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Let $F=(F,+,.)$ be a field, then $F$ must satisfy all field axioms, namely both $+$ and $.$ are associative, commutative, invertible, and unital, also $.$ is distributive over $+$.

Let $A$ be a one-dimensional vector space of $F$ where both addition and multiplication are defined. Call $A$ an one-dimensional $F$ algebra.

How many such algebras are there up to isomorphisms?

Clearly $F$ is a one-dimensional algebra over itself, but that is the only one I can come up with.

I guess I have to clarify something. An algebra $A$ of $F$ is a vector space over the field $F$ where multiplication is defined. For example, $\mathbb{C}$ is an algebra of $\mathbb{R}$, since $\mathbb{C}$ is a 2-dimensional vector space of $\mathbb{R}$ AND multiplication is defined as $(x_1,y_1).(x_2,y_2) = (x_1x_2-y_1y_2,x_1y_2+x_2y_1)$.

So the question really is, how many possible ways up to isomorphism are there to define multiplication in $F$.

Let me add that the algebra does not have to be unital.

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  • $\begingroup$ Well, a homomorphism of $F$ algebras is $F$ linear by definition, so trivially that is the only one. It sounds a little bit like you might not be assuming $F$ linearity though, just a ring homomophism between the two. Please clarify. $\endgroup$
    – rschwieb
    Commented Jun 30, 2017 at 2:19
  • $\begingroup$ If $A$ contains $F$ as a subalgebra and $A$ is one-dimensional, then $A=F$. $\endgroup$
    – anon
    Commented Jun 30, 2017 at 2:23
  • $\begingroup$ Could it be the case that $A$ is a subset of $F$? $\endgroup$
    – Sid Caroline
    Commented Jun 30, 2017 at 2:24
  • $\begingroup$ @SidCaroline Up to isomorphism, vector spaces are uniquely determined by their dimension, i.e. if $V$ is a 1-dimensional $F$-vector space, then $V\cong F$, where we consider $F$ as a 1-dimensional vector space over itself. $\endgroup$
    – Patrick Schackmar
    Commented Jun 30, 2017 at 10:46

1 Answer 1

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An algebra $A$ over a commutative ring $R$ is a ring equipped with a ring homomorphism $\varphi\colon R\to A$ such that the image of $\varphi$ is contained in the center of $A$.

Clearly, this defines a structure of $R$-module on $A$.

If $R=F$ is a field, then $\varphi$ is necessarily injective, so $F$ embeds in $A$. If $A$ is one-dimensional vector space over $F$ (that is, $F$-module), then $\varphi$ is surjective.

(Note: all rings are supposed to have an identity and the ring homomorphisms to be unital.)

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  • $\begingroup$ I have two questions: 1. why it it the case that "this clearly defines a structure of $R$-module on $A$? 2. Why is $\phi$ necessarily injective if $R$ is a field? $\endgroup$
    – Sid Caroline
    Commented Jun 30, 2017 at 23:31
  • $\begingroup$ @SidCaroline Define, for $r\in R$ and $a\in A$, $ra=\varphi(r)a$ (in the right-hand side multiplication in $A$ is used); just check the axioms are satisfied. A field $F$ has only $(0)$ and $F$ as ideals, so the kernel of $\varphi$ is $(0)$. $\endgroup$
    – egreg
    Commented Jun 30, 2017 at 23:40

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