1. ## Ideals

I'm new to this topic, so any help would be appriciated!

Note: these are all in a communitive ring $\displaystyle R$.

1. $\displaystyle a \mid b \Longleftrightarrow (b) \subseteq (a)$.

2. $\displaystyle u \in R$ is a unit $\displaystyle \Longleftrightarrow (u) = R$.

3. $\displaystyle a$ and $\displaystyle b$ are associative $\displaystyle \Longleftrightarrow (a) = (b)$.

4. $\displaystyle p$ is prime $\displaystyle \Longleftrightarrow ab \in (p)$ implies that either $\displaystyle a \in (p)$ or $\displaystyle b \in (p)$.

2. 1. $\displaystyle (a) = \{ ra : r \in R \}$
If $\displaystyle a \mid b$ then $\displaystyle b =qa$ for some $\displaystyle q \in R$ Do you see why $\displaystyle (b) \subseteq (a)$ ?
2. If $\displaystyle u \in R$ is a unit then there exist $\displaystyle u^{-1} \in R$ such that $\displaystyle uu^{-1} = 1$.
$\displaystyle (u) = \{ ru : r \in R \}$ If $\displaystyle r = qu^{-1} , q \in R$ then we see that it is all the multiple of 1 and thus the ring itself.
3. I don't know what associative is sorry.
4. $\displaystyle (p) = \{ rp : r \in R \}$
$\displaystyle ab \in (p) \implies ab = rp$ for some $\displaystyle r \in R$. Since p is prime, either a or b has to have p in its prime factorization. It follows that $\displaystyle a \in (p)$ or $\displaystyle b \in (p)$

3. for #1.

$\displaystyle (a) = \{ ra : r \in R \}$

Since $\displaystyle b = qa , \; (b) = \{ (rq)a : r \in R \}$

So my question is do we know $\displaystyle rq \in R$?

4. $\displaystyle (a) = \{ ra : r \in R \}$ This is all the multiples of a.
$\displaystyle a \mid b$ So b is a multiple of a.
$\displaystyle (b) = \{ rb : r \in R \}$ Multiples of b or multiples of multiples of a. So yes we know that $\displaystyle rq \in R$. You could also simply use the fact that a ring is closed under multiplication.

5. ah yes, I forgot about closure!