Let $A$ be a finite dimensional $\mathbb{C}$-algebra such that for any $0 \neq a \in A,$ there exists $b\in A$ such that $ab = ba =1.$ I want to show that $A = \mathbb{C}.$

First, let $U$ be a nonzero $A$-submodule of $A$. (So considering $A$ as a module over itself.) Take $0 \neq u \in U,$ then there exists $v \in A$ such that $uv = vu = 1,$ so $1 \in A.u \subseteq U,$ implying $A = U.$ In other words, $A$ is a simple $A$-module.

Now consider the $A$-endomorphism of $A$, i.e. a map $f \colon A \to A.$ By Schur's Lemma, $f \equiv \lambda \mathrm{Id}_A$ for some $\lambda \in \mathbb{C}.$ Hence $\mathrm{End}_A(A) \cong \mathbb{C}.$ But $A \cong (\mathrm{End}_A(A))^\mathrm{op} \cong \mathbb{C}^\mathrm{op} \cong \mathbb{C}$ as $\mathbb{C}$ is commutative (where $A^\mathrm{op}$ is the opposite algebra). Is this reasoning correct?

  • $\begingroup$ You may want to use the Artin-Webberburn theorem at some point. en.m.wikipedia.org/wiki/Artin–Wedderburn_theorem $\endgroup$ – Chickenmancer May 15 '17 at 15:16
  • $\begingroup$ @Justin The version of the theorem I learned was that if $A$ is a finite dimensional $\mathbb{C}$-algebra, then $A$ is a direct product of matrix rings over $\mathbb{C}.$ But $A$ is simple, so $A$ is just a direct sum of a single matrix ring, in particular $M_1(\mathbb{C})$ as simple $A$-modules must be $1$-dimensional over $\mathbb{C}$ by Schur's Lemma. Is this what you were implying? $\endgroup$ – dhk628 May 15 '17 at 15:52

By Schur's lemma, $f=\lambda \mathrm{Id}_A$ fr some $\lambda \in \mathbb C$.

No, it seems to me that in this step you are begging the question by referencing a piece of Schur's lemma that depends on what you are trying to prove. Schur's lemma definitely says that this is true for some $\lambda \in End_A(A)$, but then you need the result you are proving to conclude that this is equal to $\mathbb C$.

This should not be a big deal to prove anyway, since you presumably have the fundamental theorem of algebra at hand. Let $\alpha\in A\setminus\mathbb C$ and consider the field extension $\mathbb C\subseteq \mathbb C[\alpha]$. Because the extension is finite dimensional, $\alpha$ satisfies a minimal polynomial over $\mathbb C$. But the FTA says this polynomial factors into linear factors, one of which is $x-\alpha$. So the minimal polynomial must be $x-\alpha$ itself, and so $\alpha\in \mathbb C$ and we have reached a contradiction. So the conclusion is that $A=\mathbb C$ already.

The reasoning here applies more generally to finite-dimensional division ring extensions of algebraically closed fields. To properly extend an algebraically closed field, you will have to use something that is transcendental over the algebraically closed field.

Hitting it with Artin-Wedderburn unfortunately does not solve the problem. You've been told it is a division ring, so in particular its Wedderburn decomposition cannot be a product of more than one matrix ring, and it has to be a matrix ring of side-length $1$, because otherwise there would be zero divisors. At this point you've discovered that your ring is a finite dimensional division ring extension of $\mathbb C$: and you're right back at the problem you need to solve: why is a finite dimensional division ring extension of $\mathbb C$ equal to $\mathbb C$?


I know this has been answered but I can't help giving another proof as it is purely topological.

The map $A^* \to A^*, z \mapsto z^2$ is a covering of degree $2$, where $A^* := A \backslash \{0\}$. Now the total space of this covering is connected so this covering can't be trivial. But the base space is simply connected so this covering should be trivial, contradiction.

  • $\begingroup$ "But the base space is connected so this covering should be trivial" - simply connected? Also, I'm not sure how you prove that $z \mapsto z^2$ is a covering space for an arbitrary finite-dimensional $\mathbb C$-algebra $A$. $\endgroup$ – Dustan Levenstein May 16 '17 at 11:09
  • $\begingroup$ Sure I mean simply connected ! My argument needs more justification. First, I don't assume that $z \mapsto z^2$ is surjective but I believe it's enough to restrict to the image. The preimage of this map have always cardinality two (if restricted to the image and without $0$) and the square root is smooth so this gives local trivialisation of the map which is a covering. $\endgroup$ – user171326 May 16 '17 at 13:12
  • $\begingroup$ I don't see why $A^*$ (or the image of the square map) is simply connected. Indeed $\mathbb{C}^*$ is not simply connected... And $z \mapsto z^2$ is a nontrivial covering $\mathbb{C}^* \to \mathbb{C}^*$. $\endgroup$ – Najib Idrissi May 16 '17 at 13:15
  • $\begingroup$ $A^*$ is simply connected, as I assumed $A$ was a non-trivial extension, so as a vector space we have $A \cong \Bbb R^{2k}$ with $k > 1$. And $\Bbb R^{2k}$ minus a point is simply connected if $k > 1$. But I don't know how to prove it is surjective, so maybe indeed there is a problem showing the image is simply connected. $\endgroup$ – user171326 May 16 '17 at 13:18
  • $\begingroup$ It sounds like the same argument would imply that a finite dimensional division algebra over $\mathbb{R}$ must have dimension $1$ or $2$, and the quaternions give a counterexample. $\endgroup$ – Julian Rosen May 16 '17 at 13:38

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