# Ideals. Exercises about rings

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# Recall

💍 A ring R is a non-empty set with two binary operations, addition (a + b) and multiplication (ab), such that ∀ a, b, c ∈ R:

1. Both operations are closed: a + b ∈ R, a·b ∈ R.
2. Commutative under addition: a + b = b + a.
3. Associative under addition: (a + b) + c = a + (b + c).
4. There is an additive identity 0 ∈ R such that a + 0 = a, ∀ a ∈ R.
5. There are inverse elements for addition,, ∃-a ∈ R: a + (-a) = (-a) + a = 0.
6. Associative under product: a(bc) = (ab)c.
7. Multiplication is distributive over addition: a(bc) = ab + ac, (b + c)a = ba + ca.

A subring A of a ring R is called a (two-sided) ideal of R if it absorbs multiplication from the left and right from R, that is, ∀r ∈ R, a ∈ A, ra ∈ A and ar ∈ A. In other words, ∀r ∈ R, rA = {ra | a ∈ A} ⊆ A and Ar = {ar | a ∈ A} ⊆ A.

Let A ⊂ R, A ≠ R, R be a ring, A is an ideal of R if the following conditions are satisfied:

1. a - b ∈ A, ∀a, b ∈ A.
2. ∀r ∈ R, ∀a ∈ A, ra ∈ A and ar ∈ A.

# Exercises

• Let Z[i] = {a + bi|a, b ∈ ℤ} be the Gaussian integers. I = ⟨2 + 2i⟩ ⊆ ℤ[i] is not a prime ideal.

2·(1 + i) = (2 + 2i) ∈ I. If 2 ∈ I = ⟨2 + 2i⟩, ∃a, b ∈ ℤ such that 2 = (2 +2i)(a + bi) = (2a -2b) + (2a +2b)i ⇒ 2 = 2a -2b, 0 = 2a + 2b ⇒[0 = a + b ⇒ a = -b] 2 = 2a -2b = -4b ⇒ b = 2/-4 = -1/2, b ∉ ℤ ⊥ Similarly, 1 + i ∉ I.

• Let R be a ring with unity. If u ∈ R is a unit, then so is -u.

Recall that we have already shown that (-1)2 = (-1)(-1) = 1 and let’s prove that -a = (-1)a = a(-1).

a + (-1)a = 1a + (-1)a =[Distributivity] (1 + (-1))a = 0·a = 0 ⇒ -a = (-1)a
Futhermore, a + a(-1) = a·1 + a(-1) = a(1 + (-1)) = a·0 = 0 ⇒ -a = a(-1).

Suppose u ∈ R is a unit ⇒ ∃u-1 ∈ R, u·u-1 = 1 ⇒ (-u)(-u-1) =[-a = (-1)a = a(-1)] u(-1)(-1)u-1 =[(-1)2=1] u·1·u-1 = u·u-1 = 1. So we have demonstrated that (-u)(-u-1) = 1, hence (-u)-1 = -u-1 and hence -u is a unit ∎

• Let R be a commutative ring with additive identity 0. An element a ∈ R is nilpotent if xn = 0 for some n ∈ ℕ, n > 0, e.g., 2 ∈ ℤ4 is nilpotent since 22 = 4 ≡ 0 (mod 4).
1. If a and b are nilpotent, a + b is nilpotent. ∃n, m ∈ ℕ: an = 0, bm = 0. Let t = n*m + 1 ⇒[By the Binomial Theorem] (a + b)t = ${t \choose 0}a^tb^0 + {t \choose 1}a^{t-1}b^1 +···+ {t \choose t}a^0b^t$ where each term is of the form ${t \choose k}a^{t-k}b^k$, where t -k ≥ n (and therefore at-k = 0) or k ≥ m (and therefore bk = 0), that is, equal to zero, so (a + b)t = 0 and (a + b) is nilpotent. Suppose for the sake of contradiction t - k < n and k < m ⇒ t = t -k + k < n + m < n * m + 1 = t ⊥
2. If a is nilpotent, then ∀r ∈ R, r·a and a·r are nilpotent. Suppose a is nilpotent, then ∃n ∈ ℕ: an = r. (r·a)n =[R is a commutative ring] rn·an = rn·0 = 0, (a·r)n =[R is a commutative ring] an·rn = 0·rn = 0 ⇒ r·a and a·r are nilpotent.
3. If a is nilpotent, then a = 0 or a is a zero divisor. If a = 0, we are done. Suppose a ≠ 0, and let n ∈ ℕ be the smallest natural number such that an = 0. Please notice that n > 1, otherwise a1 = a = 0 ⊥ ⇒ (an-1)a = 0, but a ≠ 0 and an-1 ≠ 0 (n > 1 ⇒ n -1 > 0, an-1≠0 by the minimality of n), hence a is a zero divisor.
4. If R is a ring with unity, a is nilpotent, then 1 + a is a unit. Let n ∈ ℕ be the smallest natural number such that an = 0 ⇒ 1 = 1 + an = (1 + a)(1 -a + a2 -a3··· (-1)n-1an-1) ⇒ Therefore, 1 + a is a unit with multiplicative inverse (1 + a)-1 = (1 -a + a2 -a3··· (-1)n-1an-1).

Definition. Let R be a commutative ring. Given any subset A ⊆ R, the annihilator of A is defined as Ann(A) = {r ∈ R| ∀a ∈ A, r·a = 0}.

Proposition. The annihilator of A is an ideal

1. Subring. ∀r, s ∈ Ann(A), ∀a ∈ A, (r-s)a =[Distributivity] ra - sa =[r, s ∈ Ann(A)] 0 - 0 = 0 ⇒ r - s ∈ Ann(A). (rs)a =[Associativity] r(sa) =[s ∈ Ann(A)] r·0 = 0 ⇒ rs ∈ Ann(A).
2. Absorption. ∀b ∈ R, ∀r ∈ Ann(A), rb ∈ Ann(A)? ∀a ∈ A, (rb)a =[By assumption, R is a commutativity ring] (br)a =[Associativity] b(ra) =[r ∈ Ann(A)] b·0 = 0 ⇒ rb ∈ Ann(A).

Definition. Let R be a commutative ring. Given any ideal I ⊆ R, the radical of I is defined as rad(I) = $\sqrt{I}$ = {r ∈ R | ∃n ∈ ℕ: rn ∈ I}. In particular, the nilradical is the radical of the trivial ideal, NR = Nil({0}) = {r ∈ R | ∃n ∈ ℕ: rn = 0}, e.g., N12 = {r ∈ ℤ12 | ∃n ∈ ℕ: rn = 0} = {0, 6} = ⟨6⟩. The nilradical of R is the intersection of all prime ideals of R (Pr∞fWiki, Equivalence of Definitions of Nilradical of Ring).

Examples: Let R = ℤ27, rad(⟨3⟩) = {a ∈ ℤ27| ∃n ∈ ℕ: an ∈ ⟨3⟩} = ⟨3⟩, rad(⟨9⟩) = {a ∈ ℤ27| ∃n ∈ ℕ: an ∈ ⟨9⟩} = ⟨3⟩, N27 = ⟨3⟩. Let R = ℤ36, rad(⟨4⟩) = {a ∈ ℤ36| ∃n ∈ ℕ: an ∈ ⟨4⟩} = ⟨2⟩, rad(⟨6⟩) = {a ∈ ℤ36| ∃n ∈ ℕ: an ∈ ⟨6⟩} = ⟨6⟩, N36 = ⟨6⟩. More generally (it is left as an exercise), in ℤn, rad(⟨m⟩) = ⟨rad(gcd(m, n)⟩ where rad(⟨n⟩) = $\prod_{p|n, p~ prime} p$

Proposition. The radical of I is an ideal

1. Subring. ∀r, s ∈ $\sqrt{I}$ ⇒ ∃m, n ∈ ℤ: rm ∈ I, sm ∈ I, (r -s)mn = $\sum_{k=0}^{mn} (-1)^k{mn \choose k}r^{mn-k}s^k$ [mn -k ≥ max(m, n) or k ≥ max(m, n) and therefore rmn-k ∈ I or sk ∈ I, hence by the absorption property of the ideal, every term will an element of the ideal I] ∈ I ⇒ r - s ∈ $\sqrt{I}$. (rs)mn = (rm)n(sn)m ∈ I because both terms belong to I, and therefore rs ∈ $\sqrt{I}$.

Suppose for the sake of contradiction, mn -k < max(m, n) or k < max(m, n), then mn = mn -k + k < 2·max(m, n), i.e., m·n < 2·max(m, n) ⊥

2. Absorption. ∀r ∈ $\sqrt{I}$, ∃n ∈ ℤ, rn ∈ I, ∀a ∈ R, (ar)n =[R is a commutative ring] anrn ∈ [rn ∈ I and by the absorption property of the ideal] I, and therefore ar ∈ $\sqrt{I}$.

Remark: If ⟨n⟩ is a principal ideal of the ring ℤ, n > 1, such that n = p1k1 p2k2··· prkr, where p1, p2, ···, pr are distinct prime numbers, then $\sqrt{⟨n⟩} = ⟨p_1·p_2···p_r⟩$. In ℤ, $\sqrt{⟨12⟩}$ = {r ∈ ℤ | ∃n ∈ ℕ: rn ∈ ⟨12⟩} =[12 = 22·3] ⟨2·3⟩ = 6, $\sqrt{⟨36⟩}$ =[36 = 22·32] ⟨2·3⟩ = 6, $\sqrt{⟨8⟩}$ =[36 = 23] ⟨2⟩, and $\sqrt{⟨360⟩}$ =[360 = 23325] ⟨2·3·5⟩ = ⟨30⟩.

• Find all maximal ideals of ℤ8 ⊕ ℤ30 and ℤ15 ⊕ ℤ12.

Recall:

1. The ideal of ℤn are additive subgroups of ℤn, hence they all have the form ⟨d⟩, where d divides n (d | n). Besides, ⟨d⟩ is the ideal generated by d ⇒ the ideals of ℤn are ⟨d⟩, where d divides n (d | n).
2. n is a field if and only if n is prime.
3. Theorem, let R be a commutative ring with unity, A ideal ⇒ R/A is a field ↭ A is maximal
4. ℤ/nℤ ≋ ℤn, hence we can conclude that ⟨n⟩ = nℤ is a maximal ideal precisely when n is prime.

Maximal Ideals of R ⊕ S are either of the form A ⊕ S, where A is maximal in R, or of the form R ⊕ B, where B is maximal in S. First, consider that the ideals of R ⊕ S are all of the form I1 ⊕ I2 where I1 ⊆ R and I2 ⊆ S are ideals of R and S respectively.

Suppose for the sake of contradiction, U is a maximal ideal such that both I1 and I2 are different from R and S respectively, then U ⊂ R ⊕ I2 and U ⊂ I1 ⊕ S are both ideals properly containing U, and therefore U is not a maximal ideal ⊥

The obvious choices are to take a maximal ideal of one component and direct product it with the other component. The only maximal ideal of ℤ8 is ⟨2⟩ -2|8, 2 is prime- and ℤ30 has three maximal ideals, namely ⟨2⟩, ⟨3⟩, and ⟨5⟩ -2|30, 3|30, 5|30, 2, 3 and 5 are prime numbers-. Combined, that gives us four maximal ideals: ⟨2⟩ ⊕ ℤ30, ℤ8 ⊕ ⟨2⟩, ℤ8 ⊕ ⟨3⟩, and ℤ8 ⊕ ⟨5⟩.

Similarly, the maximal ideals of ℤ15 ⊕ ℤ12 are ⟨3⟩ ⊕ ℤ12, ⟨5⟩ ⊕ ℤ12, ℤ15 ⊕ ⟨2⟩, and ℤ12 ⊕ ⟨3⟩.

• Let R be the ring of all continuous functions from ℝ to ℝ.
1. Is the ideal I = {f ∈ ℝ | f(0) = f(1) = 0} a prime ideal? Recall, prime ideals, ∀f, g ∈ R, f·g ∈ I ⇒ f ∈ I or g ∈ I. Consider f(x) = x, g(x) = 1 -x, f·g ∈ I (f·g(0) = f·g(1) = 0), but f ∉ I and g ∉ I.
2. Is the ideal I = {f ∈ ℝ2 | f(1, 0) = f(0, 1) = 0} a prime ideal? Consider f(x, y) = x -1, g(x, y) = y -1, f·g ∈ I (fg(1, 0) = fg(0, 1) = 0), but f ∉ I and g ∉ I.

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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