The Kernel of a Ring Homomorphism

# The Kernel of a Ring Homomorphism

Definition: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be rings with additive identities $0_R$ and $0_S$ respectively. If $\phi$ is a homomorphism from $R$ to $S$ then the Kernel of $\phi$ is defined as $\ker (\phi) = \{ a \in R : \phi (a) = 0_S \}$. |

Note that the kernel of an homomorphism $\phi : R \to S$ is a subset of the domain of $\phi$ and it is exactly the set of elements in $R$ that are sent to the additive identity $0_S$ in $S$.

On the Basic Properties Regarding Ring Homomorphisms we have proven that if $R$ and $S$ are homomorphism then:

(1)\begin{align} \quad \phi (0_R) = 0_S \end{align}

Therefore the kernel of a homomorphism $\phi$ is never empty.

Furthermore, if $\phi$ is an injective homomorphism, then the kernel of $\phi$ contains only $0_S$.

We will now state some basic properties regarding the kernel of a ring homomorphism.

Theorem 1: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be homomorphic rings with homomorphism $\phi : R \to S$. Then:a) For all $x, y \in \ker (\phi)$, $(x +_1 y) \in \ker (\phi)$.b) For all $x, y \in \ker (\phi)$, $(x *_1 y) \in \ker (\phi)$. |

**Proof of a)**Let $x, y \in \ker (\phi)$. Then $\phi (x) = 0_S$ and $\phi (y) = 0_S$. So:

\begin{align} \quad \phi (x +_1 y) = \phi (x) +_2 \phi(y) = 0_S + 0_S = 0_S \end{align}

- Hence $(x +_1 y) \in \ker (\phi)$. $\blacksquare$

**Proof of b)**Let $x, y \in \ker (\phi)$. Then $\phi (x) = 0_S$ and $\phi (y) = 0_S$. So:

\begin{align} \quad \phi (x *_1 y) = \phi (x) *_2 \phi (y) = 0_S * 0_S = 0_S \end{align}

- Hence $(x *_1 y) \in \ker (\phi)$. $\blacksquare$

Theorem 2: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be homomorphic rings with homomorphism $\phi : R \to S$. Then $R$ and $S$ are further isomorphic with isomorphism $\phi$ if and only if $\phi (R) = S$ and $\ker (\phi) = \{ 0_R \}$. |

**Proof:**$\Rightarrow$ Suppose that $R$ and $S$ are isomorphic with isomorphism $\phi : R \to S$. Then $\phi$ is bijective. Since $\phi$ is surjective, $\phi (R) = S$. And since $\phi$ is injective, the only element in $R$ that gets mapped to $0_S$ is $0_R$, so $\ker (\phi) = \{ 0_R \}$.

- $\Leftarrow$ Suppose that $\phi (R) = S$ and $\ker (\phi) = \{ 0_R \}$. Since $\phi (R) = S$, $\phi$ is surjective.

- Let $x, y \in R$ and suppose that $\phi (x) = \phi (y)$. Then $\phi (x) - \phi(y) = 0$. So $\phi (x - y) = 0$. Hence $(x - y) \in \ker (\phi)$. So $x - y = 0$ and hence $x = y$. Therefore $\phi$ is injective.

- So $\phi$ is bijective, and hence $R$ and $S$ are isomorphic with isomorphism $\phi$. $\blacksquare$