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

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In summary, a commutative ring with prime proper ideals is a ring where every proper ideal is prime, and a field is a commutative ring where every nonzero element has a multiplicative inverse. This concept is important in abstract algebra and has applications in number theory and algebraic geometry. It can be proven that a commutative ring with prime proper ideals is a field by using the concept of maximal ideals. An example of a commutative ring with prime proper ideals that is not a field is the ring Z/6Z.
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Euge
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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.
 
  • #3
This week's problem was solved by Olinguito, Ackbach, and castor28. You can read Olinguito's solution below.
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 $A$.)
 

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

What is the definition of a commutative ring with prime proper ideals?

A commutative ring with prime proper ideals is a ring in which every proper ideal is prime. This means that the product of any two elements in the ring will belong to the ideal, and if an element is not in the ideal, then neither are its multiples.

What does it mean for a commutative ring to be a field?

A field is a commutative ring in which every nonzero element has a multiplicative inverse. This means that every element in the ring can be multiplied by another element to equal 1.

How can you prove that a commutative ring with prime proper ideals is a field?

This can be proven using the concept of maximal ideals. If a commutative ring has prime proper ideals, then it must also have maximal ideals. And if a commutative ring has maximal ideals, then it is a field. Therefore, a commutative ring with prime proper ideals is also a field.

Can you provide an example of a commutative ring with prime proper ideals that is not a field?

Yes, the ring Z/6Z, also known as the integers modulo 6, has prime proper ideals but is not a field. This is because 2 and 3 are both prime proper ideals in this ring, but neither has a multiplicative inverse.

Why is it important to understand the concept of commutative rings with prime proper ideals?

Understanding this concept is important in abstract algebra, as it helps to classify different types of rings and understand their properties. It also has applications in number theory and algebraic geometry. Additionally, knowing that a commutative ring with prime proper ideals is a field can be useful in solving certain mathematical problems and proofs.

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