$F(\sqrt d)$ is an ordered field

Question

I'm trying to solve this exercise: (hartshorne Euclid and beyond ex 15.3)

Let $F$ be an ordered field, let $d>0$, and suppose that $d$ does not have a square root in $F$. Let $F(\sqrt d)$ denote the set of all $a + b\sqrt d$, with $a, b \in F$, where $\sqrt d$ is a square root in some extension field of F. Show how to define an ordering on $F(\sqrt d)$, with $\sqrt d > 0$, such that it becomes an ordered field.

I tried a lot of orderings on $F(\sqrt d)$ (also $a-b\sqrt d$ iff $a^2\geq db^2$) but I always ended up in a lot of counts and cases. does anyone have a simpler solution? If somehow one shows that $F(\sqrt d)\subset \mathbb{R}$ then it is obviously ordered but I can't show it...

(ordering in the sense of the book)

Answer 1

If $a\ge 0,b\ge 0$ then the sign of $a+b\sqrt{d}$ is clear,

while $a-b\sqrt{d}\ge 0$ iff $a\ge b\sqrt{d}$ iff $a^2\ge b^2 d$.

It remains to check that this satisfies the ordered field axioms.

Answer 2

@reuns has certainly given the canonical answer (+1), but here's a solution that avoids any casework. A field admits an ordering if and only if it is formally real, ie if and only if a sum of squares is never equal to $-1$; so we need only show that $F(\sqrt{d})$ has that property. Suppose otherwise for contradiction; then there are $a_i,b_i\in F$ such that $$(a_1+b_1\sqrt{d})^2+\dots+(a_n+b_n\sqrt{d})^2=-1.$$ Expanding out yields $$a_1^2+\dots+a_n^2+d(b_1^2+\dots+b_n^2)+1=-2(a_1b_1+\dots+a_nb_n)\sqrt{d}.$$ Since $F$ is ordered and $d$ is positive, the left hand side is a sum of non-negative elements with $1$ and hence non-zero, so $-2(a_1b_1+\dots+a_nb_n)$ is also non-zero. But that means we can divide out by it and obtain $\sqrt{d}\in F$, giving a contradiction.