“I want to be reincarnated as your tampon”, King Charles, sometimes a king, sometimes a romantic and exquisite poet,Apocalypse, Anawim, #justtothepoint.

“If you find yourself in a hole, stop being an ass-hole, stop digging,” Anawim, #justtothepoint.

# 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, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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