Okay now for some weird tricks: take any inner product $\langle .,. \rangle$ on $\Bbb R^2$, then define a new inner product by setting $(v,w):= \langle v,w\rangle + \langle Av,Aw\rangle$. Note that since $A^2=I$, we get $(Av,Aw)=(v,w)$. Thus with respect to this new inner product, $A$ is orthogonal! Thus after choosing an orthonormal basis, $A$ will be a rotation or a rotation composed with a reflection.
But since $A^2=I$, the only possibilities are rotation by $180°$, but that's $-1$, so the last remaining possibility is just a reflection, i.e. $A$ is similar to $\begin{pmatrix}-1&0\\ 0&1\…

@MikeMiller anyway, the thing I was actually using the orientation-preserving thing for: if $f: \Bbb{CP}^n \to \Bbb{CP}^n$ is orientation-preserving, then it has a fixed point, right? Since if we take a generator $x$ of $H^2(\Bbb{CP}^n)$, then we get $H^2(f)x=ax$ for some $a$. But since $x$ generates the cohomology ring, we get that $H^{2k}(f)x^k=a^kx^k$. Now if $n$ is odd, then $a$ must be positive for this to be orientation preserving.
So the Lefschetz number is $1+a+a^2+\dots+a^n$ and this is always non-zero since either $n$ is even or $a$ is positive

In Riemann surfaces land - every map of Riemann surfaces locally look like the map $z\mapsto z^n$. When $n>1$ this map is ramified. Intuitively away from the point $0$, each point hss $n$ distinct preimages - but they come together as you approach the origin.

In algebraic geometry land - the projection $y=x^2$ to the axis $y=0$ is ramified at the point $(0,0)$. Draw the picture - and you see that generically each point has two distinct preimages which are coming together as you approach $(0,0)$. In a sense you can say that your map is ramified if you have less than your expected number of …

> I'm using the fundamental duality in Galois theory
The ramified function is from Spec(R') to Spec(R)
it's a map from a small disc to a small disc
and the question is whether it's a local isomorphism near the origin
and the weird thing is that if we make the residue field bigger
then it satisfies the "e=1" criterion for being a local isomorphism
or at least being unramified
the field got bigger
so Grothendieck said that an extension of fields must be unramified
(strictly speaking a separable extension)

I think intuitively you should be thinking of them as neighbourhoods about the origin for the map $k[x] \rightarrow k[y]$ mapping $x\mapsto y^2$, but I'm slightly hesitant about this because I think there are these things called formal schemes for these things which I've never read.

But of course you can choose some actual DVRs, for example $\mathrm{Spec} k[x,y]/(y-x^2)_{(x,y)} \rightarrow \mathrm{Spec} k[x]_{(x)}$, and you have this problem with both spaces having one closed point and one generic point (because they are DVRs). Morally your generic point of the DVR is including some data a…