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.

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 $K=I\cdot J$.

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 $$I=\left\{\sum_{k=1}^n\lambda_ka_k\mu_k \mid \lambda_k,\mu_k\in R^1\right\}\,.$$

• Now when $A$ is a finite set of generators, $(A)_R$ and $(A)_{R^1}$ are the same right? Since we can achieve multiplication with elements of $\mathbb Z$ by repeated addition. – Christoph Jan 22 '14 at 22:14
• On another note: I think we need $\sum_{k=1}^s \lambda_k a_{i_k} \mu_k$, so we can repeat generators. Otherwise $\lambda a_1 \mu+\lambda' a_1\mu'$ wouldn't be in the set, would it? – Christoph Jan 22 '14 at 22:30