# Is there an explicit description of the fields laws on this field?

I'm working on a fairly simple problem about a field, but I want to know if the operations can be explicitly described.

Suppose $c$ is not a quadratic residue modulo $p$, and consider the quotient ring $\mathbb{F}_p[X]/(x^2-c)$. Now $x^2-c$ is irreducible over $\mathbb{F}_p$, so it generates a maximal ideal, and thus $\mathbb{F}_p[X]/(x^2-c)$ can be viewed as a 2-dimensional vector space over $\mathbb{F}_p$, and thus has order $p^2$. If I take $a$ to be a root of $x^2-c$ in some extension field, then I can view the elements of the field as $0,a,c,ac,c^2,ac^2,c^3,\dots,ac^{(p^2-3)/2},c^{(p^2-1)/2}$, for a total of $p^2$ elements.

However, I don't know how to actually state what addition and multiplication look like in this field. Is there a clever way to describe the operations explicitly? Thanks.

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I think you meant is irreducible :) – Zev Chonoles Nov 11 '11 at 4:55
Those are not all the elements of the field; you have a lot of repeats, because you know for sure that $c^p=c$ (and possibly smaller powers too). – Arturo Magidin Nov 11 '11 at 5:14
Zev describes below how to do arithmetic in this field. I want to add that your description is incorrect. Because $c$ is an element of the prime field, we have $c^{p-1}=1$ by little Fermat. Thinks may go wrong even earlier. Just because $c$ is not a quadratic residue, it doesn't mean that it would also be primitive. For a most striking example $c=-1$ is not a square in $GF(3)$ or $GF(7)$, but we always have $(-1)^2=1$. – Jyrki Lahtonen Nov 11 '11 at 5:16
Oops, I think I slipped up since the unit group of fields of prime order are cyclic, but that doesn't work here. – Vika Nov 11 '11 at 5:39
@Vika: The unit group of any finite field is cyclic. But you have to look harder to find a generator. – Jyrki Lahtonen Nov 11 '11 at 6:09

## 2 Answers

I would recommend thinking about it as just addition and multiplication of polynomials in $\mathbb{F}_p[x]$, modulo $x^2-c$. Given some polynomial $f\in\mathbb{F}_p[x]$, by the division algorithm we have that $$f=(x^2-c)q+r\quad\text{ where }q,r\in\mathbb{F}_p[x], \text{ and }\deg(r)<2.$$ So for any $f\in\mathbb{F}_p[x]$, we have $$f\equiv a_0+a_1x\pmod{x^2-c}$$ for some $a_0,a_1\in\mathbb{F}_p$, and so the elements of $\mathbb{F}_p[x]/(x^2-c)$ are the $p^2$ different equivalence classes we get as $a_0$ and $a_1$ range over their $p$ possible values. We can think of $a_0+a_1x\pmod{x^2-c}$ as representing "$a_0+a_1\sqrt{c}$ ". Then the operations are just as we would expect: $$(a_0+a_1\sqrt{c})+(b_0+b_1\sqrt{c})=(a_0+b_0)+(a_1+b_1)\sqrt{c}$$ and $$(a_0+a_1\sqrt{c})(b_0+b_1\sqrt{c})=(a_0b_0+a_1b_1c)+(a_1b_0+a_0b_1)\sqrt{c},$$ because $$(a_0+a_1x)+(b_0+b_1x)\equiv(a_0+b_0)+(a_1+b_1)x\pmod{x^2-c}$$ and \begin{align*}(a_0+a_1x)(b_0+b_1x)&\equiv a_0b_0+(a_1b_0+a_0b_1)x+a_1b_1x^2 \\ &\equiv a_0b_0+(a_1b_0+a_0b_1)x+a_1b_1(x^2-0)\\ & \equiv a_0b_0+(a_1b_0+a_0b_1)x+a_1b_1(x^2-(x^2-c))\\ &\equiv (a_0b_0+a_1b_1c)+(a_1b_0+a_0b_1)x\hskip1in\pmod{x^2-c}\end{align*}

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Thanks Zev, (chars). – Vika Nov 11 '11 at 21:32

The elements you list need not be all distinct, so your list need not be the full list of elements of your field.

For example, consider $\mathbf{F}_7$; the squares are $1$, $2$, and $4$. In particular, $-1$ is not a square, so you can take $c=-1$; then your list of elements consists only of $0$, $a$, $-1$, $-a$, and $1$, which is only $5$ elements, not the required $49$.

In fact, your list can never be all the elements, because $c^p = c$ holds, so you are repeating a lot of elements, and not getting all of them.

(Added. The best you can hope for is if $c$ is a primitive element of $\mathbf{F}_7$; that is, it generates the group of units. You still get only $2p-1$ of the required $p^2$ elements, though.)

Instead, you want to think of the elements as being the result of evaluating polynomials at $a$; since every polynomial can be written as a multiple of $x^2-c$ plust a remainder, the elements of the field will be of the form $ra+s$, with $r,s\in\mathbb{F}_p$. You add them the usual way, $(ra+s) + (ta+u) = (r+t)a+(s+u)$, with addition on the right being addition modulo $p$, and you multiply them by using the fact that $x^2=c$, $$(ra+s)(ta+u) = (ru+st)a + (su+rtc).$$

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Thanks Arturo (extra chars). – Vika Nov 11 '11 at 21:32