# Notes - Rings and Modules HT24, Ideals

### Flashcards

Suppose:

- $R$ is a ring
- $I \subseteq R$

What does it mean for $I$ to be an ideal, i.e. $I \trianglelefteq R$?

It is closed under addition and $R I = I$.

Suppose:

- $R$ is a ring
- $I, J$ ideals in $R$
- $X$ any subset of $R$

Quickly prove $I + J$, $I \cap J$ and $IX$ are ideals and that $IJ \subseteq I \cap J$ and $I, J \subseteq I + J$.

- $I + J$ ideal: $I + J$ is closed under addition and if $r \in R$, then $r(i + j) = ri + rj \in I + J$.
- $I \cap J$ ideal: $I \cap J$ is closed under addition and if $i \in I \cap J$, clearly $r i \in I$ and $ri \in J$, so $ri \in I \cap J$.
- $IX$ ideal: if $i _ 1 x _ 1 + i _ 2 x _ 2 = i _ 1’ + i _ 2’ \in I$, so closed under addition and if $i x \in IX$, then $r ix = (ri)x = i’x \in IX$.
- $IJ \subseteq I \cap J$ since if $ij \in I$ and $ij \in J$
- $I, J \subseteq I + J$ immediate

What is true about the intersection of any set of ideals?

It is also an ideal.

Suppose:

- $R$ is a ring
- $T \subseteq R$

How is $\langle T \rangle$ defined, and what is this equivalent to?

“smallest ideal containing $T$”.

Suppose:

- $R$ is a ring
- $T \subseteq R$

Quickly prove that

\[\bigcap_{T \subseteq I, I \text{ ideal}\\,} I = RT\]

Clearly $RT$ is an ideal containing $T$, so $\bigcap _ {T \subseteq I, I \text{ ideal}\,} I \subseteq RT$. Now suppose $J$ is an ideal containing $T$. Then for any $t \in J$, $rt \in J$ for any $r$ hence $RT \subseteq J$. But then it is in the intersection, and we are done.

Suppose:

- $R$ is a ring
- $I$ is an ideal

What does it mean for $I$ to be a principal ideal?

It is generated by a single element, i.e.

\[I = \langle a\rangle = aR\]Suppose:

- $R$ is an integral domain
- $I$ is a principal ideal
- $I = \langle a \rangle$

Quickly prove that:

- $\langle a \rangle = \langle b \rangle \iff$ $a$ and $b$ are associates

If $I = \{0\} = \langle 0 \rangle$, then since the only associates of $0$ is $0$, the result is immediate. So we can assume that $I$ is non-zero.

For the forward direction, suppose $a, b \in R$ nonzero. Then $a \in \langle b \rangle$, so $a = rb$ for some $r$, and $b \in \langle b \rangle$, so $b = sa$ for some $s$. But then $a = rb = r(sa)$ means $a(1-rs) = 0$, and since $R$ is an integral domain, $rs = 1$ and hence $r$ and $s$ are units. Then $a = rb$, so differ by a unit.

For the backward direction, suppose $b = u a$ for some unit $u$. Then $b \in \langle a \rangle$, which implies $\langle b \rangle \subseteq I$. For the other inclusion, if $x \in I$, then $x = ra = ru^{-1}b$, so $x \in \langle b \rangle$.

Suppose:

- $\phi : R \to S$ is a surjective ring homomorphism
- $I \trianglelefteq R$
- $J \trianglelefteq S$

what can you then deduce about $\phi(I)$ and $\phi^{-1}(J)$?

They are both still ideals.

What result links PIDs and nested ascending chains of ideals?

Suppose:

- $R$ is a PID
- $I _ 1 \subseteq I _ 2 \subseteq \cdots$ is a sequence of ideals

Then:

- $I := \bigcup _ {n \ge 0} I _ n$ is an ideal
- $\exists N \in \mathbb N$ such that $\forall n \ge N$, $I _ n = I _ N = I$.

Suppose you are asked to comment on the existence of an ideal $I \trianglelefteq R$ such that $R / I \cong S$. How can you translate this into a question about the existence of homomorphisms between $R$ and $S$?

$I$ exists $\iff$ $\exists \phi : R \to S$ surjective homomorphism with kernel $I$