“If you are going through hell, be tough and keep going, but if you are in a big, deep, ugly shithole, stop digging,” Matthew retorted, but what he meant, not even he knew for sure, but it sounded cool and super positive though, Apocalypse, Anawim, #justtothepoint.
A ring homomorphism Φ from a ring R to a ring S is a structure-preserving function or mapping between two rings. More explicitly, it is a function Φ: R → S such that Φ is addition, multiplication, and unit (multiplicative identity) preserving, Φ(a + b) = Φ(a) + Φ(b), Φ(ab) = Φ(a)Φ(b), Φ(1_{R}) = 1_{S}, ∀a, b ∈ R. Figure 1.a.
💣In our original definition, a ring does no need to have a multiplicative identity, other authors define a ring to have a multiplicative identity and without the requirement for a multiplicative identity is instead called a rng, a non-unital ring or pseudoring. Φ(1_{R}) = 1_{S} only applies when a ring do have a multiplicative identity.
Theorem. There is a unique or canonical ring homomorphism from the ring of integers to any ring. Let R be a ring with unity or multiplication identity 1 (or just a ring depending on your definition, ring versus rng - a non-unital ring or pseudoring-).
Proof.
Theorem. Every ring with unity has a subring isomorphic to either ℤ or ℤ_{n}. Let R be a ring (or a ring with unity depending on your definition). If Char(R) = n, then R contains a subring isomorphic to ℤ_{n}. If Char(R) = 0, then R contains a subring isomorphic to ℤ.
Proof.
Let 1 be the unity of R. The previous result shows that Φ: ℤ → R, Φ(n) = n·1_{R} is a homomorphism ⇒ [By the First Isomorphic Theorem for Rings] ℤ/Ker(Φ) ≋ Φ(ℤ).
Recall Let Φ: R → R' be a ring homomorphism and S be a subring of R ⇒ Φ(S) is a subring of R'. In particular, Φ(R) is a subring of R’.
Ker(Φ) = ⟨n⟩ where n is the additive order of 1_{R}, that is, Φ(n) = n·1_{R} = 1_{R} + 1_{R} + ··_{n times}·· + 1_{R} = 0_{R} ⇒ If Char(R) = n, Φ(ℤ) ≋ ℤ/⟨n⟩ ≋ ℤ_{n}, hence R contains a subring, namely Φ(ℤ), isomorphic to ℤ_{n}. Otherwise, if Char(R) = 0, Φ(ℤ) ≋ ℤ/⟨0⟩ ≋ ℤ.
Recall, by Characteristic of a Ring with Unity. If 1 has infinite order under addition, then char(R) = 0. If 1 has order n under addition, then char(R) = n.
Φ(0) =[Φ is a ring homomorphism] 0.
0 = Φ(n + (-n)) = Φ(n) + Φ(-n) = n + Φ(-n) ⇒ n + Φ(-n) = 0 ⇒ Φ(-n) = -n.
1 = Φ(1) = Φ(1/n + ··_{n times}·· + 1/n) = Φ(1/n) + ··_{n times}·· + Φ(1/n) = nΦ(1/n) ⇒ nΦ(1/n) = 1 ⇒ Φ(1/n) = 1/n.
∀m/n ∈ ℚ, Φ(m/n) = Φ(m·1/n) = Φ(m)Φ(1/n) = m·1/n = m/n, Φ is the identity. Therefore, a ring homomorphism over rational numbers is either the trivial homomorphism or the identity.
Let’s set Φ(1) = a ⇒ a^{2} = a·a = Φ(1)·Φ(1) = Φ(1·1) = Φ(1) = a ⇒ a^{2} = a ⇒[An idempotent element of a ring is an element a such that a^{2} = a, e.g., 0, 1, 3, and 4 are idempotent elements of ℤ/6ℤ. A trivial idempotent refers to either of the elements 0 and 1, which are always idempotent.] 0, 1, 8, and 21 are idempotent elements of ℤ_{28}, so a = Φ(1) = 0, 1, 8, or 21.
Recall (group abstract algebra and group homomorphisms) that |Φ(1)| | |1| and we are talking about the additive order (inside the additive group) of Φ(1) -ℤ_{28}- and 1 -ℤ_{12}- respectively ⇒ |Φ(1)| = |a| | 12 (= |1| in ℤ_{12}) ⇒ |a| ∈ {1, 2, 3, 4, 6, 12}.
Besides, recall (cyclic groups) that in ℤ_{n}, |k| = $\frac{n}{gcd(n, k)}$, and therefore n = 28, (a = 0, 1, 8 and 21), |0|_{ℤ28} = 1, |1|_{ℤ28} = 28 ∉ {1, 2, 3, 4, 6, 12}, |8|_{ℤ28} = $\frac{n}{gcd(n, k)}= \frac{28}{gcd(28, 8)} = \frac{28}{4} = 7$ ∉ {1, 2, 3, 4, 6, 12}, and finally |21|_{ℤ28} = $\frac{n}{gcd(n, k)}= \frac{28}{gcd(28, 21)} = \frac{28}{7} = 4$, so the only possibilities are a = 0 or 21.
Corollary. ℤ_{n} is a homomorphic image of ℤ.
Proof.
There are two different approaches.
If Φ: R → S ring homomorphism is also onto, then Φ is called an epimorphism and S is called a homomorphic image of R.
Corollary. Every field has a subfield isomorphic either to ℤ_{p} or to ℚ. Let F be a field. If Char(F) = p, then F contains a subfield isomorphic to ℤ_{p}. If Char(F) = 0, then F contains a subfield isomorphic to ℚ.
Proof:
Recall that a field is a commutative ring with unity such that each nonzero element has a multiplicative inverse, e.g., every finite integral domain, ℤ_{p} (p prime), ℚ, ℝ, and ℂ. A field has characteristic zero or characteristic p with p prime.
By the previous result, if Char(F)=p, then F contains a subring isomorphic to ℤ_{p} and we are done.
If Char(F) = 0, then F has a subring S isomorphic to ℤ. Φ: ℤ → F, Φ(n) = n·1_{F}, Φ is a ring homomorphism, and by the first isomorphism theorem for rings, ℤ/⟨0⟩ ≋ ℤ ≋ Φ(ℤ), i.e., S = Φ(ℤ) is a subring of F isomorphic to ℤ. Therefore, S = {n·1_{F}: n ∈ ℤ} ⊆_{subring of} F.
Let T =[Since a field must have all multiplicative inverses of n·1_{F}, n ≠ 0] {ab^{-1} | a, b ∈ S, b ≠ 0}. Claim: T is a subfield of F (it is left as an exercise) and T ≋ ℚ.
Recall that a field F is a commutative (the multiplication is commutative) division ring.
Let Φ be the mapping: T → ℚ, Φ(ab^{-1}) = a/b. More precisely, it would be (m·1_{F})(n·1_{F})^{-1} → m/n.
Φ is well defined. Similar argument Φ is injective.
Φ is a ring homomorphism. Φ(ab^{-1} + cd^{-1}) =[Since (F,·) is commutative, ab^{-1} + cd^{-1} = adb^{-1}d^{-1} + bcb^{-1}d^{-1}] Φ((ad + bc)(bd)^{-1}) = (ad + bc)/bd = ad/bd + bc/bd = a/b + c/d = Φ(ab^{-1}) + Φ(cd^{-1}).
Φ((ab^{-1})(cd^{-1})) =[Since (F,·) is commutative] Φ(acb^{-1}d^{-1}) = Φ((ac)(bd)^{-1}) = ac/bd = (a/b)(c/d) = Φ(ab^{-1})Φ(cd^{-1})
Φ is injective. Suppose Φ(ab^{-1}) = Φ(cd^{-1}) ⇒ a/b = c/d ⇒ ad = cb ⇒ adb^{-1} = cbb^{-1} ⇒[Since (F,·) is commutative] ab^{-1}d = c ⇒ ab^{-1} = cd^{-1}.
By construction, Φ is onto ∎
Since the intersection of all subfields of a field is a subfield, too, every field has a smallest subfield called the prime subfield of the field.
If Char(F)=p, the prime subfield is isomorphic to ℤ_{p}. Otherwise, if Char(F) = 0, the prime subfield is isomorphic to ℚ.
Third ring isomorphism theorem. Let R be a ring with ideals I ⊆ J ⊆ R. Then, $\frac{R/I}{J/I} ≋ R/J$.
Proof:
Let’s define a map Φ: R/I → R/J given by Φ(r + I) = r + J.