Solve equations in a field with characteristic p. Let $p$ be a prime and $K$ a field with characteristic $p$. How to solve the equation $x^2+2=0$ in the field $K$? Here $x, 2, 0 \in K$. Is it equivalent to solve the equation $x^2+2 = 0 \pmod p$? What are the solutions of $x^2+2 = 0 \pmod p$? Thank you very much.
 A: The answer to your question depends on the field.
First of all, note that it is equivalent to the question, "does $K$ contain a square root of $-2$?"
Now, if $x^2+2=0$ does have a solution mod $p$ (i.e. if $-2$ is a square mod $p$), then this solution will also exist in the field $K$.  This is because $K$ contains $\mathbb{F}_p$. This in turn is because $\mathbb{F}_p$ is generated additively by $1$. For example, let $K=\mathbb{F}_3(t)$, the field of rational functions over $\mathbb{F}_3$. Then $1,2$ are both solutions to $x^2+2=0$ mod $3$, thus they are solutions in $\mathbb{F}_3$, and $K$ contains $\mathbb{F}_3$ so they are the solutions there.
On the other hand, if $-2$ is not a square mod $p$, then $K$ may or may not still contain a square root of $-2$. It definitely won't lie in the subfield $\mathbb{F}_p$, but it may lie elsewhere in the field. As examples of both possibilities, let $p=7$. First let $K=\mathbb{F}_7(t)$, the field of rational functions over $\mathbb{F}_7$. Now $x^2+2=0$ has no solutions in $\mathbb{F}_7$ because the only squares are $1,2,4$ and none of these $=-2=5$. And passing to the function field by adjoining $t$ doesn't help, because any nonconstant rational function will have a nonconstant square; in particular, its square will not equal $-2$. So in this case the equation has no solutions in $K$.
On the other hand, let $L= \mathbb{F}_7[t]/(t^2+2)$. This field was constructed precisely by adjoining a square root of $-2$ to $\mathbb{F}_7$. So now the equation definitely has solutions. In fact, the whole point of describing $L$ in this way is that $t^2+2=0$ in this field; thus $t$ is a solution; and because $(-1)^2=1$, $-t$ is the other solution.
Two more general comments:
One, you can determine whether $-2$ is a square mod $p$ using the supplemental theorems of quadratic reciprocity and the fact that the squares form an index 2 subgroup in $\mathbb{F}_p$, thus $-2$ is a square if (and only if) either both $-1$ and $2$ are squares, or neither is a square.
Two, if a field contains solutions to $x^2+2=0$, then what they look like will depend on how the field was expressed. I expressed $L$ as the adjunction of $\mathbb{F}_7$ by a square root of $-2$, so the solution was simply the thing that I adjoined, and its negative. It is less obvious that $\mathbb{F}_7[t]/(t^2+1)$ also contains square roots of $-2$, but actually it does. (In fact, it is an isomorphic field, but this is not the point I'm making.) By construction in this field $t^2+1=0$, thus $t$ is a square root of $-1$; and in $\mathbb{F}_7$, $2$ is already a square: $3,4$ are its square roots. Thus $3t,4t$ are square roots of $-2$ in this field: e.g. $(3t)^2=2t^2=2(-1)=-2$.
