It's a silly example, maybe I miss something.

I will begin with a theorem in basic algebraic geometry that states:

Let $f:X\to Y$ be a finite morphisms of affine varieties with $Y$ normal. Thus, for each $y\in Y$, we have $|f^{-1}(y)|\le \deg(f)$.



$Y$ normal means every element of $K[Y]$ contains every element of $K(Y)$ which is integral over $K[X]$

$y\in Y$ is a ramification point if $|f^{-1}(y)|\lt \deg(f)$


$$f:\mathbb A^1\to \mathbb A^1$$

$$t\mapsto t^2$$

I didn't understand why:

If $char(k)\neq 2$, then there is only one ramification point $(t=0)$.

If $char(k)=2$, then every point of $\mathbb A^1$ is a ramification point.

Thanks in advance

  • $\begingroup$ What is the derivative of $f$ if $k$ has characteristic $2$? $\endgroup$ – Ted Shifrin Dec 15 '13 at 18:46
  • $\begingroup$ @TedShifrin It's zero everywhere, but what this has to do with the question? I haven't studied yet the concept of derivative in this context. Thanks $\endgroup$ – user42912 Dec 15 '13 at 18:53
  • 3
    $\begingroup$ Ah, my fault. One standard definition of ramification point in this context is a point where the derivative vanishes. Note that in characteristic $2$, $(t-a)^2 = t^2 - a^2$, so any point in the image is a double root. I.e., $|f^{-1}(a^2)| = 1<2$. $\endgroup$ – Ted Shifrin Dec 15 '13 at 18:57
  • 1
    $\begingroup$ Did you calculate the $deg(f)$ for $char(k)\neq 2$? You should get $2$. In that case, it is easy to see that the only ramification point can be $0$ since it is the only point with exactly one point in the preimage. $\endgroup$ – user113529 Dec 15 '13 at 19:01
  • 1
    $\begingroup$ @user42912: $deg(f) = 2$ because you are not looking at $K(\mathbb{A}^1)$ embedded into $K(\mathbb{A}^1)$ in the trivial way by the identity map, but rather under the embedding $f(t) \mapsto f(t^2)$. Thus $1$ and $t$ are basis for $K(\mathbb{A}^1) \cong K(t)$ over the embedded copy $K(\mathbb{A}^1) \cong K(t^2)$. $\endgroup$ – Michael Joyce Dec 15 '13 at 19:20

We have to be maximally careful with the way function fields embed, so let me translate your map to a map on rings. The corresponding map is given by:

$$\begin{eqnarray*} \varphi : &k[y]& \to k[x] \\ &y& \mapsto x^2. \end{eqnarray*}$$

The first $k[y]$ is the coordinate ring of $Y = \Bbb{A}^1 $, and the second of $X = \Bbb{A}^1$. Taking into account this embedding, the degree of $[K(X):K(Y)] = [k(x):k(x^2)] =2$. Note the degree is not $1$, it depends very much on the embedding!

Let me describe a corresponding geometric set up. We recall that the degree of a projective variety $X$ of dimension $m$ is the number of points of intersection of $X$ with a "generic" linear subspace of codimension $m$. So for example, the degree of a line is $1$ and so is that of a hyperplane. Now embed $\Bbb{P}^1 \to \Bbb{P}^2$ via the Veronese embedding that sends $$[x:y] \mapsto [x^2 : xy:y^2].$$ You can check that the image of $\Bbb{P}^1$ is the hypersurface $V(Y^2 - XZ)$. Is the degree of $\Bbb{P}^1$ via this embedding now $1$? No! It's a conic, and Bezout's theorem tells you that the generic number of points of intersection of a conic and a line is $2$. So the degree of this variety is now $2$.

Summary: The degree of a projective variety depends on the embedding.

Thus we see: The algebraic analogue of this is exactly what's going on in your example above. You must not forget the way the function field is embedded!

With this, we just need to calculate the cardinality of the fiber, depending on the characteristic. If $\operatorname{char} k \neq 2$, then the cardinality of the fiber is always $2$ (think of $\Bbb{C}$, a non-zero complex number always has two distinct square roots) and so the only ramification point is $0$.

If $\operatorname{char} k = 2$, note that $-1 = 1$, so the cardinality of the fiber is $1$. Hence every point is a ramification point.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.