# In what way is a quadratic extention to a finite field isomorphic to a finite field of higher order?

I have read (I don't remember where) that a finite field that is quadratically extended, say $\mathbb F_p[\sqrt 3]$ for example, is isomorphic to the finite field $\mathbb F_{p^2}$ (assuming the fields are not of characteristic 2). Is this correct, and if so, what is the nature of the isomorphism? I have not been able to find a good explanation for this, and I'm new to Galois Theory and Field Theory in general, so please try to keep explanations geared toward the layman and include examples. I'm also interested in any references that describe this well.

There are a few things you need for this. First, finite fields are uniquely characterized by their size, so any two finite fields of the same size are isomorphic. This can be proved by showing that any finite field of size $p^n$ is isomorphic to $\mathbb{F}_p[x]/(x^{p^n} - x)$.
Now, what you wrote isn't always true. For example, when $p = 11$, $\mathbb{F}_p[\sqrt{3}] = \mathbb{F}_p[x]/(x^2-3) = \mathbb{F}_p$ since $x^2 - 3 \equiv (x-5)(x+5)\mod 11$. What we do have is that $\mathbb{F}_p[\sqrt{3}]$ is always a field that contains $\mathbb{F}_p$. From here you can check that whenever you have one field contained in another, the larger field also satisfies the axioms for being a vector space over the smaller field. Also in our situation we know that as a vector space, $\mathbb{F}_p[\sqrt{3}]$ is spanned by $\{1, \sqrt{3}\}$, since every element is of the form $a + b\sqrt{3}$ for $a,b \in \mathbb{F}_p$. So from linear algebra, the dimension of this vector space is at most 2. From this we can conclude the size: $$| \mathbb{F}_p[\sqrt{3}]| = |\mathbb{F}_p|^{d} = p^{d}$$ where $d$ is the dimension of the vector space we described, which is either 1 or 2. So $\mathbb{F}_p[\sqrt{3}]$ is isomorphic either to $\mathbb{F}_{p^2}$ or so $\mathbb{F}_p$, depending on whether $x^2 - 3 = 0$ has a solution in $\mathbb{F}_p$.