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

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.

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.

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 $\mathbb{Q}$.

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).