# Build field extension and solve equation

Build quadratic extension of field that contains $5$ elements. And solve $x^2+x+2=0$ in this field.

As I understand we need to build $\mathbb{F}_{5^{2}}$.

Field $\mathbb{F}_5$ contains $\{0,1,2,3,4\}.$

And as I understand field $\mathbb{F}_{5^{2}}$ contains $\{0,1,2,3,4\}$ and polynomials $\{x+1,x+2,x+3,x+4,2x+, \cdots, 4x+4\}$. $20$ polynomials $ax+b$, where $a,b \in \{0,1,2,3,4\}$. Totally we have $25$ objects in this field.

And I have no idea how to solve equation. Maybe we can try every element, but this is too difficult and here must be simpler way. Thanks for help.

• Find something interesting: if we let $t$ be the root of $x^2+x+2=0$ than $4t+4$ will be the root too! We get that $t^2+t+2=0$. If we put $x=4t+4$ than get: $(4t+4)^2+4t+4+2=16t^2+36t+22(mod5)=t^2+t+2=0$ – Giuseppe Baldinini Jun 29 '14 at 23:03

## 2 Answers

The easiest description of the quadratic extension is that it has the same $25$ elements as yours, with addition defined in the obvious way, and $x^2$ being defined as $-x-2$, that is, $4x+3$. Then $x$ is a solution of your quadratic equation.

Once we have defined $x^2$, the product $ax+b)(cx+d)$ can be defined in the "natural" way: Multiply as usual, and replace $x^2$ by $4x+3$.

An alternative to the trick in André's answer (+1) would be to use the usual quadratic formula. That works over any field where you can complete the square, which is the case whenever you can divide by two. This in turn is always possible, when $2\neq0$, i.e. when the characteristic is not equal to two.

Here we get that $x^2+x+2=0$, iff $$x=\frac{-1\pm\sqrt{1^2-4\cdot2}}2=\frac{-1\pm\sqrt{-7}}2.$$ In order to use this we need to make sense of that $\sqrt{-7}$. In $\Bbb{F}_5$ we have $-7=3$ as $-7\equiv3\pmod5$. Thus we need to locate something that we can call $\sqrt3$. Quick testing reveals that no element $a\in\Bbb{F}_5$ satisfies the equation $a^2=3$, so we need to extend the field.

The way to introduce $\sqrt3$ is the usual. For the given reason $x^2-3$ is irreducible, and thus $\Bbb{F}_5[x]/\langle x^2-3\rangle$ is a field with 25 elements. It has the element $\alpha=x+\langle x^2-3\rangle$ that satisfies $\alpha^2=3$. So in the field $\Bbb{F}_{25}=\Bbb{F}_5[\alpha]$ the original equation has the solutions $(-1\pm\alpha)/2=2\pm3\alpha$. The last step follows from the fact that $2\cdot3=6=1\in\Bbb{F}_5$, so $3=\frac12$ here.

The field $\Bbb{F}[\alpha]$ is, of course, isomorphic to the field André used. After all, there is only one field of 25 elements up to isomorphism. An explicit isomorphism can be given by mapping his (coset of) $x$ to what I called $2\pm3\alpha$. Either sign works.

Note that by creating $\sqrt3$ we also created $\sqrt2$. This is because in all extension fields of $\Bbb{F}_5$ containing a $\sqrt3$ $$2=12=4\cdot3=2^2\cdot3,$$ so $$\sqrt2=\sqrt{2^2\cdot3}=\pm2\sqrt3.$$ Thus $\pm2\alpha$ serve in the role of square roots of two in the field $\Bbb{F}_5[\alpha]$. This is yet another manifestation of the fact that there is (up to isomorphism) only one field of 25 elements.