Sums and Products of Ideals in a Commutative Ring Let $R$ be a commutative unitary ring, $I$ and ideal and $a,b$ two elements of the ring. It says in my lecture notes :
$$ (I+Ra)(I+Rb)=I^2+aI+bI+Rab $$
I'm not disagreeing to this but I'd like to know if it's hard to prove because the only trivial statement I can make is that the left side is a subset of the right side.
$$ \begin{array}{rcl}
(I+Ra)(I+Rb) & = & \lbrace (i_1+r_1a)(i_2+r_2b) \mid i_1,i_2 \in I \quad r_1,r_2 \in R \rbrace \\
& = &  \lbrace i_1i_2 + i_2r_1a + i_1r_2b + r_1r_2ab \mid i_1,i_2 \in I \quad r_1,r_2 \in R \rbrace \\
& \subset & \lbrace i_1i_2 + i_3a + i_4b + rab \mid i_1,i_2,i_3,i_4 \in I \quad r \in R \rbrace \\
& = & I^2+aI+bI+Rab \end {array}  $$
Obviously my lecturor's claim is that my $\subset$ symbol can actually be replaced by an $=$ sign.
It seems to me like $I^2+aI+bI+Rab$ is a lot bigger than $(I+Ra)(I+Rb)$ since it a priori has more freedom as my equation seems to suggest. For instance, when $i_1$, $i_2$ and $i_3$ are chosen, why should there be an $r_1$ such that $i_2r_1a=i_3a$ when we know nothing about divisibility or invertibility of the elements ?
 A: In the language of rings (and modules), whenever $AB$ is found, it usually means the set of all sums of elements of the form $ab$, with $a\in A$ and $b\in B$.
Instead of $A\{b\}$ one usually writes $Ab$, but in this case sums are not necessary, because $a_1b+\dots+a_nb=(a_1+\dots+a_n)b$, provided $A$ is closed under sums.
If $I$ is an ideal, then clearly it is closed under sums; hence
$$
Ia=\{xa:x\in I\}
$$
However, if $I$ and $J$ are ideals, then
$$
IJ=
\Bigl\{\sum_{k=1}^n x_ky_k:n\ge0, x_k\in I, y_k\in J, k=1,\dots,n\Bigr\}
$$
not to be confused with the set $\{xy:x\in I,y\in J\}$ which is generally not even closed under sums.
It's easy to see that $IJ$ is the ideal generated by the products $xy$, with $x\in I$ and $y\in J$. In the case of $Ra$, this is exactly the set of products $ra$, for $r\in R$, because $R$ is closed under sums.
With this interpretation (which is the one implicitly used in ring theory, unless the contrary is explicitly mentioned), if $A$, $B$ and $C$ are ideals, then
$$
(A+B)C=AC+BC
$$
as you can easily prove. In particular
$$
(I+Ra)(I+Rb)=I^2+RaI+IRb+RaRb
$$
Now $RaI=aI$, because
$$
\sum_{k=1}^n(r_ka)x_k=a\sum_{k=1}^n(r_kx_k)\in aI
$$
and the converse inclusion is obvious (if $R$ has identity).
You can prove similarly that $(Ra)(Rb)=Rab$. So we get
$$
(I+Ra)(I+Rb)=I^2+aI+bI+Rab
$$
as claimed.
A: Careful; the product of two ideals does not consist of all products of elements of the two ideals, but of all sums of products of elements of the two ideals. That is, it is the ideal generated by all products of elements of the two ideals. So $z\in (I+Ra)(I+Rb)$ if and only if there exist $n\in\Bbb{N}$ and $x_1,\ldots,x_n\in I+Ra$ and $y_1,\ldots,y_n\in I+Rb$ such that $z=\sum_{k=1}^nx_ky_k$.
