# Isomorphism of sets

What is an isomorphism of sets?

I know in general an isomorphism is a structure-preserving bijective map between two algebraic structures. But what algebraic structure does a set have? Does a function between sets needs to be anything more than bijective to be an isomorphism of sets?

The reason I ask is that the term is used on PlanetMat's Finite Field page in explaining why a field $F$ of characteristic $p$ has cardinality $p^r$, where $r$ is the degree of the extension $F/ \mathbb{F}_p$. It says "Since $F$ is an $r$-dimensional vector space over $\mathbb{F}_p$ for finite $r$, it is set isomorphic to $\mathbb{F}_p^r$ and so has cardinality $p^r$."

-
An "isomorphism of sets" (that is, an isomorphism in the category of $\mathbf{Sets}$) is just a bijection. So what they are saying is that $F$ can be bijected with $\mathbb{F}_p^r$ (since they are both vector spaces over $\mathbb{F}_p$ of the same dimension), but that bijection need not respect the algebraic structure of $F$. –  Arturo Magidin Dec 12 '11 at 21:18
I think that "set isomorphic" in that sentence is just noise. $F$ is isomorphic as a vector space to $\mathbb{F}_p^r$, though not canonically, and hence has the same cardinality. –  lhf Dec 12 '11 at 21:30
Ok, that makes sense. What exactly is meant by "canonically"? I can see $F$ is isomorphic as a vector space to $\mathbb{F}_p^r$ by mapping basis vectors to basis vectors and extending linearly. –  Jonathan Dec 12 '11 at 21:31
@Jonathan: In this case, the fact that the isomorphism is "not canonical" means that (i) there are lots of bijections; and (ii) there is no way to specify a 'prefered' bijection in a way that makes things "easy" or "nice". –  Arturo Magidin Dec 12 '11 at 21:36

An "isomorphism of sets" (that is, an isomorphism in the category of Sets) is just a bijection. So what they are saying is that $F$ can be bijected with $\mathbb{F}^r_p$ (since they are both vector spaces over $\mathbb{F}_p$ of the same dimension), but that bijection need not respect the algebraic structure of $F$.