Notes - Rings and Modules HT24, Ring quotients

• The kernel of a ring homomorphism is always an ideal
• Given an ideal $I$, we can construct the quotient map $q : R \to R/I$ which is a ring homomorphism with kernel $I$

Suppose:

• $R, S$ are rings
• $\phi : R \to S$ any ring homomorphism
• $I \subseteq \ker \phi$
• $I \trianglelefteq R$
• $q : R \to R/I$ quotient homomorphism

Then:

• $\exists ! \overline \phi : R/I \to S$ such that $\phi = \overline \phi \circ q$
• $\ker \overline \phi = \ker \phi / I = \{ m + I \mid m \in \ker \phi \}$

Existence: Since $\phi = \overline{\phi} \circ q$, we must have that $\overline{\phi}(m + I) := \phi(m)$, so we need to show that this map is well defined. Note that if $r + I = r’ + I$ then $r - r’ \in I$, and since $I \subseteq \ker(\phi)$ then $\phi(r) = \phi(r’)$ and hence $\phi$ is constant on the cosets of $I$. So the above is indeed well-defined. Then this is a ring homomorphism, which can be checked by considering representatives $C _ 1, C _ 2 \in R / I$ and working out $\overline\phi(C _ 1 C _ 2)$ and $\overline{\phi}(C _ 1 + C _ 2)$.

Uniqueness: Note that since $q$ is surjective and $\overline\phi(q(r))$ = $\phi(r)$, then $\overline \phi$ is completely determined by $\phi$, hence is unique.

Kernel property: Finally note that for $\ker \overline \phi$, we have $(r + I) = \phi(r) = 0 \iff r \in \ker \phi$, so we are done.

The units are the elements of $R$ coprime to $n$. By Bezout’s lemma, $a$ is coprime to $n$ iff $\exists s, t$ such that

\[as + nt = 1\]

But since $nt = 0$ in $\mathbb Z / n \mathbb Z$, we have

\[as = 1\]

If $n = p _ 1^{q _ 1} \cdots p _ r^{q _ r}$, then the nilpotent elements of $R$ are:

\[N = \\{k (p_1 \cdots p_r) \mid k \in \mathbb Z\\}\]

By the Chinese remainder theorem,

\[\frac{\mathbb Z}{n\mathbb Z} \cong \bigoplus^r_{i = 1}\frac{\mathbb Z}{p_i^{q_i} }\]

so $k^m = 0$ for some $m$ iff

\[p_i^{q_i} \mid k^m\]

For each $i$. This implies

\[N = \\{k (p_1 \cdots p_r) \mid k \in \mathbb Z\\}\]