$\Bbb R$-algebras of rank $1$

Question

I am trying to understand the definition of an associative unital $R$-algebra (without assuming commutativity). Let $R=\Bbb R$ be the real numbers. Then I first want to understand $\Bbb R$-algebras with low rank.

To define an $\Bbb R$-algebra $A$ of rank $n$, I have to give an $\Bbb R$-basis to $A$, where I treat $A$ as an $\Bbb R$-module. Say I choose $\{x_1,\dots,x_n\}$. I do not assume that $A$ is commutative, so next I have to define all of the pair-wise products $x_ix_j=c_{ij}\in A$ for $i,j\in\{1,\dots,n\}$.

If I try to do this in rank $1$, I take $A$ to be generated by $x$ where $x$ is a formal symbol not in $\Bbb R$. Then all the elements are of the form $\{ax\mid x\in\Bbb R\}$. I also must $x\cdot x\in A$, so I choose $x\cdot x = qx$. Then if I want to check if this is just isomorphic as an $\Bbb R$-algebra to $\Bbb R$, I claim that I can define the map by $x\mapsto 1$ and use $\Bbb R$-linearity. $$q=q\varphi(x)=\varphi(qx)=\varphi(x\cdot x)=\varphi(x)\varphi(x)=1,$$ so this is not an isomorphism unless $q=1$. Then there is an isomorphism class of $\Bbb R$-algebras of rank $1$ for every $q\in\Bbb R$?

Or I am doing something wrong, and there is only one $\Bbb R$-algebra of rank $1$ up to isomorphism for some reason (which is definitely true at the level of the module structure).

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edit: I just realised I forgot, I have to send $1_A$ to $1_{\Bbb R}$. So I determine what $1_A$ is. Let $1_A=ix$, then this satisfies $ax\cdot ix = aiqx\implies i=q^{-1}$. So that my $\Bbb R$-linear map just send $q^{-1}x$ to $1$. Then $$q^{-1}\varphi(x)=\varphi(q^{-1}x)=1\implies \varphi(x)=q.$$ Then $\varphi(ax)=aq$ and this is a bijective ring homomorphism as long as $q\in\Bbb R-\{0\}$. So maybe there are two isomorphism classes? One with $q=0$ and one with $q\ne 0$?

Answer 1

There is a unique associative unital $\Bbb{R}$-algebra of rank $1$.

Proof.

Let $A$ be an $\Bbb{R}$-algebra of rank $1$.

Since $\Bbb{R}$ is a field, the map $\Bbb{R}\to A$ defined by $r\mapsto r\cdot 1$ is a unital ring homomorphism and therefore injective (since $A\ne 0$, because it has rank 1). However, since $\Bbb{R}$ and $A$ both have dimension $1$, it must also be surjective. Thus it is an isomorphism.

In general, let $k$ be a field. All rank $n$ algebras over $k$ are quotients of $k\langle x_1,\ldots,x_{n-1}\rangle $. ($n \ne 0$)

Proof.

Let $A$ be a rank $n$ $k$-algebra. Since $A\ne 0$, $1\ne 0$ in $A$.

Extending $\{1\}$ to a basis, we let $1,X_1,\ldots,X_{n-1}$ be a basis for $A$. Then define $\phi:k\langle x_1,\ldots,x_{n-1}\rangle \to A$ by $x_i\mapsto X_i$. Since $1,X_1,\ldots,X_{n-1}$ are in the image of $\phi$, $\phi$ is surjective. Thus $A$ is a quotient of $k\langle x_1,\ldots,x_{n-1} \rangle$ by the first isomorphism theorem.

The reason this is a generalization is that when $n=1$, we have that every rank $1$ algebra is a quotient of $k$, which immediately tells us that the unique rank $1$ algebra over any field is just $k$.

Answer 2

It's pretty easy to show, directly from definitions and first principles, that any $1$-dimensional $\Bbb R$-algebra having a non-trivial multiplication operation is isomorphic to $\Bbb R$ itself. In fact, we may replace $\Bbb R$ by any field $F$ and obtain the analogous result; this we shall do below.

We recall that the multiplication of an algebra $\mathcal A$ is trivial provided the product of any two elements is zero,

$\forall a, b \in \mathcal A, \; ab = 0; \tag 1$

likewise, the multiplication is non-trivial when

$\exists a, b \in \mathcal A, \; ab \ne 0.\tag 2$

Now an arbitrary finite-dimensional algebra $\mathcal A$ over any field $F$ has a basis of $n$ elements, where

$n = \dim_F \mathcal A;\tag 3$

if we denote the elements of a basis by

$x_1, x_2, \ldots, x_n \in \mathcal A, \tag 4$

then the multiplication on all of $\mathcal A$ may be specified by defining it on the $n^2$ basis products $x_ix_j$, and since

$\mathcal A = \text{span} \{ x_1, x_2, \ldots, x_n \}, \tag 5$

for every $1 \le i, j \le n$, we may find

$c_{ij}^k \in F, \; 1 \le i, j, k \le n, \tag 6$

such that

$x_i x_j = \displaystyle \sum_{k = 1}^n c_{ij}^k x_k; \tag 7$

in this way we define a product on all of $\mathcal A$, since the $e_i$ are a basis.

We may apply this program in the case

$\dim_F \mathcal A = 1, \tag 8$

a particularly simple undertaking since then there is only the coefficient $c_{11}^1$.

In light of (8), we may pick any

$0 \ne x \in \mathcal A, \tag 9$

and then (7) becomes

$x^2 = \beta x, \; \beta \in F; \tag{10}$

it is clear that the multiplication so specified is trivial if and only if $\beta = 0$; therefore we take

$\beta \ne 0. \tag{11}$

We may now consider the possible existence of a multiplicative unit in $\mathcal A$; such a unit must be of the form $\alpha x$, $\alpha \in F$, and satisfy

$(\alpha x)(\gamma x) = \gamma x \tag{12}$

for any $\gamma \in F$; then since

$\gamma x = (\alpha x)(\gamma x) = \alpha \gamma x^2 =\alpha \gamma \beta x, \tag{13}$

taking

$\gamma \ne 0, \tag{14}$

we see that (13) forces

$\alpha \beta = 1; \tag{15}$

that is,

$\alpha = \beta^{-1}; \tag{16}$

we may check this against (12):

$(\alpha x)(\gamma x) = \alpha \gamma x^2 = \alpha \gamma \beta x = \gamma x, \tag{17}$

verifying that $\alpha x$ is indeed a unit for $\mathcal A$.

Having established that $\mathcal A$ is unital, we search for multiplicative inverses; for

$\gamma x \ne 0, \tag{18}$

we seek $\delta \in F$ such that

$(\gamma x)(\delta x) = \alpha x, \tag{19}$

whence

$\gamma \delta \beta x = \gamma \delta x^2 = \alpha x; \tag{20}$

according to (16) this yields

$\gamma \delta \alpha^{-1}= \alpha, \tag{21}$

or

$\delta = \gamma^{-1} \alpha^2; \tag{22}$

we check:

$(\gamma x)(\delta x) = (\gamma x)( \gamma^{-1} \alpha^2 x^2) = \alpha^2 \beta x = \alpha x, \tag{23}$

the unit of $\mathcal A$, and we have established the existence of inverses for the non-$0$ elements of $\mathcal A$, now seen to be a field. We may thus present an isomorphism

$\theta: F \cong \mathcal A, \theta(1_F) = \alpha x, \tag{24}$

and extending to all of $F$ by linearity:

$\theta(\gamma1_F) = \gamma\theta(1_F) = \gamma \alpha x; \tag{25}$

the reader may easily verify $\theta$ is an isomorphism, for example

$\theta(\gamma \delta) = \gamma \delta \alpha x = \gamma \delta \alpha x \alpha x = (\gamma \alpha x)(\delta \alpha x) = \theta(\gamma)\theta(\delta); \tag{26}$

the other axioms for field isomorphisms may also be readily validated for $\theta$.

We have established that every $1$-dimensional $F$-algebra with a non-trivial multiplication is isomorphic the the field $F$ itself. As for $1$-d algebras over $F$ with trivial multiplication, any two $\mathcal A_1$ and $\mathcal A_2$ are isomorphic in a number of ways; indeed, for any $x \in \mathcal A_1$ and $y \in \mathcal A_2$ we may set

$\phi: \mathcal A_1 \to \mathcal A_2, \; \phi(x) = y, \tag{27}$

and extend $\phi$ to all of $\mathcal A_1$ via

$\phi(\gamma x) = \gamma \phi(x) = \gamma y; \tag{28}$

such $\phi$ are also additive, since

$\phi(\gamma_1x + \gamma_2x) = \phi((\gamma_1 + \gamma_2)x) =(\gamma_1 + \gamma_2) \phi(x) =\gamma_1 \phi(x) + \gamma_2 \phi(x) = \gamma_1y + \gamma_2y ; \tag {29}$

as far as concerns multiplication, we have

$\phi((\gamma_1 x)(\gamma_2 x)) = \phi(0) = 0 = \phi(\gamma_1 x) \phi (\gamma_2 x); \tag{30}$

we thus see that any such $\phi$ as in (28) is an isomorphism of the $F$-algebras $\mathcal A_1$ and $\mathcal A_2$.

We remark that a one-dimensional $F$-algebra with non-trivial multiplication may also be examined in terms of the presence of non-vanishing idempotent elements; that is, we seek

$0 \ne \alpha \in F, \; (\alpha x)^2 = \alpha x, \tag{31}$

from which

$\alpha^2 \beta x = \alpha^2 x^2 = (\alpha x)^2 = \alpha x; \tag{32}$

thus again as above,

$\alpha \beta = 1, \tag{32}$

which as we have seen above ((12)-(17)) shows that $\alpha x$ is the multiplicative identity of $\mathcal A$. Similarly, in the event that $\mathcal A$ is multiplicatively trivial,

$x^2 = 0, \; \forall x \in \mathcal A; \tag{33}$

every element of $\mathcal A$ is nilpotent.

In summary, we see that $\mathcal A$ is always isomorphic to $F$ when considered as a mere $F$-vector space; when considered as an $F$-algebra, however, $\mathcal A$ may either be isomorphic to $F$ or to the multiplicatively trivial algebra in which every product is $0$. There are thus precisely two isomorphism classes of $F$-algebras satisfying (8).