Let $R$ be a commutative ring, and $j \in R$. We wish to prove that $JR \equiv \{ jr : r \in R \}$ is an ideal of $R$. To prove that $jR$ is an ideal, we need to verify that $(jR, 0, +)$ is an abelian group, and that $jR$ is closed under multiplication by the entire ring.
This seems elementary, but I have this paranoid feeling that I'm missing something, so I'd appreciate it if someone verified this for me!
To prove that $(jR, 0, +)$ an abelian group:
- $0 = j0 \in jR$, hence identity is satisfied
- for all $r_1, r_2 \in R$, $jr_1 + jr_2 = j(r_1 + r_2) \in jR$, hence closure is satisfied
- for all $r \in R, jr \in jR \implies j(-r_1) \in jR$, hence $jR$ closed under inverses.
- commutativity of $+$ follows since the ring $(R, +)$ is a commutative ring.
To prove that it's closed under multiplication under the entire ring:
- Let $jr \in jR$. We pick any $s \in R$ and show that multiplying this element keeps us in $jR$: $s(jr) = (jr) s = j (rs) \in jR$.
Hence, $jR$ is indeed an ideal of $R$.
Another way to show that $(jR, 0, +)$ forms a group which is often used is the general observation that for any group $G$ and non-empty subset
$S \subset G, \tag 1$
$S$ is a subgroup of $G$ if and only if
$a, b \in S \Longrightarrow ab^{-1} \in S; \tag 2$
it is easy to see that this holds if $S$ is a subgroup, since then $b^{-1} \in S$ etc; going the other way, (2) yields
$a \in S \Longrightarrow e = aa^{-1} \in S, \tag 3$
so $S$ contains the identity element; then
$e, a \in S \Longrightarrow a^{-1} = ea^{-1} \in S, \tag 4$
so $S$ contains inverses of all its elements; finally,
$a, b \in S \Longrightarrow b^{-1} \in S \Longrightarrow ab = a(b^{-1})^{-1} \in S, \tag 5$
and $S$ is closed under the group operation. So we see that $S$ must be a subgroup of $G$.
We may apply this to $Rj$, since
$aj, bj \in Rj \Longrightarrow aj - bj = (a - b)j \in Rj, \tag 6$
and thus $(Rj, 0, +)$ is an (abelian) subgroup of $(R, 0, +)$.
$OE\Delta$.