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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?

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  • $\begingroup$ 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? $\endgroup$
    – mr_e_man
    Jul 15, 2020 at 14:15
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    $\begingroup$ 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? $\endgroup$ Jul 15, 2020 at 15:59
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    $\begingroup$ @JasonDeVito Giving non-unital counterexamples is cheating! $\endgroup$
    – Earthliŋ
    Jul 15, 2020 at 16:33
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    $\begingroup$ @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.) $\endgroup$ Jul 15, 2020 at 16:34
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    $\begingroup$ $\Bbb{Q}(\sqrt2)$ and $\Bbb{Q}(\sqrt3)$ are not isomorphic as $\Bbb{Q}$-algebras in spite of both being 2-dimensional. $\endgroup$ Jul 16, 2020 at 21:36

6 Answers 6

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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.

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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.

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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 $\mathbb{Q}$.

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    $\begingroup$ 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. $\endgroup$
    – tomasz
    Jul 15, 2020 at 0:51
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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).

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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

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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 $n$.

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  • $\begingroup$ $\mathbb{C}[x]/\{x^2\}$ is finite dimensional and commutative, but not isomorphic to $\mathbb{C}\times \mathbb{C}$. $\endgroup$
    – tkf
    Jul 15, 2020 at 0:15
  • $\begingroup$ @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. $\endgroup$
    – tomasz
    Jul 15, 2020 at 0:18
  • $\begingroup$ I think you just need to add the condition that it does not contain nilpotent elements. $\endgroup$
    – tkf
    Jul 15, 2020 at 0:19
  • $\begingroup$ @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... $\endgroup$
    – tomasz
    Jul 15, 2020 at 0:35
  • $\begingroup$ @tkf: I meant over an algebraically closed field. $\endgroup$
    – tomasz
    Jul 15, 2020 at 15:34

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