2

I encountered the following exercise (Exercise 3.9, page 24) in the book A Course in Galois Theory by D. J. H. Garling (Cambridge Univ. Press) :

Assume that $R$ is an infinite commutative ring with a multiplicative identity for which $R/I$ is a finite ring for every non-zero ideal $I$ in $R$. Prove that $R$ must be an integral domain.

Not even sure how to approach this! I don't have a feeling why this assertion should be true.

user26857
  • 53,190
student
  • 1,472
  • 6
    Where did this question come from? When you pose the task as a command "Prove that..." it makes the problem sound like homework. – KCd Aug 29 '20 at 02:49
  • 1
    Sorry! It is an exercise in A Course in Galois Theory by D. J. H. Garling. Chapter 3. I am trying to work the book out on my own-so it is not homework but effectively is I guess! – student Aug 29 '20 at 13:30
  • I assumed that $R$ had non-zero zero divisors $a$ and $b$ with $ab=0$ despite being an infinite ring and satisfying the condition that all quotient rings modulo non-zero ideals are finite. We know that $R/(ab)$ is infinite but this contains $R/a$ and $R/b$. I was trying to see if I can find an infinite subring of $R/a$ and thereby get a contradiction. – student Aug 29 '20 at 14:56
  • 1
    The ring $R/(ab)$ does not contain $R/(a)$. Would you say $\mathbf Z/(15)$ contains a subring $\mathbf Z/(5)$? Or that $\mathbf Z$ contains a subring $\mathbf Z/(5)$? This exercise from Garling has nothing to do with Galois theory, so if you get really stuck just move on and do not worry about it. – KCd Aug 29 '20 at 17:18
  • 1
    @student If you have no idea how to approach a problem, try an easy case. In this case, let $R=\mathbb Z$ and try to prove $\mathbb Z$ is an integral domain using the hypothesis. We know $\mathbb Z$ is infinite and we know $\mathbb Z/(n)$ is finite whenever $n \ne 0$. – user5826 Aug 29 '20 at 19:07
  • 2
    I voted to reopen because the OP has shown effort in the comment section. – J. De Ro Aug 29 '20 at 19:11
  • KCd you raised an issue that exposes my ignorance! Obviously one cannot conclude that ${\bf Z}$ contains ${\bf Z}/(5)$ as a subring although $(0)\subset (5)$ in the ring ${\bf Z}$!! I was not thinking clearly at all... – student Aug 29 '20 at 20:16

1 Answers1

3

Hint: If $ab=0$, think about the ideal $I=\{x\in R:ax=0\}$.

More details are hidden below.

If $b\neq 0$, then $I$ is a nonzero ideal, so $R/I$ is finite. But $I$ is the kernel of the map $R\to R$ given by multiplication by $a$, so $R/I$ is isomorphic as an $R$-module to the image of this map, which is just the ideal $J$ generated by $a$. If $J$ were a nonzero ideal, then $R/J$ would be finite, but then $R$ would be finite since both $J\cong R/I$ and $R/J$ are finite. Thus $J$ is the zero ideal, so $a=0$.

Eric Wofsey
  • 342,377