Generators of Principal Ideal Suppose $R$ is a commutative ring with $1\neq0$. Let $P_{1}, P_{2},\dots,P_{n}$ be $n$ principal ideals of $R$.
Assume $a_{1}, a_{2},\dots,a_{n}$ be the generator of $P_{1}, P_{2},\dots,P_{n}$ respectively. Then, what is the generator of the ideal $P_{1}P_{2}\dotsm P_{n}$?
My Intuition:
$a_{1}a_{2}\dotsm a_{n}$ is the generator of the ideal $P_{1}P_{2}\dotsm P_{n}$.
Reason for intuition:
Any element of any element of the ideal $P_{1}P_{2}\dotsm P_{n}$ is of the form $b_{1}b_{2}\dotsm b_{n}$ such that $b_{i}\in P_{i}$ for $i=1,2,\dots,n$.
But each $b_{i}$ is of the form $c_{i}a_{i}$ in the ideal $P_{i}$ for some $c_{i}\in R$.
Therefore, each element of the ideal $P_{1}P_{2}\dotsm P_{n}$ is of the form $\prod_{i=1} ^{i=n} c_{i}a_{i}$.
Now, $R$ is commutative ring. Therefore, we have $(\prod_{i=1} ^{i=n} c_{i})(\prod_{i=1} ^{i=n} a_{i})$.
Now, Let $c=\prod_{i=1} ^{i=n} c_{i}$ for some $c\in R$ because $R$ is a ring. Therefore $\prod_{i=1} ^{i=n} a_{i}$ is the generator of $\prod_{i=1} ^{i=n} P_{i}$.
Another question:
What happens when $R$ is non-commutative with $1\neq 0$?
Please give me hints.
 A: If $I,J$ are two ideals in a commutative ring, the elements of the product $IJ$ aren't necessarily of the form $xy$ where $x\in I$, $y\in J$. Instead, the elements are sums of such products: thus each element of $IJ$ has the form $\sum_{i=1}^n x_iy_i$ where $x_i\in I$ and $y_i\in J$.
You're right that if $I,J$ have given generating sets, say $I=(x_1,\ldots,x_m)$ and $J=(y_1,\ldots,y_n)$, then $IJ$ is generated (as an ideal) by products $x_iy_j$ where $1\leq i\leq m$ and $1\leq j\leq n$. Try to prove this!
So to show that a product of principal ideals is again principal: your proof isn't quite correct, but it's almost correct. Do you see how to fix it?

For noncommutative rings:
I don't think a product of principal ideals is even finitely-generated, let alone principal, even if you only work with one-sided ideals. Keep in mind that if $I$ is the two-sided ideal of $R$ generated by some element $x$, then it's still not true that elements of $I$ have the form $rxs$ for $r,s\in R$ --- the elements of $I$ are sums of such products. If you want to work with left ideals only, even then the elements of the product $RxRy$ have the form $rxsy$ for $r,s\in R$. There's a pesky "$s$" in the middle of that monomial.
For example, let $R=\mathbf{C}\langle x,y\rangle$ be a noncommutative polynomial ring in two variables $x,y$, and consider the left ideals $I=Rx$ and $J=Ry$. Then $IJ=RxRy$ is not finitely-generated as a left ideal: I think this might not be so easy to see, and it involves some combinatorics that I will sweep under the rug.

Claim: The left ideal $IJ$ is not finitely-generated.

Proof. Let $I_k$ be the left ideal generated by $\{xmy : |m|\leq k\}$ for $k\geq 0$ --- what this notation means is that you have $x$, followed by a monomial $m$ of length $\leq k$, then ending in a $y$. So for example, $I_3$ contains the monomial $x^3y^2=x(x^2y)y$, but it does not contain the monomial $xy^7=x(y^6)y$ because the "sandwiched" $y^6$ is too long.
Then the $I_k$'s form a strict ascending chain $I_1\subsetneq I_2\subsetneq I_3\subsetneq \cdots$, and the product $IJ$ is obtained as the union: $IJ=\cup_{k\geq 0} I_k$. So $IJ$ cannot be finitely-generated. QED
Think of it like this: $I$ is the set of polynomials all of whose monomials end in $x$. Similar for $J$. And then the product $IJ$ is the set of polynomials all of whose monomials have at least one $x$, and end in $y$ --- but what could be in between that $x$ and the last $y$ could be arbitrarily large.
