# Every $R$-module is free $\implies$ $R$ is a division ring

From Grillet's Abstract Algebra, section VIII.5.

Definitions. A division ring is a ring with identity in which every nonzero element is a unit. A vector space is a unital module over a division ring.

Theorem 5.2. Every vector space has a basis.

Exercises.

(7.) Show that $R$ is a division ring if and only if it has no left ideal $L \neq 0, R$.

(*9.) Prove that $R$ is a division ring if and only if every left $R$-module is free.

I'm trying to solve exercise 9. I suspect it should be formulated:

Let $R$ be a unital ring. Then $R$ is a division ring $\iff$ every unital left $R$-module is free.

Attempt of proof: $(\implies)$: Theorem 5.2.

$(\impliedby)$: By exercise 7, it suffices to show that every non-zero left ideal of $R$ is equal to $R$. Let $I\!\neq\!0$ be a left ideal. As a $R$-module, $I$ has a basis $B$.

How can I show that $I=R$?

-
Why $b_2b_1+(-b_1)b_2=0$? This happens iff $b_1$ commute with $b_2$, or I missing something? –  Frank Murphy Oct 25 '11 at 23:28
Uf, you're right @Frank. I'll delete that part, thanks. –  Leon Oct 26 '11 at 0:09

## 5 Answers

Here is a different answer, but it is much less elementary. If every module is free, then every module is projective, so every module is semi-simple, and the ring is an Artinian semi-simple ring, so a direct product of matrix rings over division rings. Clearly any ring direct summand is a projective non-free module, so we have a matrix ring over a division ring. However, the natural module is a projective non-free module unless the matrix ring is of degree 1, that is, unless it is already a division ring. In other words, apply Artin–Wedderburn theory.

-
You probably mean "...so every module is semi-simple, and the ring is semi-simple". Anyway, why is every projective module semisimple? Can this be found in some textbook? Furthermore, we have $R\!\cong\!R_1\!\times\!\ldots\!\times\!R_k$, where $R_i\!=\!M_{n_i}(D_i)$. Then $R_i$ is an ideal of $R$, so a left $R$-module. How do I know it's not free? Uniqueness from the Artin-Wedderburn theorem? If $R\!\cong\!M_n(D)$, how do I know $n\!=\!1$? –  Leon Oct 26 '11 at 3:13
@Leon: "Artinian semisimple" is the unambiguous form of "semisimple ring" since some people consider the integers to be a semisimple ring (while I would call it a J-semisimple ring). If every module is projective, then every module is semisimple ( A ≤ B implies B/A is projective implies A is a direct summand of B). The key thing about the ring direct factors is they are two-sided ideals, so the earlier argument works: they have a nontrivial annihilator. Finally: D^n is an Mn(D) module, but its D-dimension is wrong to be free. –  Jack Schmidt Oct 26 '11 at 4:38
Here is a clarified solution of Jack, coherent with definitions and propositions in Grillet's book: If every left $R$-module is free, then every left $R$-module is projective, so $R$ is semisimple. Therefore by Artin-Wedderburn theorem, $R\!\cong\!R_1\!\times\!\ldots\!\times\!R_k$ where $R_i\!=\!M_{n_i}(D_i)$ and $D_i$ are division rings. Every $0\!\times\!\ldots\!\times\!R_i\!\times\!\ldots\!\times\!0\equiv R_i$ is an ideal of $R$, so a submodule of $_RR$, thus $R_i$ is free by assumption. But if $k\!\geq\!2$, then $R_i$ is not a free $R$-module: $\mathrm{Ann}_R(R_i)\!\neq\!0$ since –  Leon Oct 26 '11 at 16:27
$(r_1,\ldots,r_{i-1},0,r_{i+1},\ldots,r_k)R_i\!=\!0$, $\rightarrow\leftarrow$. This proves $k\!=\!1$, so $R\!\cong\!M_n(D)$. Furthermore, if $n\!\geq\!2$, then $D^n$ is a $M_n(D)$-module which is not free: if $D^n\!\cong\!\bigoplus_{i\in I}\!M_n(D)$, then they are also isomorphic as $D$-modules, yet $\dim_DD^n\!=\!n\!\neq\!(n^2)^{|I|}\!=\!\dim_D\bigoplus_{i\in I}\!M_n(D)$, $\rightarrow\leftarrow$. This proves $n\!=\!1$, which means $R\!\cong\!M_1(D)\!\cong\!D$. $\blacksquare$ –  Leon Oct 26 '11 at 16:27
Thank you very much for a creative solution, and also for your counterexample. I originally hoped for a more elementary solution, but I like yours very much, since it shows how to use different fields of algebra in an elegant way; everything is just a part of one beautiful theory. I have to tick your answer as the accepted one. :) –  Leon Oct 26 '11 at 16:35

I will paraphrase Pete Clark's "Commutative algebra" notes (pp. 24-25), available here.

As Julian's answer and Amitesh's comment point out, if $R$ is a commutative ring, then if $R$ was not a field, there would exist a two-sided proper ideal $I$. Then $R/I$ would be a nontrivial $R$-module with $0\not=I=ann(R/I)$, whence $R/I$ would be a nonfree $R$-module.

For the noncommutative case: a ring with no nonzero proper twosided ideals may admit a free module. Prof. Clark constructs an example: I quote,

Noncommutative Remark: If $R$ is a non-commutative ring such that every left $R$-module is free, then the above argument shows R has no nonzero proper twosided ideals, so is what is called a simple ring. But a noncommutative simple ring may still admit a nonfree module. For instance, let $k$ be a field and take $R = M_2 (k)$, the $2\times 2$ matrix ring over $k$. Then $k\oplus k$ is a left R-module which is not free. However, suppose $R$ is a ring with no proper nontrivial one-sided ideals. Then $R$ is a division ring – i.e., every nonzero element of $R$ is a unit – and every $R$-module is free.

At the end he asserts what you have been trying to prove. As mentioned by him in the comments, this is expanded in his noncommutative algebra notes, p.6:

Every left $R$-module is free $\Rightarrow$ $R$ is a division ring:

(We follow an argument given by Manny Reyes on MathOverflow.) Let $I$ be a maximal left ideal of $R$ and put $M = R/I$. Then $M$ is a simple left $R$-module: it has no nonzero proper submodules. By assumption $M$ is free: choose a basis $\{x_i\}_{i\in I}$ and any one basis element, say $x_1$. By simplicity $Rx_1 = M$. Moreover, since $x_1$ is a basis element, we have $Rx_1 \cong R$ as $R$-modules. We conclude that as left $R$-modules $R\cong M$, so $R$ is a simple left $R$-module. This means it has no nonzero proper left ideals and is thus a division ring.

-
Well, I have to give this +1. But let me also mention that in $\S 1.6$ of my (not nearly so voluminous) non-commutative algebra notes -- math.uga.edu/~pete/noncommutativealgebra.pdf -- I discuss this sort of thing in some detail (including proofs, of course). –  Pete L. Clark Oct 26 '11 at 4:11
@Pete I wonder how many times your notes have been mentioned on this website because I have certainly lost count! We should perhaps add the following to the FAQ: if your question is on commutative algebra, then look at Pete's notes before asking. –  Amitesh Datta Oct 26 '11 at 7:08
@Pete: I second Amitesh's comment, for sure, and I thank you for a wonderful set of notes! I didn't know the noncommutative algebra notes; I looked up the argument there and expanded the answer (which, in fact, was a bit confusing, as I myself was confused by the matter of sidedness). Thank you. –  lentic catachresis Oct 26 '11 at 12:10
The quotient by maximal left ideal argument is simple and clear, I like it much more than my Artin–Wedderburn round-a-bout. –  Jack Schmidt Oct 26 '11 at 15:15
Thank you Bruno, Pete, and Amitesh. This is the elegant elementary solution I had hoped for from the start. The only reason I didn't accept it is the creativity of Jack's solution. Again, thanks everyone. –  Leon Oct 26 '11 at 16:42

Look at the $R$-module $R/I$. Show that it can't have a basis.

-
Trying to prove $R/I$ is a $R$-module without a basis: by a similar argument as in the opening post, if $R/I$ had a basis, then it'd consist of a single element, so $R/I$ is cyclic, i.e. $R/I=R\!\cdot\!(r_0\!+\!I)$. Uhm, what now... –  Leon Oct 25 '11 at 21:07
@Leon Let $x$ be a non-zero element of $R/I$. Prove that the annihilator $\text{Ann}_{R}(x)=\{r\in R:rx=0\}\neq 0$ if $I\neq 0$. In particular, $\{x\}$ is not an $R$-linearly independent tuple for any such $x$. Of course, an $R$-basis of $R/I$ would have to contain a one-element $R$-linearly independent tuple, unless it is empty, in which case $R/I=0$. –  Amitesh Datta Oct 25 '11 at 21:42
@AmiteshDatta: How do you prove that annihilator inequality for non-commtuative rings? What if x is not in the center of R? It doesn't seem clear to me that I has to do any annihilating at all. –  Jack Schmidt Oct 25 '11 at 21:45
@Leon: there is no reason that b*r should be in I. I is a left ideal, not a right ideal. –  Jack Schmidt Oct 25 '11 at 22:25
@Leon: Counterexample found. This is not a sufficient argument. Take a ring lacking IBN, in particular $R=End(k^\infty)$, then $R$ contains a nonzero proper left ideal $I$ such that $R/I$ is free (of finite rank of your choice). –  Jack Schmidt Oct 26 '11 at 1:01

Note: As Jack Schmidt points out below, this doesn't actually work for a noncommutative ring:

Alternatively: assume that $R$ is not a division ring. Select a nonzero non-unit $a\in R$ and adjoin an inverse of $a$ to $R$. Then $R[a^{-1}]$ is a non-free $R$-module.

-
Does this work for non-commutative rings? How do you define R[a^-1] so that R[a^-1] = R when a is a unit? –  Jack Schmidt Oct 25 '11 at 21:39
I'm assuming that $a$ is not a unit, so I don't really care what happens to $R[a^{-1}]$ when it is. For a non-commutative ring, possibly the inverse being adjoined should be specifically a right inverse (or a left one? figure it out!) and it may be necessary to choose $a$ as an element that does not already have such a one-sided inverse. –  Henning Makholm Oct 25 '11 at 21:57
sorry, I should be more direct. There is no such things as R[a^-1]. You cannot invert elements of non-commutative domains in general. –  Jack Schmidt Oct 25 '11 at 22:24
What I'm imagining is the ... hmm, I see. The argument I had in mind doesn't quite work in the non-commutative case. –  Henning Makholm Oct 25 '11 at 22:41

Cite planetmath:

Assume now that every left $R$-module is free. In particular every left $R$-module is projective, thus $R$ is semisimple and therefore $R$ is Noetherian. This implies that $R$ has invariant basis number. Let $I\subseteq R$ be a nontrivial left ideal. Thus $I$ is a $R$-module, so it is free and since all modules are projective (because they are free), then $I$ is direct summand of $R$. If $I$ is proper, then we have a decomposition of a $R$-module $$R\simeq I\oplus I',$$ but rank of $R$ is $1$ and rank of $I\oplus I'$ is at least $2$. Contradiction, because $R$ has invariant basis number. Thus the only left ideals in $R$ are $0$ and $R$. Now let $x\in R$. Then $Rx=R$, so there exists $\beta\in R$ such that $$\beta x=1.$$ Thus every element is left invertible. But then every element is invertible. Indeed, if $\beta x=1$ then there exist $\alpha\in R$ such that $\alpha\beta =1$ and thus $$1=\alpha\beta=\alpha(\beta x)\beta=(\alpha\beta)x\beta=x\beta,$$ so $x$ is right invertible. Thus $R$ is a divison ring. $\square$

-