# The Fundamental Theorem of Ring Homomorphisms

On the Quotient Rings page said that if $(R, +, *)$ is a ring and $(I, +, *)$ is an ideal, then the quotient of $R$ by $I$ is the set $R / I$ of equivalence classes $a + I = \{ a + i : i \in I \}$ with the operations of addition and multiplication defined for all $(a + I), (b + I) \in R / I$ by:

(1)So in order to define a quotient ring we need a ring $(R, +, *)$ and an ideal $(I, +, *)$. Now suppose that $R$ and $S$ are homomorphic rings with homomorphism $\phi : R \to S$. Then $\ker (\phi)$ is a subring of $R$ (this is relatively easy to show). We would ultimately like $\ker (\phi)$ to be an ideal of $R$ to define the quotient $R / \ker (\phi)$.

This is not true in general though. Recall that when we defined a ring, we required that there exists an element $1 \in R$ such that $a * 1 = a$ and $1 * a = a$ for all $a \in R$, that is, we required the existence of a multiplicative identity. However, by definition, $\ker (\phi)$ in general cannot be a subring of $R$ because if $1$ is the identity of $R$, then by definition of a ring homomorphism, $\phi(1)$ is mapped to the multiplicative identity of $S$ and not to the additive identity of $S$.

To establish a fundamental theorem of ring homomorphisms, we make a small exception in not requiring that $\ker (\phi)$ is an ideal for the quotient $R / \ker (\phi)$ to be defined.

Theorem 1 (The Fundamental Theorem of Ring Homomorphisms): Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be homomorphic rings with ring homomorphism $\phi : R \to S$. Then $R / \ker (\phi) \cong \phi (R)$. |

**Proof:**Let $K = \ker (\phi)$ and let $\psi : R / K \to \phi (R)$ be defined for all $(a + K) \in R$ by:

- Then $\psi$ is well-defined, for if $a + K = b + K$ then $a = b + k$ for some $k \in K$ and so:

- Now let $(a + K), (b + K) \in R / K$. Then:

- We lastly show that $\psi$ is bijective.

- Let $(a + K), (b + K) \in R / K$ and suppose that $\psi (a + K) = \psi (b + K)$. Then $\phi(a) = \phi(b)$. So $\phi(a - b) = 0$. So $a - b \in K$. So $a + K = b + K$. Hence $\psi$ is injective.

- Furthermore, for all $a \in \phi (R)$ we have that $(a + I) \in R / K$ is such that $\psi (a + K) = a$. So $\psi$ is surjective.

- Hence $\psi$ is bijective and so $\psi$ is an isomorphism from $R / K$ to $\phi (R)$, that is:

For example, consider the rings $\mathbb{Z}$ and $\mathbb{Z}_n$. We can define a homomorphism $\phi : \mathbb{Z} \to \mathbb{Z}_n$ for all $z \in \mathbb{Z}$ by:

(8)Indeed $\phi$ is a homomorphism, as you should verify. Note that:

(9)So by the fundamental theorem of ring homomorphisms we have that:

(10)