2
$\begingroup$

Exercise 2.5 says (in part): "Let $F$ be an irreducible polynomial in $k[X,Y]$ [$k$ algebraically closed], and suppose $F$ is monic in $Y$: $F = Y^n + a_1(X)Y^{n-1} + \cdots$, with $n>0$. Let $V=V(F)\subset\mathbb{A}^2$. Show that the natural homomorphism from $k[X]$ to $\Gamma(V) = k[X,Y]/(F)$ is one-to-one".

Surely this is not a sharp condition; for example, if $F=XY-1$, then $\Gamma(F) \cong k[X,1/X]$, and the natural homomorphism is one-to-one.

What is an example of an irreducible $F$ for which this statement does not hold? It seems to me that if some polynomial $g\in k[X]$ maps to zero in $\Gamma(V)$, then it must be a multiple of $F$, which never happens.

$\endgroup$
  • 2
    $\begingroup$ As you may have guessed, the reason for assuming $F$ is monic in $Y$ is the second part of the exercise: "show that the residues $1, \overline{Y}, \ldots, \overline{Y}^{n-1}$ generate $\Gamma(V)$ over $k[X]$ as a module." This is an example of "integral + finite type $\implies$ finite" as mentioned here. $\endgroup$ – André 3000 Jun 23 '18 at 22:56
2
$\begingroup$

The condition of being one-to-one or injective is equivalent to the map is dominant, i.e., the closure of the image is the whole target variety (see Georges' answer in this thread for example). In the particular case we're dealing with you have the following composition of ring maps: $$ k[x]\hookrightarrow k[x,y]\twoheadrightarrow k[x,y]/(f), $$ which, in turn, geometrically corresponds to: $$ V(f)\hookrightarrow\mathbb{A}_{x,y}^2\twoheadrightarrow\mathbb{A}^1_x, $$ where the first map is the embedding of the curve and the last map is the projection onto the $x$-coordinate.

With this in mind your question translates to asking for a curve $V(f)$ in $\mathbb{A}^2$ such that its projection map onto the $x$-axis is not dominant, this is, when is/isn't the closure of the image all of $\mathbb{A}^1_x$? Now, since $\mathbb{A}^1$ has dimension 1 the closure of such a set can either be the whole space or a 0-dimensional subspace, this is, a finite union of closed points. Can you finish the argument from here?

$\endgroup$

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.