# For any two Ideals $A$ and $B$,$A+B=\langle A \cup B \rangle$

Below is the proof of : Prove that for any two ideals $A$ and $B$ of ring $R$,$A+B=\langle A \cup B~\rangle$ .

Proof:

By theorem (for any two ideals of a ring $R$ ,then the set $A+B$ is an ideal of $R$) such that ,$A \subseteq A+B$ and $B \subseteq A+B$ , so $A\cup B \subseteq A+B.$

Let $X$ be any ideal of $R$ such that $A\cup B \subseteq X.$ If $z \in A+B$ then $z = a+b$, $a \in A$,$b \in B$. Now $a \in A\cup B~~$,$~~b \in A\cup B \implies a,b \in X \implies z \in X.$

Hence,$A+B \subseteq X$.Consequently ,by definition $A+B=\langle A \cup B \rangle$

The thing I can't understand is where did we use the fact that $X$ is an ideal.Won't the proof work if $X$ is just a subset of $R$ such that $A\cup B \subseteq X.$?

You concluded that $a,b \in X$, and therefore $z = a+b \in X$. This would not be true if $X$ were an arbitrary subset.
For instance, in $\Bbb Z$, look at the ideals $(2)$ and $(3)$. $(2) + (3) = \Bbb Z$, but we could take $X = (2) \cup (3)$, which is not an ideal and does not contain $(2) + (3)$ (in particular, $1 = 3 - 2 \not\in X$.
The proof you gave did, however, show that this holds whenever $A$ and $B$ are ideals, and $X$ is any additive subgroup containing $A$ and $B$ (ie scaling didn't come into play).
• I agree with you .but what if $X$ is just a subring in $R$,and not an ideal... – spectraa Sep 23 '14 at 6:32
• @spectraa An "additive subgroup" is just a subset $S \subseteq X$ that is closed under addition -- $a + b \in S$ whenever $a$ and $b$ are both elements of $S$. As Nimda mentioned, any subring must be an additive subgroup, simply by the definition of ring. So the statement I mentioned can be read as "Whenever $S$ is a subset of $X$ that contains $A$ and $B$, and is closed under addition (in particular $S$ could be a subring of $X$ containing $A$ and $B$), $S$ must contain the ideal $A + B$." – Andrew Maurer Sep 24 '14 at 5:20