# Notes - Rings and Modules HT24, Ring quotients

### Flashcards

Quickly prove that if $R$ is a ring, then the ideals in $R$ are exactly the kernels of ring homomorhisms from $R$.

- 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 $I$

Can you state the “universal property of quotients” for a given ring $R$?

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 $\phi = \overline \phi \circ q$
- $\ker \overline \phi = \ker \phi / I = \{ m + I \mid m \in \ker \phi \}$

Quickly prove the universal property of quotients, i.e. that if

- $R, S$ 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.

Suppose $R = \mathbb Z / n \mathbb Z$. State the units of $R$, and quickly prove your description is true.

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

\[as = 1\]Suppose $R = \mathbb Z / n\mathbb Z$. State the nilpotent elements of $R$ and quickly prove your description is true.

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

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