The Fundamental Theorem of Ring Homomorphisms

# 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)
\begin{align} \quad (a + I) + (b + I) = (a + b) + I \end{align}
(2)
\begin{align} \quad (a + I) * (b + I) = (a * b) + I \end{align}

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:
(3)
\begin{align} \quad \psi (a + K) = \phi (a) \end{align}
• Then $\psi$ is well-defined, for if $a + K = b + K$ then $a = b + k$ for some $k \in K$ and so:
(4)
\begin{align} \quad \psi (a + K) = \psi ((b + k) + K) = \psi ((b + K) + K) = \psi (b + K) \end{align}
• Now let $(a + K), (b + K) \in R / K$. Then:
(5)
\begin{align} \quad \psi ((a + K) + (b + K)) = \psi ((a + b) + K) = \phi (a + b) = \phi(a) + \phi(b) = \psi (a + K) + \psi (b + K) \end{align}
(6)
\begin{align} \quad \psi ((a + K) * (b + K)) = \psi ((a * b) + K) = \phi (a * b) = \phi (a) * \phi (b) = \psi (a + K) * \psi (b + K) \end{align}
• 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:
(7)
\begin{align} \quad R / \ker (\phi) \cong \phi (R) \quad \blacksquare \end{align}

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)
\begin{align} \quad \phi (z) = z \mod n \end{align}

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

(9)
\begin{align} \quad \ker \phi (z) = \{ 0, \pm n, \pm 2n, .. \} = n \mathbb{Z} \end{align}

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

(10)
\begin{align} \quad \mathbb{Z} / n\mathbb{Z} \cong \mathbb{Z}_n \end{align}