Mixing
Why is this function regular?
$begingroup$
In this English translation of Serres FAC, on page 62 he is giving the structure of $mathbb{P}_{r}(K)$.
He sets $t_i$ to be the $i$-th coordinate function on $K^{r+1}$, and defines $$V_i = { x in K^{r+1} | t_i(x) neq 0}.$$ Then $U_i$ is the projection of $V_i$ from $K^{r+1}$ to $mathbb{P}_{r}(K)$, taking the latter to be defined in the usual way by forming the quotient with the equivalence relation.
I do not understand the assertion "the function $t_j/t_i$ is regular on $V_i$ and invariant for $K^{*}$, thus defines a function on $U_i$.”
In order to check this function is regular we use the definition Serre gives on page page 39 that a function $phi colon U to V$ is regular on $U$ if
$U, V$ are both locally closed subspaces of $K^{r}$.
$phi$ is continuous.- If $x in U$ and $f in mathscr{O}_{phi(x),V}$ then $f circ phi in mathscr{O}_{x,U}$
Can someone please help me verify the claims made in the quoted sentence above? I am not even sure what we are taking the codomain of $t_j/t_i$ to be, I am assuming $K$? I am sure these are trivial matters, but I do not understand.
algebraic-geometry sheaf-theory
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
$endgroup$
add a comment |
$begingroup$
In this English translation of Serres FAC, on page 62 he is giving the structure of $mathbb{P}_{r}(K)$.
He sets $t_i$ to be the $i$-th coordinate function on $K^{r+1}$, and defines $$V_i = { x in K^{r+1} | t_i(x) neq 0}.$$ Then $U_i$ is the projection of $V_i$ from $K^{r+1}$ to $mathbb{P}_{r}(K)$, taking the latter to be defined in the usual way by forming the quotient with the equivalence relation.
I do not understand the assertion "the function $t_j/t_i$ is regular on $V_i$ and invariant for $K^{*}$, thus defines a function on $U_i$.”
In order to check this function is regular we use the definition Serre gives on page page 39 that a function $phi colon U to V$ is regular on $U$ if
$U, V$ are both locally closed subspaces of $K^{r}$.
$phi$ is continuous.- If $x in U$ and $f in mathscr{O}_{phi(x),V}$ then $f circ phi in mathscr{O}_{x,U}$
Can someone please help me verify the claims made in the quoted sentence above? I am not even sure what we are taking the codomain of $t_j/t_i$ to be, I am assuming $K$? I am sure these are trivial matters, but I do not understand.
algebraic-geometry sheaf-theory
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
$endgroup$
add a comment |
$begingroup$
In this English translation of Serres FAC, on page 62 he is giving the structure of $mathbb{P}_{r}(K)$.
He sets $t_i$ to be the $i$-th coordinate function on $K^{r+1}$, and defines $$V_i = { x in K^{r+1} | t_i(x) neq 0}.$$ Then $U_i$ is the projection of $V_i$ from $K^{r+1}$ to $mathbb{P}_{r}(K)$, taking the latter to be defined in the usual way by forming the quotient with the equivalence relation.
I do not understand the assertion "the function $t_j/t_i$ is regular on $V_i$ and invariant for $K^{*}$, thus defines a function on $U_i$.”
In order to check this function is regular we use the definition Serre gives on page page 39 that a function $phi colon U to V$ is regular on $U$ if
$U, V$ are both locally closed subspaces of $K^{r}$.
$phi$ is continuous.- If $x in U$ and $f in mathscr{O}_{phi(x),V}$ then $f circ phi in mathscr{O}_{x,U}$
Can someone please help me verify the claims made in the quoted sentence above? I am not even sure what we are taking the codomain of $t_j/t_i$ to be, I am assuming $K$? I am sure these are trivial matters, but I do not understand.
algebraic-geometry sheaf-theory
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
$endgroup$
In this English translation of Serres FAC, on page 62 he is giving the structure of $mathbb{P}_{r}(K)$.
He sets $t_i$ to be the $i$-th coordinate function on $K^{r+1}$, and defines $$V_i = { x in K^{r+1} | t_i(x) neq 0}.$$ Then $U_i$ is the projection of $V_i$ from $K^{r+1}$ to $mathbb{P}_{r}(K)$, taking the latter to be defined in the usual way by forming the quotient with the equivalence relation.
I do not understand the assertion "the function $t_j/t_i$ is regular on $V_i$ and invariant for $K^{*}$, thus defines a function on $U_i$.”
In order to check this function is regular we use the definition Serre gives on page page 39 that a function $phi colon U to V$ is regular on $U$ if
$U, V$ are both locally closed subspaces of $K^{r}$.
$phi$ is continuous.- If $x in U$ and $f in mathscr{O}_{phi(x),V}$ then $f circ phi in mathscr{O}_{x,U}$
Can someone please help me verify the claims made in the quoted sentence above? I am not even sure what we are taking the codomain of $t_j/t_i$ to be, I am assuming $K$? I am sure these are trivial matters, but I do not understand.
algebraic-geometry sheaf-theory
algebraic-geometry sheaf-theory
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
edited Jan 29 at 22:50
Prince M
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
asked Jan 29 at 21:04
Prince MPrince M
2,0761521
2,0761521
add a comment |
add a comment |
1 Answer
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$begingroup$
In our case we have a map $p:V_ito K$, with coordinates $t=(t_1,...,t_{r+1})$ on $V_i$ and $z$ on $K$. Now take $x=(x_1,...,x_{r+1})in V_i$ and an element $fin mathcal{O}_{p(x),K}$, which is nothing else but an element of $k(z)$ of the form $frac{g(z)}{h(z)}=frac{Sigma lambda_m z^m}{Sigma mu_n z^n}$ with $h(p(x))neq 0$.
Now consider $(fcirc p)(t)=frac{gcirc p}{hcirc p}(t)=frac{Sigma lambda_m (frac{t_j}{t_i})^m}{Sigma mu_n (frac{t_j}{t_i})^n}$. Note that $gcirc p$ and $hcirc p$ are not polynomials in $t_1,..., t_{r+1}$, but we can multiply by a sufficently big $t_i^k$ giving us a representation of $fcirc p(t)=frac{t_i^kcdot gcirc p(t)}{t_i^kcdot hcirc p(t)}$ as quotient of polynomials. Now we only have to show that $(t_i^kcdot hcirc p)(x)=x_i^kcdot h(p(x))neq 0$. But we know that $h(p(x))neq 0$ by the assumption that $f$ is regular on $p(x)$ and $x_i^kneq 0$ by definition of $V_i$.
For continuity you might want to apply a similar idea. If $f(z)in k[z]$ and Z=V(f), you want $p^{-1}(Z)=V(g)$ for some polynomial $gin k[t_1,...,t_{r+1}$], so you can consider $g=fcirc p$ and then...
answered Jan 30 at 10:46
NotoneNotone
8181413
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In our case we have a map $p:V_ito K$, with coordinates $t=(t_1,...,t_{r+1})$ on $V_i$ and $z$ on $K$. Now take $x=(x_1,...,x_{r+1})in V_i$ and an element $fin mathcal{O}_{p(x),K}$, which is nothing else but an element of $k(z)$ of the form $frac{g(z)}{h(z)}=frac{Sigma lambda_m z^m}{Sigma mu_n z^n}$ with $h(p(x))neq 0$.
Now consider $(fcirc p)(t)=frac{gcirc p}{hcirc p}(t)=frac{Sigma lambda_m (frac{t_j}{t_i})^m}{Sigma mu_n (frac{t_j}{t_i})^n}$. Note that $gcirc p$ and $hcirc p$ are not polynomials in $t_1,..., t_{r+1}$, but we can multiply by a sufficently big $t_i^k$ giving us a representation of $fcirc p(t)=frac{t_i^kcdot gcirc p(t)}{t_i^kcdot hcirc p(t)}$ as quotient of polynomials. Now we only have to show that $(t_i^kcdot hcirc p)(x)=x_i^kcdot h(p(x))neq 0$. But we know that $h(p(x))neq 0$ by the assumption that $f$ is regular on $p(x)$ and $x_i^kneq 0$ by definition of $V_i$.
For continuity you might want to apply a similar idea. If $f(z)in k[z]$ and Z=V(f), you want $p^{-1}(Z)=V(g)$ for some polynomial $gin k[t_1,...,t_{r+1}$], so you can consider $g=fcirc p$ and then...
answered Jan 30 at 10:46
NotoneNotone
8181413
$endgroup$
add a comment |
$begingroup$
In our case we have a map $p:V_ito K$, with coordinates $t=(t_1,...,t_{r+1})$ on $V_i$ and $z$ on $K$. Now take $x=(x_1,...,x_{r+1})in V_i$ and an element $fin mathcal{O}_{p(x),K}$, which is nothing else but an element of $k(z)$ of the form $frac{g(z)}{h(z)}=frac{Sigma lambda_m z^m}{Sigma mu_n z^n}$ with $h(p(x))neq 0$.
Now consider $(fcirc p)(t)=frac{gcirc p}{hcirc p}(t)=frac{Sigma lambda_m (frac{t_j}{t_i})^m}{Sigma mu_n (frac{t_j}{t_i})^n}$. Note that $gcirc p$ and $hcirc p$ are not polynomials in $t_1,..., t_{r+1}$, but we can multiply by a sufficently big $t_i^k$ giving us a representation of $fcirc p(t)=frac{t_i^kcdot gcirc p(t)}{t_i^kcdot hcirc p(t)}$ as quotient of polynomials. Now we only have to show that $(t_i^kcdot hcirc p)(x)=x_i^kcdot h(p(x))neq 0$. But we know that $h(p(x))neq 0$ by the assumption that $f$ is regular on $p(x)$ and $x_i^kneq 0$ by definition of $V_i$.
For continuity you might want to apply a similar idea. If $f(z)in k[z]$ and Z=V(f), you want $p^{-1}(Z)=V(g)$ for some polynomial $gin k[t_1,...,t_{r+1}$], so you can consider $g=fcirc p$ and then...
answered Jan 30 at 10:46
NotoneNotone
8181413
$endgroup$
add a comment |
$begingroup$
In our case we have a map $p:V_ito K$, with coordinates $t=(t_1,...,t_{r+1})$ on $V_i$ and $z$ on $K$. Now take $x=(x_1,...,x_{r+1})in V_i$ and an element $fin mathcal{O}_{p(x),K}$, which is nothing else but an element of $k(z)$ of the form $frac{g(z)}{h(z)}=frac{Sigma lambda_m z^m}{Sigma mu_n z^n}$ with $h(p(x))neq 0$.
Now consider $(fcirc p)(t)=frac{gcirc p}{hcirc p}(t)=frac{Sigma lambda_m (frac{t_j}{t_i})^m}{Sigma mu_n (frac{t_j}{t_i})^n}$. Note that $gcirc p$ and $hcirc p$ are not polynomials in $t_1,..., t_{r+1}$, but we can multiply by a sufficently big $t_i^k$ giving us a representation of $fcirc p(t)=frac{t_i^kcdot gcirc p(t)}{t_i^kcdot hcirc p(t)}$ as quotient of polynomials. Now we only have to show that $(t_i^kcdot hcirc p)(x)=x_i^kcdot h(p(x))neq 0$. But we know that $h(p(x))neq 0$ by the assumption that $f$ is regular on $p(x)$ and $x_i^kneq 0$ by definition of $V_i$.
For continuity you might want to apply a similar idea. If $f(z)in k[z]$ and Z=V(f), you want $p^{-1}(Z)=V(g)$ for some polynomial $gin k[t_1,...,t_{r+1}$], so you can consider $g=fcirc p$ and then...
answered Jan 30 at 10:46
NotoneNotone
8181413
$endgroup$
In our case we have a map $p:V_ito K$, with coordinates $t=(t_1,...,t_{r+1})$ on $V_i$ and $z$ on $K$. Now take $x=(x_1,...,x_{r+1})in V_i$ and an element $fin mathcal{O}_{p(x),K}$, which is nothing else but an element of $k(z)$ of the form $frac{g(z)}{h(z)}=frac{Sigma lambda_m z^m}{Sigma mu_n z^n}$ with $h(p(x))neq 0$.
Now consider $(fcirc p)(t)=frac{gcirc p}{hcirc p}(t)=frac{Sigma lambda_m (frac{t_j}{t_i})^m}{Sigma mu_n (frac{t_j}{t_i})^n}$. Note that $gcirc p$ and $hcirc p$ are not polynomials in $t_1,..., t_{r+1}$, but we can multiply by a sufficently big $t_i^k$ giving us a representation of $fcirc p(t)=frac{t_i^kcdot gcirc p(t)}{t_i^kcdot hcirc p(t)}$ as quotient of polynomials. Now we only have to show that $(t_i^kcdot hcirc p)(x)=x_i^kcdot h(p(x))neq 0$. But we know that $h(p(x))neq 0$ by the assumption that $f$ is regular on $p(x)$ and $x_i^kneq 0$ by definition of $V_i$.
For continuity you might want to apply a similar idea. If $f(z)in k[z]$ and Z=V(f), you want $p^{-1}(Z)=V(g)$ for some polynomial $gin k[t_1,...,t_{r+1}$], so you can consider $g=fcirc p$ and then...
answered Jan 30 at 10:46
NotoneNotone
8181413
answered Jan 30 at 10:46
NotoneNotone
8181413
answered Jan 30 at 10:46
NotoneNotone
8181413
answered Jan 30 at 10:46
NotoneNotone
8181413
8181413
add a comment |
add a comment |
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