1. ## Ideals

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

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

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

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

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

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

2. 1. $(a) = \{ ra : r \in R \}$
If $a \mid b$ then $b =qa$ for some $q \in R$ Do you see why $(b) \subseteq (a)
$
?
2. If $u \in R
$
is a unit then there exist $u^{-1} \in R$ such that $uu^{-1} = 1$.
$(u) = \{ ru : r \in R \}$ If $r = qu^{-1} , q \in R$ 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. $(p) = \{ rp : r \in R \}$
$ab \in (p) \implies ab = rp$ for some $r \in R$. Since p is prime, either a or b has to have p in its prime factorization. It follows that $

a \in (p)
$
or $

b \in (p)
$

3. for #1.

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

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

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

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

5. ah yes, I forgot about closure!