multiplicative character evaluated at -1 (from Ireland and Rosen's number theory book) I'm self studying from Ireland and Rosen's  "A Classical Introduction to Number Theory" Second edition. Near the beginning (page 153 in my book) of Chapter 11 the authors discuss the number of solutions of a hypersurface in a finite field.
At one point they involve the character of order 2 on a finite field of $q^s$ elements. Let this character be denoted by X$q^s$.
They then mention that if $-1$ is not a square in the prime subfield of q elements, then then  X$q^s$(-1)=-1 for s odd and X$q^s$(-1)=1 for s even.
I don't understand why this is. 
Thanks for the help.

I don't have the book with me, but I think your problem lies in the sentence

They then mention that if -1 is not a square in the field...

which should really read

... if -1 is not a square in $\mathbf F_q$...

(in other words, in this sentence "the field" is $\mathbf F_q$, not  $\mathbf F_{q^s}$.)

On a finite field with $q^s$ elements (where $q$ is odd), there is precisely one quadratic character, the Legendre symbol:
$$\mathbf F_{q^s}^* \to \pm 1$$
$$ a \mapsto \left(\frac{a}{q^s}\right) = a^{(q^s-1)/2}.$$
Now $(-1/q^s) = (-1)^{(q^s-1)/2}$ is $+1$ if and only if $q^s \equiv 1 \mod 4$. This happens either if $q \equiv 1 \mod 4$ and $s$ is arbitrary, or $q \equiv 3 \mod 4$ and $s$ is even. In other words, if $-1$ is not a square in $\mathbf F_{q^s}$, then $s$ is odd and $-1$ is not a square in $\mathbf F_{q}$ (and conversely).
