# Boolean ring, induction

• Jan 19th 2009, 04:23 PM
zelda2139
Boolean ring, induction
A commutative ring $A$ is Boolean if $x^2 = x$ for all $x \in A$.
In a Boolean ring $A$, show that every ﬁnitely generated ideal in $A$ is principal.

I am pretty sure this proof is by induction because: $(x,y)=(x-xy+y)$ The inclusion $(\supseteq)$ is clear, and the identities $x(x-xy+y)=x$ and $y(x-xy+y)=y$ give $(\subseteq)$. How do I prove this by induction?
• Jan 19th 2009, 04:48 PM
NonCommAlg
Quote:

Originally Posted by zelda2139
A commutative ring $A$ is Boolean if $x^2 = x$ for all $x \in A$.
In a Boolean ring $A$, show that every ﬁnitely generated ideal in $A$ is principal.

I am pretty sure this proof is by induction because: $(x,y)=(x-xy+y)$ The inclusion $(\supseteq)$ is clear, and the identities $x(x-xy+y)=x$ and $y(x-xy+y)=y$ give $(\subseteq)$. How do I prove this by induction?

first of all you don't need to add "commutativity" in the definition of a Boolean ring because it's a result of the condition $x^2=x, \ \forall x \in A.$ to complete your induction just note that

$I==Ax_1 + \cdots + Ax_{n-1}+Ax_n.$ now if $J==,$ then $I=J+Ax_n==.$