MHB What is the proof that a commutative ring with prime proper ideals is a field?

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Here is this week's POTW:

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Prove that every commutative ring $A$ with unity in which every proper ideal is prime, is a field.
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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

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Hello MHB community,

In case there was any confusion with this problem, unity in a ring is assumed to be nonzero here.

 

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This week's problem was solved by Olinguito, Ackbach, and castor28. You can read Olinguito's solution below.

Spoiler

Let $0\ne x\in A$. We want to show that $x$ has a multiplicative inverse.

First we note that $A$ is an integral domain as the zero ideal $[0]$ is prime (so if $rs\in[0]$ then either $r\in[0]$ or $s\in[0]$). Consider the principal ideal $[x^2]$ generated by $x^2$. If $[x^2]=A$ then $1\in[x^2]$ and so $1=ax^2$ for some $a\in A$ $\implies$ $ax$ is the multiplicative inverse of $x$. Otherwise $[x^2]$ is a prime ideal; then, as $x^2=x\cdot x\in[x^2]$, we have $x\in[x^2]$ $\implies$ $x=bx^2$ for some $b\in A$; hence, as $A$ is an integral domain, $1=bx$ $\implies$ $b$ is the multiplicative inverse of $x$.

So every nonzero element of $A$ has a multiplicative inverse, showing that $A$ is a field. (In this case $[0]$ is the only prime ideal of $A$.)