Solves Theorem 3.2.19 in Bland's Abstract Algebra

[Math Amateur]

I am reading The Basics of Abstract Algebra by Paul E. Bland ...

I am focused on Section 3.2 Subrings, Ideals and Factor Rings ... ...

I need help with another aspect of the proof of Theorem 3.2.19 ... ... Theorem 3.2.19 and its proof reads as follows:
https://www.physicsforums.com/attachments/8270

In the above proof by Bland we read the following:"... ... Hence \(\displaystyle x = you =yxb\) which implies that \(\displaystyle yb = e\) ... ...
Can someone please explain exactly how/why \(\displaystyle x = you =yxb\) implies that \(\displaystyle yb = e\) ... ...

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***EDIT***

Is it simply because \(\displaystyle x = yxb = xyb\) since R is commutative and then

\(\displaystyle x = xyb \Longrightarrow yb = e\) ... is that correct?

But how do we know \(\displaystyle x \neq 0\) ...------------------------------------------------------------------------------------------------------

Peter

 

[castor28]

Hi Peter,

We know that $x\ne0$ because the proof says that $xR$ is a nonzero prime ideal.

To be completely correct, you should say that $x=xyb$ implies $x(1-yb)=0$, and this implies $yb=1$ because $x\ne0$ and $R$ is an integral domain.