple of a and b. Now we must show that it is the smallest common multiple. Suppose the lcm is another say m′; hence a divisor of a and b. Now we can write m′= ah′ and m′= bk′, which means m′ ∈ (a) ∩ (b). Thus m′ ∈ (m). But this would lead to a contradiction since ∃ z ∈ ℤ. m ⋅ z = m', that is m ∣ m' which is a contradiction.

4. Let (a,b) = (d). Then a,b ∈ (d) ⇒ d∣a,b, so d is a common divisor of a,b. Conversley d ∈ (a,b) so d = ra + sb, r,s ∈ ℤ, we must show now that d is the gcd. Let now c be an arbitrary common divisor, so c∣a,b ⇒ c∣ d = ra + sb. Hence d is a common divisor of a,b that is divisible by every common divisor. Therefore, by definition, d is a greatest common divisor of a,b.

   To generalize the result we use the associative law for the intersection of sets (A ∩ B) ∩ C = A ∩ (B ∩ C). We analyze first the case for 3 elements. We want to show that

   [(a) ∩ (b)] ∩ (c) = (a) ∩ [(b) ∩ (c)] = (x),   x = lcm(a,b,c)

   holds true. We already proved that (a) ∩ (b) = (m) = lcm(a,b) and (b) ∩ (c) = (m') = lcm(b,c) hence (m) ∩ (c) = (a) ∩ (m') = (x) this means x = lcm(m,c) = lcm(m',a) = lcm(lcm(a,b),c) = lcm (lcm(b,c),a) = lcm(a,b,c).

   To generalize to n elements

   (a) ∩ (b) ∩ ... ∩ (n) = (a) ∩ (b) ∩ (n) = (x),   x = lcm(a,b,..,n)

   We apply recursevely the previuos step.

   to generalize the second result let

   (a,b,..,n) = (d)   gcd(a,b,..,n).

   Then a,b,..,n ∈ (d) ⇒ d∣a,b,...,n so d is a common divisor of a,b,..,n. Conversely d ∈ (a,b,..,n) so d = ra + sb + ... + tn, r,s,t ∈ ℤ, we must show now that d is the gcd. Let now c be an arbitrary common divisor, so c∣a,b,,.n ⇒ c∣ d = ra + sb + ... + tn. Hence d is a common divisor of a,b,..,n that is divisible by every common divisor. Therefore, by definition, d is a greatest common divisor of a,b,..,n.

5. Owing to the definitions given earlier on regarding the Operations with ideals we have

   (a) + (b) = {a+b | a ∈ (a), b ∈ (b)}

   Since (d) = {dz | z ∈ ℤ} and by Bézout's Identity, d = az' + bz'' we have (d) = {(az' + bz'')z | z,z'' ∈ ℤ} = {(az' + bz'')z | z,z',z'' ∈ ℤ} = {(a ⋅ z* + b ⋅ z*') | z*, z*' ∈ ℤ}, it is readily verified that the equality holds.

   From the definitions we have

   (a) ⋅ (b) := {∑ⁿ_i=0 a ⋅ b ⋅ z_i | i ∈ ℕ, a_i ∈ (a), b ∈ (b)}

   (ab) = {∑ⁿ_i=0 ab ⋅ z_i | i ∈ ℕ, a_i ∈ (a)

   from which it is readily seen that they are equal sets.

6. The CRT can be stated as ℤ/(mn) being isomorphic to ℤ/(m) x ℤ/(n) if m,n are coprime.

   Let R = ℤ. The condition ℤ = I₁ + I₂ is satisfied by I₁ = mℤ and I₂ = nℤ with gcd(m,n) = 1. Bezout's identity gives 1 in m ℤ + nℤ, And we get all of ℤ from there since (m) + (n) = (gcd(m,n)). We observe that I₁ ∩ I₂ = m ℤ + nℤ = (mn)ℤ.

7. we already calculated this gcd when treating polynomials: −53/27, which is the polynomial (-53/27)⋅x⁰, and we make it monic by dividing by -53/27, 1x⁰ = 1. Thus I = (1) = ℝ[x].

8. The ideals of ℤ_n are, first of all, additive subgroups of ℤ_n. These we know to all have the form (d) where d divides n. But, as we know, the set (d) is the ideal generated by d. So we have just proven that

   The ideals in ℤ_n are precisely the sets of the form (d) where d divides n.

   Since we are interested in maximal ideals, and this concept is defined in terms of containment of ideals in one another, we now need to determine when we can have (d₁) ⊂ (d₂). This is the case if and only if d₁ ∈ (d₂), which is true if and only if there is an element a ∈ ℤ so that ad₂ = d₁, i.e. if and only if d₂ divides d₁. We are now ready to prove the main result: an ideal I in ℤ_n> is maximal if and only if I = (p) where p is a prime dividing n. If I has this form and J is another ideal in ℤ_n with I ⊂ J then J = (d) for some d dividing n. By our comments above this means that d divides p, i.e. d = 1 or d = p, which means that J = ℤ_n or J = I, proving that I is maximal. For the converse, suppose I = (d), (d dividing n) is maximal but d is not prime. Then d = kl with d > k, l > 1. But then I ⊊ (k) ⊊ ℤ_n. The first inequality follows from the fact that k < d implies k ∉ I. The second follows from the fact that k is a divisor of n but is not 1, therefore is not a unit in ℤ_n and so 1 ∉ (k) . But this string of inequalities implies that I is not maximal, a contradiction. Therefore d must be prime, and we are finished.

   The positive divisors of 8 are, 1,2,4, and 8, so the ideals in ℤ₈ are:

   • (1) = ℤ₈

   • (2) = {0,2,4,6}

   • (4) = {0,4}

   • (8) = {0}.

   When you're working in ℤ₈ and you write (2) it is short for ([2]₈), for example for (2), we have:

   (2) = ([2]₈) = {[2]₈ ⋅ z | z ∈ ℤ₈} = {[2]₈ ⋅ [0]₈, [2]₈ ⋅ [1]₈, ..., [2]₈ ⋅ [7]₈} = {[0]₈, [2]₈, [4]₈, [6]₈}

   Of these by inspection (2) is maximal (and therefore prime), whereas (1) and (9) are improper, so neither prime nor maximal. We see that (4) ⊂ (2), so (4) is not maximal. Also 2⋅2 ∈ (4) but 2 ∉ (4), so (4) is not prime.

   The positive divisors of 9 are 1,3,9, so the ideals in ℤ₉, are (1) = ℤ₉, (3) = {0,3,6}, (9) = {0}. Of these by inspection (3) is maximal (therefore prime).

   The positive divisors of 10 are, 1,2, 5 and 10 so the ideals in ℤ₁₀ are:

   (1) = ℤ₁₀, (2) = {0,2,4,6,8}, (5) = {0,5}, (10) = {0}.

   The maximal ideals of ℤ₂₄ are (2) and (3). The other ideals are (1) = ℤ₂₄, (4), (6), (8), (25) = {0}.

   The maximal ideals of ℤ₃₀ are (2), (3), (5). The other ideals are (1) = ℤ₃₀, (6), (10), (15), (30) = {0}.

9. M is an ideal. Suppose there exists an ideal I containing M and different from M. Then there exists a function g(x) ∈ I, g(x) ∉ M, implying g(2) = a ≠ 0. Then the function difference d(x) = g(x) − a ∈ M ⊂ I;

   Consider that I is closed (as it is an ideal) under addition. Since f ∈ I and f − a ∈ I, we find that the constant nonzero function a = f − (f − a) ∈ I. Since a ≠ 0, we also have the continuous (constant) function 1/a ∈ R, which means that I contains a(1/a) = 1, since ideals are closed under multiplication with any element of the parent ring. But this means that 1 ∈ I, so that I = R, hence M is maximal as desired.

10. I + aR is an ideal of R and contains properly I, thus if I is maximal, we have necessarily I + aR = R. Conversely, suppose I+ aR = R, ∀a ∈ R \ I. Let U any ideal such that I ⊆ U ⊆ R and U ≠ I. Then there exists a a ∈ U \ I ⇒ U ⊇ I + aR = R ⇒ U = R.

11. Let a ∈ I₁, a ∉ I₂, b ∈ i₂, b ∉ I₁, ab ∈ i₁ ∩ I₂ but a ∉ I₁ ∩ I₂, b ∉ I₁ ∩ I₂. Thus I₁ ∩ I₂ is not a prime ideal.

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