# Size of a field with field extensions is hopefully equal to $p^n-1$

I've been given some terminology that I'm not sure is correct.

I have a finite field of prime multiplicative order $$p$$. Then I "adjoin" square roots of numbers, whose square does not exist modulo $$p$$. For example, modulo $$5$$, we have

$$1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 4, 4^2 \equiv 1$$

My understanding is that I can create a second value, say $$\sqrt{2}$$, that doesn't exist modulo $$5$$, and use it like I would use the imaginary number $$i$$ in the complexes modulo $$5$$.

So what I'm trying to do is use a set of $$n$$ square roots to extend the primes to $$p^n - 1$$ numbers plus a zero value, just like with complexes there are $$p^2 - 1$$ unique values besides $$0$$. I believe that I can assume the field is closed with respect to addition, subtraction, and multiplication if the product of any of these square roots does not create an additional, "new" sqauare root.

My question is, do these fields have $$p^n$$ total elements, including zero, and what can I call them? I guess I'm sort of looking for a crash course in "field extensions".

• I don't understand either what you mean by "just like with complexes there are $p^2-1$ unique values", or by "multiplicative prime order $p$". Could you give more details about the complex case ? I think i can answer your question, but want to be sure I have a full understanding. – GreginGre Nov 13 '19 at 9:46
• There are no finite fields of order $5+1$. A finite field of prime multiplicative order must have characteristic $2$ and order $M_q+1$ where $M_q$ is a Mersenne prime. – lhf Nov 13 '19 at 10:41
• @GreginGre: Thanks for your interest. I was really trying to say that there are $p^2 - 1$ nonzero values, not unique values. The complex case was just an example. I really have numbers like $c_1+c_2 \sqrt{d_2} + c_3 \sqrt{d_3} + c_4 \sqrt{d_4} + \dots + c_n \sqrt{d_n}$, all modulo $p$, for $d_k$ fixed for any $k$, and $c_j$ varying. If my intuition is correct, there will be a total of $p^n-1$ nonzero values possible, which arise from different combinations of $c_j$. I assume that if the square roots don't exist, then they can be used as "field extensions", or whatever the term is. – Matt Groff Nov 13 '19 at 23:55
• @GreginGre: I also think that I meant to talk about multiplicative order modulo $p$, for $p$ prime. – Matt Groff Nov 14 '19 at 0:21

Let us try to add $$\sqrt 2$$ to the field $$K = \mathbb Z / 5 \mathbb Z$$, as in your example. Note that you want to have an element $$x$$ in this extension that satisfies $$x^2 -2 = 0$$. Now it turns out that the minimal way to accomplish this is to consider the field $$L = K[x]/(x^2-2),$$ where $$K[x]$$ is the polynomial ring over $$K$$ in one variable and $$x$$ is an indeterminate. Now $$L$$ is indeed a field since $$(x^2-2)$$ is a maximal ideal. You can think about this field as follows. An element is an equivalence class of polynomials, and the polynomial $$x$$ satisfies $$x^2 = 2$$. This should show you how to do arithmetic in this thing.

Alternatively, if you haven't done any ring theory, a possibly more instructive way of viewing this field (also pointed out by @GreginGre) is to have it be $$L = \{a + b\sqrt{2} \mid a,b\in K\}$$ (which is isomorphic to the previous $$L$$), which maybe makes it more obvious that $$L$$ is a vector space over $$K$$.

More generally, any field is a vector space over a subfield, so if you add things to $$\mathbb Z/p \mathbb Z$$ (for $$p$$ prime) you will get a field of $$p^n$$ elements, where $$n$$ depends on what you added. In our case we have $$n = 2$$ (because the degree of $$x^2-2$$ is $$2$$). But you see, if $$\sqrt 3$$ doesn't by chance happen to be an element of $$L$$ (it won't I guess), you will have to extend $$L$$ by $$\sqrt 3$$, yielding a field of $$(5^2)^2$$ elements.

As it turns out, I guessed wrong (see Jyrki Lahtonen's comment). The fact than all finite fields of order $$p^n$$ are isomorphic implies that whenever you add one square root to $$\mathbb Z/p \mathbb Z$$, you in fact add all the square roots.

• You seem to know what I'm trying to do, +1 for your help. I would agree that, generally, if you add $\sqrt{2}$ to $K$ to get $L$, and then add $\sqrt{3}$ to $L$, you get a vector space of ${p^2}^2$ elements. What I'm also wondering, though, is that if $\sqrt{2 \cdot 3}$ essentially already exists modulo $p$, or in $L$, then I believed that you'd get a field of ${p^2}\cdot p$ elements, instead of the full ${p^2}^2$ elements. I'm not sure that this is exactly how things work, though. – Matt Groff Nov 14 '19 at 0:09
• @MattGroff You are getting there, but there are also a few subtleties and pitfalls. Your train of thought hits one of them already. It makes no difference whatsoever whether you adjoin $\sqrt{2}$ or $\sqrt{3}$ to extend $\Bbb{F}_5$. You see, in $\Bbb{F}_5$, we have $3=-2$ (because they are congruent modulo $5$). We also have $-1=4=2^2$. So if we added $u=\sqrt{2}$, then we get $\sqrt3$ free of charge! All because $$(2u)^2=4u^2=-u^2=-2=3.$$ The same thing happens whenever you adjoin square roots to a prime field $\Bbb{F}_p$. When you add one of them, you get the rest free of charge. – Jyrki Lahtonen Nov 14 '19 at 5:56
• (cont'd) It is possible to get fields of $p^4$ elements (with $p^4-1$ non-zero). But you need an irreducible degree four polynomial to construct them. – Jyrki Lahtonen Nov 14 '19 at 5:57
• Levi, this is a fine description of a quadratic extension, but you ALWAYS get all the square roots by adjoining one that was not missing. This has to do with quadratic residue and quadratic non-residues. The ratio (or the product) of two quadratic non-residues is always a quadratic residue. Implying the result generalizing the fact $\sqrt3\in\Bbb{F}_5(\sqrt2)$ explained two comments up. – Jyrki Lahtonen Nov 14 '19 at 5:59