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

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

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

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).

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 $mathbb{Q}$.

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 $Lsupsetneq K$, and then $L$ and $K^{[L:K]}$ have the same dimension and are not isomorphic.

Even if $K$ is algebraically closed, $K^4$, $M_{2times 2}(K)$ 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 $Acong k^n$ for some $n$.

Another very familiar example: $mathbb{C}neqmathbb{R}times mathbb{R}$. 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.

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.