# Are all finite-dimensional algebras of a fixed dimension over a field isomorphic to one another?

Suppose I have a finite-dimensional algebra $$V$$ of dimension $$n$$ over a field $$\mathbb{F}$$. Then $$V$$ is an $$n$$-dimensional vector space and comes equipped with a bilinear product $$\phi : V \times V \to V$$.

Suppose now that I have another finite-dimensional algebra $$W$$ of dimension $$n$$ over $$\mathbb{F}$$ equipped with a bilinear product $$\psi: W \times W \to W$$. Certainly, $$V$$ and $$W$$ are isomorphic as vector spaces but are they isomorphic as $$\mathbb{F}$$-algebras? The question I'm really asking here is - Are all $$n$$-dimensional algebras over $$\mathbb{F}$$ isomorphic to one another?

If the answer is yes, then this is my attempt at constructing such an isomorphism. Suppose I want to define an $$\mathbb{F}$$-algebra isomorphism between $$V$$ and $$W$$.

To do this I'd need to define a map $$f : V \to W$$ such that

• $$f(ax) = af(x)$$ for all $$a \in \mathbb{F}, x \in V$$
• $$f(x+y) = f(x) + f(y)$$ for all $$x, y \in V$$
• $$f(\phi(x, y)) = \psi(f(x), f(y))$$ for all $$x, y \in V$$

If $$\{v_1, \dots, v_n\}$$ and $$\{w_1, \dots, w_n\}$$ are bases for $$V$$ and $$W$$ respectively then both $$\phi$$ and $$\psi$$ being bilinear maps are completely determined by their action on basis vectors $$\phi(v_i, v_j)$$ and $$\psi(w_i, w_j)$$ for $$1 \leq i, j, \leq n$$. It turns out that $$\phi(v_i, v_j) = \sum_{k=1}^n \gamma_{i,j,k}v_k$$ and $$\psi(v_i, v_j) = \sum_{k=1}^n \xi_{i,j,k}w_k$$ for some collection of scalars $$\gamma_{i,j,k}$$ and $$\xi_{i,j,k}$$ called structure coefficients. So then if both the $$n^3$$ collections of scalars $$\gamma_{i,j,k}$$ and $$\xi_{i,j,k}$$ are all non-zero then we can define $$f : V \to W$$ by $$f(a_1v_1 + \cdots + a_nv_n) = a_1 \frac{\xi_{i,j,1}}{\gamma_{i,j,1}}w_1 + \cdots + \frac{\xi_{i,j,n}}{\gamma_{i,j,n}}w_n$$ and it will turn out that $$f$$ is the desired isomorphism of algebras as one can then check that $$f(\phi(v_i, v_j)) = \psi(w_i, w_j) = \psi(f(v_i), f(v_j))$$ for all $$i$$ and $$j$$.

However what if it's the case that for $$\phi$$ some $$\gamma_{i, j, k}$$ is zero and the corresponding $$\xi_{i, j, k}$$ is non-zero? I don't see any way to get an isomorphism in that case. Is it still possible to construct an isomorphism in that case?

• The right side of the equation has $i,j$, not summed or quantified, while the left side has no $i,j$. What do you mean by this? – mr_e_man Jul 15 '20 at 14:15
• Everyrone is working way to hard to provide counter examples. What about the field $\mathbb{F}$ vs. the additive group of $\mathbb{F}$ with $0$-multiplication? – Jason DeVito Jul 15 '20 at 15:59
• @JasonDeVito Giving non-unital counterexamples is cheating! – Earthliŋ Jul 15 '20 at 16:33
• @Earthliŋ: Perhaps it is, but I made sure to read the question for the word "unit" before posting my comment. (I also purposefully didn't provide it as an answer.) – Jason DeVito Jul 15 '20 at 16:34
• $\Bbb{Q}(\sqrt2)$ and $\Bbb{Q}(\sqrt3)$ are not isomorphic as $\Bbb{Q}$-algebras in spite of both being 2-dimensional. – Jyrki Lahtonen Jul 16 '20 at 21:36

They will not necessarily be isomorphic. Consider $$V = \mathbb F[x] / (x^n)$$ and $$W = \mathbb F^n$$ with componentwise multiplication.These are both $$n$$ dimensional $$\mathbb F$$ algebras. However, $$V$$ contains a nilpotent element, $$x$$, whereas $$W$$ contains no nilpotent elements. Indeed, if we had an $$\mathbb F$$-algebra homomorphism $$f: V \longrightarrow W$$ then as $$0 = f(x^n) = f(x)^n$$, we'd need $$f(x) = 0$$ so any map between the two must have a nontrivial kernel.

Another very familiar example: $$\mathbb{C}\neq\mathbb{R}\times \mathbb{R}$$. The complex numbers are a field, but $$(1,0)(0,1)=(0,0)$$ in $$\mathbb{R}\times \mathbb{R}$$, so it has non-trivial zero-divisors.

In general, the answer is "no", even if one requires $$V$$ and $$W$$ to be fields.

For example, the rings $$\mathbb{Q}(\sqrt{2})$$ and $$\mathbb{Q}(\sqrt{3})$$ are two non-isomorphic fields that both have dimension $$2$$ over $$\mathbb{Q}$$.

• I guess if you require $V$ and $W$ both to be fields, then this is true iff $V$ is algebraically closed, real closed or pseudofinite. – tomasz Jul 15 '20 at 0:51

There are several answers that point out why the statement of the question cannot be true, probably the simplest example being $$\Bbbk [x] / (x^2) \not\simeq \Bbbk \times \Bbbk$$.

Classifying all finite-dimensional algebras of a given dimension is actually rather involved and very far from being just one algebra in each dimension.

Note that you can even come up with finite-dimensional noncommutative algebras. For example, from the quiver $$\bullet \to \bullet$$ you can build a noncommutative $$3$$-dimensional algebra with $$\Bbbk$$-basis $$e_1, e_2, \alpha$$, where

• $$e_1, e_2$$ are viewed as "constant paths" at the vertices, which are orthogonal idempotents, i.e. $$e_i e_j = \delta_{ij}$$
• $$\alpha$$ is viewed as corresponding to the arrow and $$e_1, e_2$$ are viewed as "identities at" the vertices, so $$e_1 \alpha = \alpha$$ and $$\alpha e_2 = \alpha$$
• the product of paths which cannot be composed are defined to be $$0$$ in this algebra, so $$e_2 \alpha = \alpha e_1 = \alpha^2 = 0$$ and extending these rules linearly gives a well-defined associative multiplication.

More generally, you can take the path algebra of any quiver and quotient by any two-sided ideal, which if you choose the ideal correctly will give a finite-dimensional algebra, which is usually non-commutative.

Finite-dimensional algebras can be studied via their categories of finite-dimensional modules (which in some cases can actually be described rather explicitly) and it turns out that the construction of finite-dimensional algebras via quivers gives all algebras up to Morita equivalence (i.e. using quivers you find the module categories of all finite-dimensional algebras).

Let $$G$$ and $$H$$ be finite groups of the same order, such that $$G$$ is abelian and $$H$$ is not.

Then the group rings $$V = \mathbb{F}G$$ and $$W = \mathbb{F}H$$ share the same dimension, but $$V$$ is commutative while $$W$$ is not.

https://en.wikipedia.org/wiki/Group_ring

No. For example, $$\mathbf Q[\sqrt n]$$ are pairwise nonisomorphic (where $$n$$ ranges over squarefree integers distinct from $$1$$), but all have dimension $$2$$ over $$\mathbf Q$$.

In general, if $$K$$ is not algebraically closed, then it admits a finite algebraic extension $$L\supsetneq K$$, and then $$L$$ and $$K^{[L:K]}$$ have the same dimension and are not isomorphic.

Even if $$K$$ is algebraically closed, $$K^4$$, $$M_{2\times 2}(K)$$ and $$K[x]/(x^4)$$ are non-isomorphic four-dimensional algebras over $$K$$.

Edit: as suggested in the comments, the counterexamples listed above are essentially all the counterexamples. More precisely, the answer is yes if you restrict yourself to finite-dimensional algebras over an algebraically closed field $$k$$ which are reduced (contain no nilpotents). In other words, the only such algebras are of the form $$k^n$$.

One can show that in this case, the algebra $$A$$ is semisimple (because it is Artinian and the Jacobson radical is zero), so by Wedderburn's theorem, it follows that it is a product of matrix rings over division algebras. Since there are no finite-dimensional division algebras over $$k$$ (because $$k$$ is algebraically closed), and no proper matrix ring is reduced (because it contains strictly upper triangular matrices, which are nilpotent), it follows that $$A\cong k^n$$ for some $$n$$.

• $\mathbb{C}[x]/\{x^2\}$ is finite dimensional and commutative, but not isomorphic to $\mathbb{C}\times \mathbb{C}$. – tkf Jul 15 '20 at 0:15
• @tkf: You're right. For some reason, I thought that such an algebra would automatically be semisimple, but your example shows that this is not the case. I removed the erroneous claim. – tomasz Jul 15 '20 at 0:18
• I think you just need to add the condition that it does not contain nilpotent elements. – tkf Jul 15 '20 at 0:19
• @tkf: I guess commutative+reduced is sufficient, because that makes the algebra the ring of regular functions of a zero-dimensional variety. But I don't see whether or not reduced finite-dimensional implies commutative... – tomasz Jul 15 '20 at 0:35
• @tkf: I meant over an algebraically closed field. – tomasz Jul 15 '20 at 15:34