I want to list all $9$-dimensional $\mathbb{R}$-algebras up to isomorphism. The Wedderburn-Artin theorem helps I guess.

Wedderburn-Artin. Let $R$ be a semi-simple $K$-algebra. There are division $K$-algebras $D_1,...,D_n$ and $m_1,...,m_n\in\mathbb{Z}_{>0}$ such that $$R\cong\text{Mat}_{m_1}(D_1)\times...\times \text{Mat}_{m_n}(D_n)$$ and the pairs $(D_i,m_i)$ are unique up to permutation.

Besides I know that every divison $\mathbb{R}$-algebra is isomorphic to either $\mathbb{R},\mathbb{C}$ or $\mathbb{H}$.

So we have the algebras

  1. $\text{Mat}_{1}(\mathbb{R})^9=\mathbb{R}^9(\cong \mathbb{R}^7\times \mathbb{C}\cong\mathbb{R}^5\times \mathbb{C}^2\cong...\cong \mathbb{R}^1\times \mathbb{C}^4 )$
  2. $\text{Mat}_{3}(\mathbb{R})\cong \text{Mat}_{3}(\mathbb{C})\cong \text{Mat}_{3}(\mathbb{H})$
  3. $\text{Mat}_{2}(\mathbb{R})^2\times\mathbb{R}\cong \text{Mat}_{2}(\mathbb{H})^2\times\mathbb{R}\cong \text{Mat}_{2}(\mathbb{H})^2\times\mathbb{R}$
  4. $\text{Mat}_{2}(\mathbb{R})\times\mathbb{H}\times\mathbb{R}$
  5. $\mathbb{R}^5\times\mathbb{H}$

So my answer is that every semisimple $\mathbb{R}$-algebra is either isomorphic to $\mathbb{R}^9,~\mathbb{R}^5\times\mathbb{H},~\text{Mat}_{3}(\mathbb{R}),~\text{Mat}_{2}(\mathbb{R})^2\times\mathbb{R}$ or $\text{Mat}_{2}(\mathbb{R})\times\mathbb{H}\times\mathbb{R}$.

Is that right?

  • 1
    $\begingroup$ Your first chain of isomorphisms only holds for $\mathbb{R}$-vector spaces, not for $\mathbb{R}$-algebras. For instance, $\mathbb{R} \times \mathbb{H}^2$ is noncommutative (as $\mathbb{H}$ is noncommutative), while $\mathbb{R}^9$ is commutative. $\endgroup$
    – Geoff
    Jan 5, 2017 at 8:49
  • $\begingroup$ You are right. So we have $\mathbb{R}^2\cong\mathbb{C}$ but $\mathbb{R}^4\ncong\mathbb{H}$? Is $\mathbb{R}^5\times\mathbb{H}\cong\mathbb{R}\cong\mathbb{H}$? $\endgroup$
    – user404105
    Jan 5, 2017 at 8:56
  • $\begingroup$ Not quite still. You still have $\mathbb{R}^2 \not\cong \mathbb{C}$ because, among other reasons, $\mathbb{R}^2$ has zero divisors (for any $0 \ne x,y \in \mathbb{R}, (x,0)(0,y) = (0,0)$). I'll write an answer for your question in a little bit that's fairly careful. $\endgroup$
    – Geoff
    Jan 5, 2017 at 9:00
  • $\begingroup$ Thank you, I am looking forward to your answer. $\endgroup$
    – user404105
    Jan 5, 2017 at 9:40

1 Answer 1


In order to classify all semisimple $\mathbb{R}$-algebras of dimension $9$, we will consider things in a few steps. Our two basic insights are as follows: (a) there is only one $\mathbb{R}$ algebra of dimension $1$ ($\mathbb{R}$) and (b) we can determine the semisimple $\mathbb{R}$-algebras of dimension $9$ by classifying the ones without nilpotent elements (the reduced algebras) and then classifying the ones with.

Step 1: We will consider first only reduced $\mathbb{R}$-algebras, i.e., algebras with no nilpotent elements. Since the product of division rings has no nilpotent elements, classifying semisimple reduced algebras is equivalent to figuring out how many different ways we can take the product of finite dimensional division rings over $\mathbb{R}$ and sum the dimensions up to $9$. Since $\mathbb{C}$ and $\mathbb{H}$ are of even dimension over $\mathbb{R}$, we will count all the products of division rings with dimension $8$, as taking the product of these algebras with $\mathbb{R}$ will give a reduced algebra of real dimension $9$.

Begin by noting that $\mathbb{R},\mathbb{C},$ and $\mathbb{H}$ are the only finite dimensional $\mathbb{R}$-division rings and $\mathbb{R}^2 \not\cong \mathbb{C}, \mathbb{R}^4 \not\cong \mathbb{H}$, and $\mathbb{C}^2 \not\cong \mathbb{H}$. Then we have that $$ \mathbb{R}^9, \mathbb{R}^7 \times \mathbb{C}, \mathbb{R}^5 \times \mathbb{C}^2, \mathbb{R}^5 \times \mathbb{H}, \mathbb{R}^3 \times \mathbb{C}^3, \mathbb{R}^3 \times \mathbb{C} \times \mathbb{H}, \mathbb{R} \times \mathbb{C}^2 \times \mathbb{H}, \mathbb{R} \times \mathbb{C}^4, \mathbb{R} \times \mathbb{H}^2 $$ are the distinct nonisomorphic classes of reduced semisimple $\mathbb{R}$-algebras of dimension $9$; we can see this is all of them by counting the distinct ways of getting a dimension eight product of division algebras out of $\mathbb{R}, \mathbb{C},$ and $\mathbb{H}$.

Step 2: We will now classify the isomorphism classes of all nonreduced semisimple $\mathbb{R}$-algebras of dimension $9$. Note that since we care about nonreduced algebras now, we will have nontrivial matrix rings appearing in the product decompositions. We now note that $$ \dim_{\mathbb{R}}(\operatorname{Mat}_{2}(\mathbb{R})) = 4, \quad \dim_{\mathbb{R}}(\operatorname{Mat}_{2}(\mathbb{C})) = 8, \quad \dim_{\mathbb{R}}(\operatorname{Mat}_{2}(\mathbb{H})) = 16$$ and $$9 = \dim_{\mathbb{R}}(\operatorname{Mat}_{3}(\mathbb{R})) < \dim_{\mathbb{R}}(\operatorname{Mat}_{3}(\mathbb{C})) < \dim_{\mathbb{R}}(\operatorname{Mat}_{3}(\mathbb{H})).$$ Since $$ 9 < \dim_{\mathbb{R}}(\operatorname{Mat}_{3}(\mathbb{C})), \dim_{\mathbb{R}}(\operatorname{Mat}_{3}(\mathbb{H})), \dim_{\mathbb{R}}(\operatorname{Mat}_{2}(\mathbb{H})),$$ no semisimple $\mathbb{R}$-algebra of dimension $9$ can have any of the above three matrix rings in their decompositions. We then determine that the nonisomorphic classes of nonreduced semisimple $\mathbb{R}$-algebras of dimension $9$ are $$ \mathbb{R} \times \operatorname{Mat}_{2}(\mathbb{C}), \mathbb{R} \times \left(\operatorname{Mat}_{2}(\mathbb{R})\right)^2, \mathbb{R} \times \operatorname{Mat}_{2}(\mathbb{R}) \times \mathbb{H}, \mathbb{R} \times \operatorname{Mat}_{2}(\mathbb{R}) \times \mathbb{C}^2, \mathbb{R}^5 \times \operatorname{Mat}_{2}(\mathbb{R}),$$ and $$ \mathbb{R}^3 \times \operatorname{Mat}_{2}(\mathbb{R}) \times \mathbb{C}, \operatorname{Mat}_{3}(\mathbb{R}). $$ Analogously to the prior case, we can find the list of all such algebras by counting how many different ways we can get an $8$-dimensional $\mathbb{R}$-algebra that contains at least one matrix ring and then dealing with any outlying cases as needed.

  • $\begingroup$ Wow, thank you! So there are 16 isomorphism clases, right? How do you get for example $\dim_{\mathbb{R}}(\text{Mat}_2(\mathbb{H}))=16(=2\cdot 2\cdot\dim_{\mathbb{R}}(\mathbb{H}))$? $\endgroup$
    – user404105
    Jan 5, 2017 at 10:47
  • $\begingroup$ Well if $R$ is a ring with identity, then as an $R$-$R$ bimodule, $\operatorname{Mat}_{n}(R)$ has rank $n^2$. Since $\operatorname{Mat}_{2}(\mathbb{H})$ has dimension $4$ as a two-sided vector space over $\mathbb{H}$, and since $\mathbb{H}$ is a $4$ dimensional algebra over $\mathbb{R}$, it follows that $\dim_{\mathbb{R}}(\operatorname{Mat}_{2}(\mathbb{H})) = 4\cdot 4 = 16$. For a more concrete reason, try to write down an $\mathbb{R}$-basis to the space of $2\times2$-matrices with entries in $\mathbb{H}$. $\endgroup$
    – Geoff
    Jan 5, 2017 at 11:45
  • $\begingroup$ If $E_{ij}$ are the matrices with the entry $1$ at $(i,j)$ and $0$ elsewhere, the matrices $aE_{ij},bE_{ij},cE_{ij},dE_{ij}$ (a,b,c,d let be a basis of $\mathbb{H}$) form a basis. Is this true? $\endgroup$
    – user404105
    Jan 5, 2017 at 12:51
  • $\begingroup$ Yes, if $\lbrace a, b, c, d \rbrace$ is an $\mathbb{R}$-basis to $\mathbb{H}$, then the set $\lbrace aE_{ij}, bE_{ij}, cE_{ij}, dE_{ij} \; | \; 1 \leq i, j \leq n \rbrace$ describes an $\mathbb{R}$-basis to $\operatorname{Mat}_{n}(\mathbb{H})$. $\endgroup$
    – Geoff
    Jan 5, 2017 at 21:30

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