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Let $X$ be a variety over $k$. Let $L$ be a linear system associated to a divisor $D$ on $X$, of dimension $n$, parametrized by a projective space $\mathbb{P}(V)\subset\mathbb{P}(L(D))$. Here $L(D)=\{f\in k(X)^*|D+div(f)\geq 0\}\cup\{ 0\}$. That means to each $f\mod k^*$ we associate divisor $D+div (f)$ and $L$ is the image of $\mathbb{P}(V)$. We take an arbitrary basis $f_0,\dots ,f_n$ of $V$ and define the map:$$\phi _L :X\rightarrow\mathbb{P}^n$$ $$x\mapsto (f_0(x),\dots f_n(x))$$ Let $\mathcal{B}(L)$ be the intersection of supports of divisors in $L$. The following is left as an exercise and I failed to solve it: Prove that $\phi_L |_{X-\mathcal{B}(L)}:X-\mathcal{B}(L)\rightarrow\mathbb{P}^n$ is a well defined morphism.

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    $\begingroup$ If $f_0(x), \dots , f_n(x)$ all vanish at a given $x \in X$, then $\phi_L$ sends $x$ to $[0: \dots : 0]$, which is not a valid point in $\mathbb P^n$! So to define a morphism, you have to avoid points where all $f_0(x), \dots , f_n(x)$ vanish, i.e. you have to avoid points in the baselocus $\mathcal B(L)$. $\endgroup$
    – Kenny Wong
    Commented Mar 16, 2017 at 12:31
  • $\begingroup$ Why is this exactly $\mathcal{B}(L)$? This is obviously the common locus of $V$, but I don't really see why it would be the intersection of supports of $D\in L$? $\endgroup$
    – ralleee
    Commented Mar 16, 2017 at 12:42
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    $\begingroup$ The "support of the divisor $f_0$" means "the locus where $f_0$ vanishes". That's just the definition! The intersection of the supports of all $\sum_i c_i f_i$ (where $c_i$ are constants) is therefore the same thing as the locus where $f_0, \dots , f_n$ all vanish. $\endgroup$
    – Kenny Wong
    Commented Mar 16, 2017 at 16:19
  • $\begingroup$ Thanks. Isn't the definition where it vanishes + where it is not defined? Does it matter? In our case we don't search for the intersection of supports of $\sum _i c_i f_i$, it is rather $D+\sum _i c_i f_i$, right? $\endgroup$
    – ralleee
    Commented Mar 16, 2017 at 19:12
  • $\begingroup$ Yes, good. The definition of "$f$ vanishing" is a bit subtle, and depends on your choice of $D$. I'll write out a fuller answer. Bear with me... $\endgroup$
    – Kenny Wong
    Commented Mar 16, 2017 at 19:35

1 Answer 1

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Suppose $D = \sum_i n_i D_i$, where each $D_i$ is an irreducible codimension-one closed subvariety on $X$. If the variety is smooth, it can be shown that there exists an open cover $\{ U_\alpha \}$ for $X$, and a collection of regular functions $g_{i\alpha}$ on each $\{ U_\alpha \}$, such that $D_i \cap U_\alpha$ is the vanishing locus of $g_{i\alpha}$, and such that $g_{i\alpha}$ vanishes precisely with multiplicity one on this locus.

[For example, if $X = \mathbb P^1$ and $D = [1:0]$, then we can pick the open sets to be $U_0 = \{ [1 : z] : z \in k \}$ and $U_1 = \{ [w: 1] : w \in k \}$. We can pick $g_0= z$, since this vanishes at $[1:0]$ with multiplicity one, and we can pick $g_{1} = 1$, since $[1:0]$ is not contained inside $U_1$ at all.]

On each $U_\alpha$, let us define the rational function $$ g_{D\alpha} = \prod_{i}g_{i\alpha}^{n_i}.$$ The purpose of defining this rational function $g_{D \alpha}$ is that the divisor of zeroes and poles of $g_{D\alpha}$ within $U_\alpha$ exactly coincides with $D \cap U_\alpha$.

Now, given a rational function $f$ on $X$, the condition that $f$ is a member of the linear system $L(D)$ is that $$D + {\rm div}(f) \geq 0.$$ If you think about it, you'll realise that this condition is equivalent to the condition that, on every $U_\alpha$, the function $$ h_\alpha : = g_{D \alpha} f$$ is a regular function.

So here is how we are going to define the morphism $\phi_L$:

If the rational functions $f_0, \dots, f_n$ form basis for $L(D)$, we define, for each open set $U_\alpha$, a corresponding collection of regular functions $$(h_0)_\alpha = g_{D \alpha} f_0 \ \ , \ \ \dots \ \ , \ \ (h_n)_\alpha = g_{D \alpha} f_n$$ as above.

Next, we define the morphism $\phi_L$ on each $U_\alpha$ separately. On the patch $U_\alpha$, we define the morphism $\phi_L$ by $$ \phi_L (x) = [(h_0)_\alpha(x) : \dots : (h_n)_\alpha(x) ].$$

But hang on a minute! Since we're defining $\phi_L (x)$ patch by patch, we'd better check that our definitions agree on the overlaps between the patches! Well, on the overlap $U_\alpha \cap U_\beta$, you can see that $$ [(h_0)_\beta(x) : \dots : (h_n)_\beta(x) ] = [\frac{g_{D \beta}(x)}{g_{D \alpha}(x)}(h_0)_\alpha(x) : \dots : \frac{g_{D \beta}(x)}{g_{D \alpha}(x)} (h_n)_\alpha(x) ]$$ Since we make the identification $[y_0 : \dots : y_n] \sim [\lambda y_0 : \dots : \lambda y_n]$ in projective space, we see that $$ [(h_0)_\beta(x) : \dots : (h_n)_\beta(x) ] = [(h_0)_\alpha(x) : \dots : (h_n)_\alpha(x) ],$$ so there is no disagreement between patches.

Finally, where do things go wrong? Things go wrong at any point $x \in U_\alpha$ where $$(h_0)_\alpha(x) = \dots = (h_n)_\alpha(x) = 0.$$ This is because $[0: \dots : 0]$ is not a legitimate point in projective space.

[It should be straightforward to check that if $x$ is both in $U_\alpha$ and in $U_\beta$, then $$(h_0)_\alpha(x) = \dots = (h_n)_\alpha(x) = 0 \iff (h_0)_\beta(x) = \dots = (h_n)_\beta(x) = 0 ,$$ so this condition is independent of the choice of open chart.]

You can also check that the condition that $(h_0)_\alpha(x) = \dots = (h_n)_\alpha(x) = 0$ is equivalent to the condition that $x$ is in the support of each of the divisors $$ D + {\rm div}(f_0) \ \ , \ \ \dots \ \ , \ \ D + {\rm div}(f_n).$$ This is the same as the statement that $x$ is in the base locus of the linear system $L(D)$.

I hope this helps.

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  • $\begingroup$ Yes it helps. You've worked with full of $L(D)$ but it's the same. $\endgroup$
    – ralleee
    Commented Mar 16, 2017 at 22:58
  • $\begingroup$ Cool - glad it was useful. $\endgroup$
    – Kenny Wong
    Commented Mar 16, 2017 at 23:03

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