# Boolean ring, induction

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

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

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

I am pretty sure this proof is by induction because: $\displaystyle (x,y)=(x-xy+y)$ The inclusion $\displaystyle (\supseteq)$ is clear, and the identities $\displaystyle x(x-xy+y)=x$ and $\displaystyle y(x-xy+y)=y$ give $\displaystyle (\subseteq)$. 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 $\displaystyle x^2=x, \ \forall x \in A.$ to complete your induction just note that

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