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Suppose $R$ is a commutative ring with $1neq0$. 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 $P_1P_2dotsm P_n$?

My Intuition:

$a_1a_2dotsm a_n$ is the generator of the ideal $P_1P_2dotsm P_n$.

Reason for intuition:

Any element of any element of the ideal $P_1P_2dotsm P_n$ is of the form $b_1b_2dotsm b_n$ such that $b_iin P_i$ for $i=1,2,dots,n$.

But each $b_i$ is of the form $c_ia_i$ in the ideal $P_i$ for some $c_iin R$.

Therefore, each element of the ideal $P_1P_2dotsm P_n$ is of the form $prod_i=1 ^i=n c_ia_i$.

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 $cin R$ 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 $1neq 0$?

Please give me hints.

If $I,J$ are two ideals in a commutative ring, the elements of the product $IJ$ aren't necessarily of the form $xy$ where $xin I$, $yin J$. Instead, the elements are sums of such products: thus each element of $IJ$ has the form $sum_i=1^n x_iy_i$ where $x_iin I$ and $y_iin J$.

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 $1leq ileq m$ and $1leq jleq n$. 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 $r,sin R$ --- 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 $r,sin R$. There's a pesky "$s$" in the middle of that monomial.

For example, let $R=mathbfClangle x,yrangle$ 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 $m$ for $kgeq 0$ --- 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 $I_1subsetneq I_2subsetneq I_3subsetneq cdots$, and the product $IJ$ is obtained as the union: $IJ=cup_kgeq 0 I_k$. 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.