MHB Solves Theorem 3.2.19 in Bland's Abstract Algebra

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SUMMARY

The discussion focuses on the proof of Theorem 3.2.19 from Paul E. Bland's "The Basics of Abstract Algebra," specifically addressing the implications of the equation $$x = yxb$$ leading to $$yb = e$$. Participants clarify that since $$x$$ is a nonzero element of a prime ideal in an integral domain, the equation $$x(1-yb)=0$$ directly leads to the conclusion that $$yb=1$$. This understanding is crucial for grasping the properties of ideals and their behavior in commutative rings.

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  • Understanding of abstract algebra concepts, particularly subrings and ideals.
  • Familiarity with integral domains and their properties.
  • Knowledge of commutative rings and their operations.
  • Ability to interpret mathematical proofs and theorems.
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  • Study the properties of prime ideals in integral domains.
  • Explore the implications of commutativity in ring theory.
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Students of abstract algebra, mathematicians focusing on ring theory, and anyone seeking a deeper understanding of theorems related to ideals and their implications in commutative rings.

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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 $$x = you =yxb$$ which implies that $$yb = e$$ ... ...
Can someone please explain exactly how/why $$x = you =yxb$$ implies that $$yb = e$$ ... ...

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***EDIT***

Is it simply because $$x = yxb = xyb$$ since R is commutative and then

$$x = xyb \Longrightarrow yb = e$$ ... is that correct?

But how do we know $$x \neq 0$$ ...------------------------------------------------------------------------------------------------------

Peter
 
Last edited:
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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.
 

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