Finite generating set of the product of finitely generated ideals. Let $R$ be a ring (not necessarilly unitary) and $I=(a_1,\dots,a_n)$, $J=(b_1,\dots,b_m)$ finitely generated ideals. By, definition
$$
I = \bigcap \left\{\ I'\subseteq R\ \text{ideal}\ \big|\ a_i\in I'\text{ for $i=1,\dots,n$}\ \right\}
$$
and $J$ similarly.
Now the product ideal is defined as
$$
I\cdot J = \left( \left\{ \ \alpha\cdot\beta \ \big|\ \alpha\in I, \beta\in J\ \right\} \right).
$$
We see that
$$
I\cdot J = \left\{\ \sum_{k=1}^s \alpha_k \beta_k \ \big|\ s\in\mathbb N, \alpha_k\in I, \beta_k\in J\ \right\}.
$$
Now I'd like to prove that $I\cdot J$ is finitely generated, namely
$$
I\cdot J = \left( \left\{\ a_i b_i \ \big|\ 1\le i\le n, 1\le j\le m\ \right\} \right).
$$
Edit: See my answer below for a counter-example in the non-commutative case and a proof in the commutative case.
 A: If $R$ doesn't have unit, the easiest is to formally adjoin a unit: let $R^1:=R\oplus\Bbb Z$ with straightforward multiplication.
Then, the finitely generated ideal $I=(a_1,\dots,a_n)$ can be written similarly as
$$I=\left\{\sum_{k=1}^n\lambda_ka_k\mu_k \mid \lambda_k,\mu_k\in R^1\right\}\,.$$
A: I figured out the proposed generating set doesn't work in the non-commutative case, even if we have a unit.
Let me investigate this in $R=\mathbb R^{2\times 2}$. Set
$$
I = \Bigg( \pmatrix{1 & 0 \\ 0 & 0} \Bigg),\quad
J = \Bigg( \pmatrix{0 & 0 \\ 0 & 1} \Bigg),\quad
$$
and
$$
K = \Bigg( \pmatrix{1 & 0 \\ 0 & 0} \pmatrix{0 & 0 \\ 0 & 1} \Bigg)
= \Bigg\{ \pmatrix{0 & 0 \\ 0 & 0} \Bigg\}.
$$
We wan't to check if $I\cdot J = K$:
$$
I\cdot J \ni
\underbrace{\pmatrix{1 & 0 \\ 0 & 0}
\pmatrix{a & b \\ c & d}}_{\in I}
\underbrace{\pmatrix{0 & 0 \\ 0 & 1}}_{\in J}
=
\pmatrix{a & b \\ 0 & 0}
\pmatrix{0 & 0 \\ 0 & 1}
=
\pmatrix{0 & b \\ 0 & 0},
$$
which is not in $K$ if $b\neq 0$.
Thus $I\cdot J\neq K$, so $(a)\cdot(b)=(ab)$ does not hold for non-commutative rings.

Let me prove it for the commutative non-unitary case. Here we have
\begin{align*}
I = (a_1,\dots,a_n) &=
\left\{\ \sum_{i=1}^n r_i a_i + \sum_{i=1}^n k_i a_i\ \Big|\ r_i\in R, k_i\in\mathbb Z\right\}, \\
J = (b_1,\dots,b_m) &=
\left\{\ \sum_{i=1}^m s_i b_i + \sum_{i=1}^m l_i a_i\ \Big|\ s_i\in R, l_i\in\mathbb Z\right\},
\end{align*}
where $l a$ with $l\in\mathbb Z$ is just a notation for $a+a+\cdots+a$ or $(-a)+(-a)+\cdots+(-a)$.
Now $$I\cdot J = ( \{\ \alpha\beta \ |\ \alpha\in I, \beta\in J\ \} )$$ by definition. Obviously $$K := ( \{\ a_i b_j\ |\ 1\le i\le n, 1\le j\le m\} ) \subseteq I\cdot J,$$ since we just picked a subset of the generators of $I\cdot J$. All we need to do is to see that all generators of $I\cdot J$ lie in $K$. So we start with
\begin{align*}
\alpha\beta &=
\left(\sum_{i=1}^n r_i a_i + \sum_{i=1}^n k_i a_i\right)
\left(\sum_{i=1}^m s_i b_i + \sum_{i=1}^m l_i b_i\right).
\end{align*}
Now we have the following types of terms in the product:
\begin{align*}
(r_i a_i)(s_j b_j) &= (r_i s_j) a_i b_j \in K, \\
(r_i a_i)(l_j b_j) &= l_j (r_i (a_i b_j)) \in K, \\
(k_i a_i)(s_j b_j) &= k_i (s_j (a_i b_j)) \in K, \\
(k_i a_i)(l_j b_j) &= (k_i l_j)(a_i b_j) \in K.
\end{align*}
Thus, $\alpha\beta\in K$ and $K=I\cdot J$.
