Let the commutative ring $R$ and its element $a$ and $b$
$\langle a\rangle +\langle b\rangle =\langle a,b\rangle $
$\langle a\rangle \langle b\rangle =\langle ab\rangle $
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Q1) What is the inclusion relation between $\langle a\rangle $, $\langle a,b\rangle $ and $\langle ab\rangle $?
Q2) Please tell me what point was I've wrong.
Here is my thought.
Let $r_1$ and $r_2$ be elements of the ring $R$.
It looks obvious that $\langle ab\rangle \subseteq \langle a\rangle $ since $a\mid ab$
An the element form of $\langle a\rangle +\langle b\rangle $ and $\langle a\rangle $ are $r_1a + r_2b$ and $r_1a$ respectively.
Hence $\langle a\rangle +\langle b\rangle \subseteq \langle a\rangle $ since $0 \in R$
But $\langle a,b\rangle = \langle d\rangle $ (here $d$ means $\gcd(a,b)$)
Therefore $\langle a\rangle \subseteq \langle d\rangle $ since $d\mid a$
So $\langle a\rangle = \langle a,b\rangle $ ?? (Surely It is a false statement.)
Any help would be appreciated. Thanks.
A couple of things. In arbitrary commutative rings, there may be no such thing as a $\gcd$. Second, $\langle a, b \rangle $ is defined to be smallest ideal that contains both $a$ and $b$. In particular, unless $b \in \langle a \rangle$, you'll have $\langle a \rangle \subsetneq \langle a, b \rangle $. There are plenty of elements of the form $r_1a+r_2b$ that aren't multiples of $a$.
For any $r \in R, (ab)r=a(br) \in \langle a \rangle$ so $\langle ab \rangle \subseteq \langle a \rangle$ and it's obvious that $\langle a \rangle \subseteq \langle a, b \rangle$. Some examples in $\Bbb Z$ should convince you that both inclusions can be proper.