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💍 A ring R is a non-empty set with two binary operations, addition (a + b) and multiplication (ab), such that ∀ a, b, c ∈ R:
Let R be a ring and let A ⊆ R be a subring. In a ring, cosets are always respect to addition, r ∈ R, r + A = {r + a | a ∈ A}.
Notice that since (R, +) is an Abelian group, if A is a subgroup under addition ⇒ A is an Abelian subgroup under addition, and therefore, A is a normal subgroup of R, i.e., A ◁ R and we can form a quotient group (R/A, +). The quotient group is an Abelian group with respect to addition.
What condition do we need to satisfy on the subring A so that R/A = {r + A | r ∈ R} is a ring?
Basically, given a, b ∈ R, we need (a + A)(b + A) = ab + A to be well defined, that is, this operation does not depend on the chosen representatives.
Let a + S = c + S and b + S = d + S, (a + S)(b + S) = ab + S = cd + S or equivalently ab -cd ∈ S?
a + S = c + S ⇒ a -c ∈ S ⇒ ∃s_{1} ∈ S, a - c = s_{1} ⇒ a = c + s_{1}. Similarly, ∃s_{2} ∈ S, b = d + s_{2}.
ab = (c + s_{1})(d + s_{2}) = cd + cs_{2} +s_{1}d + s_{1}s_{2} ⇒ ab -cd = cs_{2} +s_{1}d + s_{1}s_{2}. So we need ab -cd ∈ S ↭ cs_{2} +s_{1}d + s_{1}s_{2} ∈ S ↭[s_{1}s_{2} ∈ S, S is a subring] cs_{2}, s_{1}d ∈ S. This is called the absorption property.
A subring A of a ring R is a left ideal of R if it absorbs multiplication from the left from R, that is, ∀r ∈ R, ∀a ∈ A, ra ∈ A.
A subring A of a ring R is a right ideal of R if it absorbs multiplication from the right from R, that is, ∀r ∈ R, ∀a ∈ A, ar ∈ A.
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:
Let R be a commutative ring with unity. A principal ideal is an ideal generated by a single element a of R through multiplication by every element of R, ⟨a⟩ = Ra = {ra | r ∈ R} = {ar | r ∈ R}., e.g., ⟨x⟩ = ⟨x, x^{2}⟩ is a principal ideal of ℤ[x], but ⟨x, 3⟩ is not.
Proof. ⟨a⟩ is an ideal.
∀i ∈ ⟨a⟩ ⇒ ∃r’ ∈ R such that i = r’a. ∀r ∈ R, ir = (r’a)r = [R is a commutative ring] (ar’)r = a(r’r) ∈ ⟨a⟩.
Remark. If R is non-commutative or does not contain 1, the situation is most complicated, the closest concept would be $\bigcap_{I~ ideal, a ∈ I} I$.
Suppose m ∈ ℤ, a ∈ nℤ ⇒ ∃a’ ∈ ℤ: a = na’ ⇒ ma = m(na’) = n(ma’) ∈ nℤ ∎
Therefore, the principal ideals in ℤ are of the form ⟨n⟩ = nℤ.
Take f(x) ∈ ℤ[x], g(x) ∈ I, deg(f(x)·g(x)) = deg(f(x)) + deg(g(x)) ≥ deg(g(x)) ≥ 2 ⇒ deg(f(x) + g(x)) ≥ 2 ⇒ f(x) + g(x) ∈ I ∎
For R = ℤ, the ideal ⟨4, 6⟩ = {4a + 6b | a, b ∈ ℤ} = {2a | a ∈ ℤ} = ⟨2⟩
It’s not an ideal because it’s not true that ri ∈ I for every r ∈ R and i ∈ S. Consider $r(x) = \begin{cases} 1, &x ∈ ℚ \\ -1, &x ∉ ℚ \end{cases}$ and i = e^{x}, $ri(x) = \begin{cases} e^x, &x ∈ ℚ \\ -e^x, &x ∉ ℚ \end{cases}$ is not differentiable. In fact it’s not even continuous ⇒ ri ∉ I ⇒ I is not an ideal of R. However, it’s a subring since it’s nonempty and the product and difference of differentiable functions are differentiable functions, too.
Proposition. Suppose R is a commutative ring with unity. ⟨a⟩ = R ↭ a ∈ R is a unit.
Proof:
⇒) Suppose that ⟨a⟩ = R ⇒[By assumption, R is a commutative ring with unity] 1 ∈ ⟨a⟩ ⇒ ∃r∈R: ar = 1 ⇒ a is a unit (it has a multiplicative inverse, namely r).
⇐) Suppose a ∈ R is a unit ⇒ ∃r∈R: ar = 1 ⇒ 1 ∈ ⟨a⟩ ⇒ ∀b ∈ R, b = b·1 ∈ ⟨a⟩ ⇒ R = ⟨a⟩.
Corollary. The only ideals of a field F are the zero ideal and the unit ideal.
Proof.
Let I be an ideal of F. If I is the zero ideal, we are done.
Suppose I is a non-zero ideal of F ⇒ ∃a ∈ I, a ≠ 0 (the zero identity -addition- of F) ⇒[F is field and in a field, all non-zero elements are units] a is a unit ⇒[Proposition. Suppose R is a commutative ring with unity. ⟨a⟩ = R ↭ a ∈ R is a unit.] I = F, i.e., I is the unit ideal.
Proposition. Every ideal of ℤ is a principal ideal.
Proof: Suppose that I ⊆ ℤ is an ideal of ℤ.
Let R be a ring and let I be an ideal of R. In particular, R is an Abelian group under addition, and I is a normal subgroup of R, I ◁ R, therefore we could form the factor group R/I = {r + I | r ∈ R}.
As in the groups case the cosets partition the ring into disjoint subsets. The element r is called the coset representative of the coset. Besides, r_{1} and r_{2} represent the same coset ↭ r_{1} - r_{2} ∈ I. The ideal I itself is a coset, 0 + I.
Theorem. Let R be a ring and let I be a subring of R. The set of cosets R/I = {r + I | r ∈ R} is a ring under the operations (s + I) + (t + I) = (s + t) + I and (s + I)(t + I) = st + I iff I is an ideal of R.
Proof.
⇐) We only need to check that these operations are well-defined. Suppose that I is an ideal.
Let’s suppose s + I = s’ + I, t + I = t’ + I ⇒[s - s’ ∈ I, t - t’ ∈ I] s = s’ + a, t = t’ + a’, where a, a’ ∈ I.
(s - s’) + (t - t’) = (s + t) - (s’ + t’) ∈ I (I is closed under addition and s - s’ ∈ I, t - t’ ∈ I) ⇒[aH = bH ↭ a^{-1}b ∈H ↭_{Additive notation} a-b ∈ H] (s + t) + I = (s’ + t’) + I.
st + I = (s’+a)(t’ + a’) + I = s’t’ + at’ + s’a’ + aa’ + I =[I is an ideal, a, a’ ∈ I, and therefore it absorbs at’ + s’a’ + aa’] s’t’ + I , hence st + I = s’t’ + I. We already know that the set of cosets forms a group under addition and it is trivial to prove that the multiplication is associative and distributive over addition.
⇒) Suppose for the sake of contradiction, R/I is a ring and yet I is not an ideal of R ⇒ ∃a ∈ I, r ∈ R: ar ∉ I or ra ∉ I. Without loss of generality, let’s suppose that ar ∉ I, a ∈ I ⇒[a + H = H = 0 + H ↭ a ∈ H] a + I = 0 + I.
(a + I)(r + I) = ar + I, (0 + I)(r + I) = 0r + I = 0 + I = I. Then, ar + I = I ⊥ ar ∉ I, so the multiplication is not well-defined.
(2 + 4ℤ) + (3 + 4ℤ) = 5 + 4ℤ = 1 + 4 + 4ℤ =[aH = H ↭ a ∈ H. In addition notation, a + H = H ↭ a ∈ H] 1 + 4ℤ. (2 + 4ℤ)(3 + 4ℤ) = 6 + 4ℤ = 2 + 4 + 4ℤ = 2 + 4ℤ.
0 + ⟨3⟩ = {0, 3} = [0]
1 + ⟨3⟩ = {1, 4} = [1]
2 + ⟨3⟩ = {2, 5} = [2]
(1 + ⟨3⟩) + (1 + ⟨3⟩) = (1 + 1) + ⟨3⟩ = 2 + ⟨3⟩
(1 + ⟨3⟩)(1 + ⟨3⟩) = (1 · 1) + ⟨3⟩ = 1 + ⟨3⟩
ℤ[x]/I where I = {a_{2}x^{2} + a_{3}x^{3}+···+a_{n}x^{n}| n ≥ 2, a_{i} ∈ ℤ}, e.g., (x^{4}+2x^{2}+4x+2) + I =[x^{4}+2x^{2} ∈ I] (4x+2) + I. In general, all cosets have the form (ax + b) + I = [Some authors will use another notation] $\overline {ax + b}$. ((ax +b) + I)((cx +d) + I) = (acx^{2} + (ad + bc)x + bd) + I = [acx^{2} ∈ I] ((ad + bc)x + bd) + I.
Let R = M_{2}(ℤ) = {$(\begin{smallmatrix}a_1 & a_2\\ a_3 & a_4\end{smallmatrix})$ | a_{i} ∈ ℤ}, and let I be the subset of R with even entries. I is an ideal of R (it is a similar reasoning behind 2ℤ being an ideal of the ring ℤ).
Subring. Let x, y ∈ I : x = $(\begin{smallmatrix}2a_1 & 2a_2\\ 2a_3 & 2a_4\end{smallmatrix})$, y = $(\begin{smallmatrix}2b_1 & 2b_2\\ 2b_3 & 2b_4\end{smallmatrix})$. Then, x - y = $(\begin{smallmatrix}2a_1 & 2a_2\\ 2a_3 & 2a_4\end{smallmatrix}) - (\begin{smallmatrix}2b_1 & 2b_2\\ 2b_3 & 2b_4\end{smallmatrix}) = (\begin{smallmatrix}2(a_1-b_1) & 2(a_2-b_2)\\ 2(a_3-b_3) & 2(a_4-b_4)\end{smallmatrix})$∈ I, x · y = $(\begin{smallmatrix}2a_1 & 2a_2\\ 2a_3 & 2a_4\end{smallmatrix}) · (\begin{smallmatrix}2b_1 & 2b_2\\ 2b_3 & 2b_4\end{smallmatrix}) = (\begin{smallmatrix}4(a_1b_1+a_2b_3) & 4(a_1b_2+a_2b_4)\\ 4(a_3b_1+a_4b_3) & 4(a_3b_2+a_4b_4)\end{smallmatrix})$∈ I.
Absorption. x ∈ I, y ∈ R, xy = $(\begin{smallmatrix}2b_1 & 2b_2\\ 2b_3 & 2b_4\end{smallmatrix})(\begin{smallmatrix}a_1 & a_2\\ a_3 & a_4\end{smallmatrix}) = (\begin{smallmatrix}2(b_1a_1 + b_2a_3) & 2(b_1a_2 + b_2a_4)\\ 2(b_3a_1 + b_4a_3) & 2(b_3a_2 + b_4a_4)\end{smallmatrix})$ ∈ I. Analogously, yx ∈ I (this step is absolutely necessarily given that R is a non-commutative ring), therefore I is indeed an ideal of R.
Futhermore, every member of R can be written in the form $(\begin{smallmatrix}2q_1 +r_1 & 2q_2 + r_2\\ 2q_3 + r_3 & 2q_4 + r_4\end{smallmatrix})$. Therefore, $(\begin{smallmatrix}2q_1 +r_1 & 2q_2 + r_2\\ 2q_3 + r_3 & 2q_4 + r_4\end{smallmatrix})+I = (\begin{smallmatrix}r_1 & r_2\\ r_3 & r_4\end{smallmatrix})+(\begin{smallmatrix}2q_1 & 2q_2\\ 2q_3 & 2q_4\end{smallmatrix})+I = (\begin{smallmatrix}r_1 & r_2\\ r_3 & r_4\end{smallmatrix})+ I$ since an ideal absorbs its own elements.
In other words, R/I = {$(\begin{smallmatrix}r_1 & r_2\\ r_3 & r_4\end{smallmatrix}) + I|r_i∈${0, 1}}, |R/I| = 2^{4} = 16, e.g., $(\begin{smallmatrix}3 & 6\\ 5 & -3\end{smallmatrix})+I = (\begin{smallmatrix}1 & 0\\ 1 & 1\end{smallmatrix}) + (\begin{smallmatrix}2 & 6\\ 4 & -4\end{smallmatrix}) + I = (\begin{smallmatrix}1 & 0\\ 1 & 1\end{smallmatrix}) + I$ since an ideal absorbs its own elements as it was stated previously.
Proposition. Given a ring R with ideals I, J ⊆ R, I + J = {i + j | i ∈ I, j ∈ J}, IJ = {i_{1}j_{1} + ··· + i_{n}j_{n} | i_{k} ∈ I, j_{k} ∈ J, ∀k: 1 ≤ k ≤ n}, I ∩ J, I + J, IJ are all ideals.
Proof.
It is left as an exercise to prove that I ∩ J, I + J, IJ are all subrings.
∀r ∈ R, a ∈ I ∩ J ⇒[a∈ I, a∈ J, I and J ideals] ra ∈ I, ar ∈ I, ra ∈ J, ar ∈ J ⇒ ra ∈ I ∩ J and ar ∈ I ∩ J ⇒ I ∩ J is an ideal of R.
∀r ∈ R, a ∈ I + J ⇒ a = i + j, i ∈ I, j ∈ J ⇒ ra = r(i + j) =[Distributivity] ri + rj ∈ I + J, ri ∈ I, rj ∈ J. Mutatis mutadis, ar ∈ I + J, and hence I + J is an ideal.
∀r ∈ R, a ∈ IJ ⇒ a = i_{1}j_{1} + ··· + i_{n}j_{n}, i_{k} ∈ I, j_{k} ∈ J, ∀k: 1 ≤ k ≤ n ⇒ ra = r(i_{1}j_{1} + ··· + i_{n}j_{n}) =[Distributivity] r(i_{1}j_{1}) + ··· + r(i_{n}j_{n}) =[Associativity] (ri_{1})j_{1} + ··· + (ri_{n})j_{n} ∈ IJ because ri_{k} ∈ I (By assumption, I is a ring), j_{k} ∈ J, ∀k: 1 ≤ k ≤ n. Mutatis mutandis, ar ∈ IJ.
Proposition. Let m, n ∈ ℕ, l = lcm(m, n), d = gcd(m, n). Then, mℤ ∩ nℤ = lℤ, mℤ + nℤ = dℤ ↭ mℤ ∩ nℤ = ⟨m⟩ ∩ ⟨n⟩ = ⟨l⟩, ⟨m⟩ + ⟨n⟩ = ⟨d⟩.
Proof.
Suppose, a ∈ mℤ ∩ nℤ ⇒ a ∈ mℤ and a ∈ nℤ ⇒ ∃x, y ∈ ℤ such that a = mx = ny ⇒ m | a and n | a ⇒[l = lcm(m, n)] l | a ⇒ ∃z ∈ ℤ, a = lz ⇒ a ∈ ⟨l⟩ = lℤ ⇒ mℤ ∩ nℤ ⊆ lℤ.
Suppose, a ∈ lℤ ⇒ ∃x ∈ ℤ, a = lx ⇒[Since l = lcm(m, n), ∃b, c ∈ ℤ: l = mb = nc] a = (mb)x = (nc)x ⇒[Associativity] a = m(bx) = n(cx) ∈ mℤ ∩ nℤ ⇒[We have already proved that mℤ ∩ nℤ ⊆ lℤ] mℤ ∩ nℤ = lℤ.
Suppose a ∈ mℤ + nℤ ⇒ ∃x, y ∈ ℤ such that a = mx + ny ⇒[d = gcd(m, n) ⇒∃z, w ∈ ℤ: m = dz, n = dw] a = (dz)x + (dw)y =[Associativity] d(zx) + d(wy) =[Distributivity] d(zx + wy) ∈ ⟨d⟩ = dℤ ⇒ mℤ + nℤ ⊆ dℤ.
Suppose a ∈ dℤ ⇒ ∃t ∈ ℤ: a = dt ⇒[d = gcd(m, n) ⇒∃x, y ∈ ℤ: d = mx + ny] a = (mx + ny)t =[Distributivity] (mx)t + (ny)t =[Associativity] m(xt) + n(yt) ∈ mℤ + nℤ ⇒[We have already proved that mℤ + nℤ ⊆ dℤ] mℤ + nℤ = dℤ ∎
Proposition. Let R be an integral domain. ⟨a⟩ = ⟨b⟩ ↭ ∃u ∈ R, u unit such that a = bu.
Proof.
⇒) Suppose ⟨a⟩ = ⟨b⟩ ⇒[R is an integral domain, i.e., a commutativity ring with unity ⇒ 1 ∈ R] a = 1·a ∈ ⟨a⟩ ⇒[By assumption, ⟨a⟩ = ⟨b⟩, hence a ∈ ⟨b⟩] ∃r ∈ R: a = br.
Similarly b ∈ ⟨b⟩ = ⟨a⟩, ∃s ∈ S: b = as ⇒ a = br = (as)r =[Associativity] a(sr) ⇒ a = a(sr) ⇒[Cancellation laws apply in integral domains] sr = 1 ⇒ s and r are units and a = br.
⇐) Suppose ∃u ∈ R, u unit such that a = bu.
∀r ∈ ⟨a⟩ ⇒ ∃x ∈ R: r = ax ⇒[a = bu] r = (bu)x =[Associativity] b(ux) ∈ ⟨b⟩ ⇒ ⟨a⟩ ⊆ ⟨b⟩.
a = bu ⇒[u is a unit ⇒ ∃u^{-1} ∈ R such that uu^{-1} = u^{-1}u = 1] b = au^{-1}. ∀s ∈ ⟨b⟩ ⇒ ∃y ∈ R: s = by ⇒[b = au^{-1}] s = (au^{-1})b =[Associativity] a(u^{-1}b) ∈ ⟨a⟩ ⇒ ⟨b⟩ ⊆ ⟨a⟩ ⇒ ⟨a⟩ = ⟨b⟩ ∎