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Let $R$ be a ring with unit element, $R$ not necessarily commutative, such that the only right-ideals of R are $(0)$ and $R.$ Prove that R is a division ring.

This was a problem from the book, "Topics in Algebra" by I.N Herstein from Chapter- 3 (Ring Theory) , Page number- 139 (2nd Edition).

So, we basically need to show that the non-zero elements of $R$ forms a group under multiplication. This means we only need to prove that each non-zero element of $R$ has an inverse.

I tried solving the problem as follows:

Given, $R$ is a ring with unit element, we consider an $a\neq 0\in R$. Then $aR=\{ar:a\in R\}$ is a right ideal of $R.$ Since, $aR\neq 0$ so, $aR=R.$ This means that every element of $R$ can be expressed as a multiple of $a.$ In particular, as $1\in R$ $\exists b\in R$ such that $ab=1.$

Now, I need to show somehow, that $ba=1$ as well because, $b$ is the inverse of $a$ iff $ab=ba=1.$ But this is the point where I am stuck. Things get difficult for ne because $R$ is not necessarily commutative. So, I do not get how to proceed further.

Thomas Finley
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  • Consider the annihilator ${ r : ar = 0 }$ and show that it's a right ideal, hence must be zero. This means that left multiplication by $a$ is an injection, which means concretely that you can cancel $a$ on the left from two sides of an equation. Now, starting from $ab = 1$ you get that $aba = a$... – Qiaochu Yuan Sep 30 '23 at 05:15
  • @QiaochuYuan Are you using a non-zero $a$ in $R$? I think a little bit of an elaboration is demanded :) – Thomas Finley Sep 30 '23 at 05:18
  • "So, we basically need to show that the non-zero elements of $R$ forms an abelian group under multiplication." Why should $R-{0}$ be abelian? – user10354138 Sep 30 '23 at 05:19
  • @Thomas: I'm using the same $a$ you are, continuing from your argument. – Qiaochu Yuan Sep 30 '23 at 05:21
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    Does this answer your question? What is true for a ring with exactly two right ideals (see solution to question); see also the solution in https://math.stackexchange.com/questions/1151319/trivial-right-or-left-ideals-of-r-imply-that-r-is-a-division-ring – leslie townes Sep 30 '23 at 05:25

1 Answers1

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Consider the ideal generated by $b (\neq 0)$

\begin{equation} (b) = R \implies \exists c \in R \text{ such that } bc = 1 \end{equation}

We have \begin{equation} ab = 1, bc = 1 \implies abc = c \implies a = c \end{equation}

Thererfore, $ba = 1$

total dependent random choice
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