Proof of the First Isomorphism Theorem for Rings

The first isomorphism theorem states that every ring homomorphism factors through its kernel, providing the canonical decomposition of homomorphisms into surjections, isomorphisms, and inclusions.

Statement

Theorem4.6First isomorphism theorem for rings

Let $\varphi: R \to S$ be a ring homomorphism. Then $\ker(\varphi)$ is an ideal of $R$ and $R/\ker(\varphi) \cong \mathrm{im}(\varphi)$ . The isomorphism is given by $\bar{\varphi}(r + \ker(\varphi)) = \varphi(r)$ .

Proof

Step 1: $\ker(\varphi)$ is an ideal.

Let $I = \ker(\varphi) = \{r \in R : \varphi(r) = 0\}$ .

$I$ is an additive subgroup: $\varphi(0) = 0$ so $0 \in I$ ; if $a, b \in I$ then $\varphi(a - b) = \varphi(a) - \varphi(b) = 0$ ; if $a \in I$ then $\varphi(-a) = -\varphi(a) = 0$ .
$I$ absorbs multiplication: if $a \in I$ and $r \in R$ , then $\varphi(ra) = \varphi(r)\varphi(a) = \varphi(r) \cdot 0 = 0$ , so $ra \in I$ . Similarly $ar \in I$ .

Step 2: $\bar{\varphi}$ is well-defined.

If $r + I = r' + I$ , then $r - r' \in I$ , so $\varphi(r - r') = 0$ , hence $\varphi(r) = \varphi(r')$ . Thus $\bar{\varphi}(r + I) = \varphi(r)$ is independent of the coset representative.

Step 3: $\bar{\varphi}$ is a ring homomorphism.

$\bar{\varphi}((r+I) + (s+I)) = \bar{\varphi}((r+s)+I) = \varphi(r+s) = \varphi(r) + \varphi(s) = \bar{\varphi}(r+I) + \bar{\varphi}(s+I)$ .

$\bar{\varphi}((r+I)(s+I)) = \bar{\varphi}(rs+I) = \varphi(rs) = \varphi(r)\varphi(s) = \bar{\varphi}(r+I)\bar{\varphi}(s+I)$ .

$\bar{\varphi}(1+I) = \varphi(1) = 1$ .

Step 4: $\bar{\varphi}$ is injective.

$\bar{\varphi}(r+I) = 0 \iff \varphi(r) = 0 \iff r \in I \iff r + I = 0 + I$ .

Step 5: $\bar{\varphi}$ is surjective onto $\mathrm{im}(\varphi)$ .

For any $\varphi(r) \in \mathrm{im}(\varphi)$ , we have $\bar{\varphi}(r + I) = \varphi(r)$ .

Therefore $\bar{\varphi}: R/I \to \mathrm{im}(\varphi)$ is a ring isomorphism. $\blacksquare$

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Applications

ExampleApplications of the first isomorphism theorem

Evaluation homomorphism: $\mathrm{ev}_a: k[x] \to k$ by $f \mapsto f(a)$ has kernel $(x - a)$ , giving $k[x]/(x-a) \cong k$ .
Complex numbers: $\mathrm{ev}_i: \mathbb{R}[x] \to \mathbb{C}$ by $f \mapsto f(i)$ has kernel $(x^2 + 1)$ , giving $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$ .
Finite fields: The map $\mathbb{F}_p[x] \to \mathbb{F}_{p^n}$ by $f \mapsto f(\alpha)$ (where $\alpha$ is a root of an irreducible polynomial $g$ of degree $n$ ) has kernel $(g)$ , giving $\mathbb{F}_p[x]/(g) \cong \mathbb{F}_{p^n}$ .
Coordinate rings: $k[x,y] \to k[t]$ by $x \mapsto t^2$ , $y \mapsto t^3$ has kernel $(y^2 - x^3)$ , so $k[x,y]/(y^2 - x^3) \cong k[t^2, t^3]$ .

RemarkThe factorization paradigm

The first isomorphism theorem implements the "factorization" of morphisms: $\varphi = \iota \circ \bar{\varphi} \circ \pi$ where $\pi: R \twoheadrightarrow R/I$ is the projection, $\bar{\varphi}: R/I \xrightarrow{\sim} \mathrm{im}(\varphi)$ is the isomorphism, and $\iota: \mathrm{im}(\varphi) \hookrightarrow S$ is the inclusion. This pattern appears throughout algebra (groups, rings, modules, algebras) and is the prototype for factorization systems in category theory.