Boolean ring, induction

**zelda2139** · January 19th 2009, 04:23 PM

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?

**NonCommAlg** · January 19th 2009, 04:48 PM

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

$I=<x_1, \cdots , x_{n-1}, x_n>=Ax_1 + \cdots + Ax_{n-1}+Ax_n.$ now if $J=<x_1, \cdots, x_{n-1}>=<x>,$ then $I=J+Ax_n=<x,x_n>=<x-xx_n+x_n>.$

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