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I corrected a mathematical typo
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Hugo Chapdelaine
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So if we use Riemann-Roch for smooth projective curves over $K$ the problem becomes easy. So without lost of generality we may assume that $C/K$ is smooth and projective since a conic admits a point over $K$ iff $C(K)$ is infinite (this is because the existence of a parametrization over $K$).

If $C(K)$ is not empty we are done. So now assume that $C(K)$ is empty. We will try to reach a contradiction. Let $Q$ be a point in $C(L)$ with minimal field of definition $L$ and let $[L:K]=m\equiv 1\pmod{2}$. Choose a point $P$$P_1$ in $C(\overline{K})$ that lives in a quadratic extension of $K$. Such a point exists since we have a conic and $C(K)$ is empty. Let $P:=P_1+P_1^{\sigma}\in Div_K(C)$. Thus we have $deg(P)=2$ and $deg(Q)=m\geq 3$. We way thus write $m=2a+1$ for some positive integer $a$. Now let $$ D:=[Q]-a[P]\in Div_K(C) $$
We have $deg(D)=1$. Now let us consider the line bundle $L_D$ on $C$ where $$ L_D=\{f\in K(C):div(f)\geq -D\} $$ By Riemann-Roch, we have that $dim_K(L_D)=2$ and thus there exists a non-constant function $f\in L_D$. Note that the map $$ C(\overline{K})\rightarrow P^1(\overline{K}) $$ given by $x\mapsto [f(x),1]$ has degree $deg(div(f)_{\infty})$. So in general, it is not an embedding. Now let us work over $\overline{K}$ so that $[Q]=[Q_1]+[Q_2]\ldots+[Q_m]$ and $[P]=[P_1]+[P_2]$. Since $$ div(f)\geq -D, $$ $f\in K(C)$ (so $deg(div(f))=0$ and $div(f)$ is $G_K$-invariant) we must have that over $\overline{K}$ $$ div(f)=a[P_1]+a[P_2]+[P_3]-[Q_1]-[Q_2]-\ldots -[Q_m] $$ where $deg([P_3])=1$. In other words we have thatThis forces $P_3$ isto be defined over $K$. This contradicts the fact that $C(K)$ was empty.

So if we use Riemann-Roch for smooth projective curves over $K$ the problem becomes easy. So without lost of generality we may assume that $C/K$ is smooth and projective since a conic admits a point over $K$ iff $C(K)$ is infinite (this is because the existence of a parametrization over $K$).

If $C(K)$ is not empty we are done. So now assume that $C(K)$ is empty. We will try to reach a contradiction. Let $Q$ be a point in $C(L)$ with minimal field of definition $L$ and let $[L:K]=m\equiv 1\pmod{2}$. Choose a point $P$ in $C(\overline{K})$ that lives in a quadratic extension of $K$. Such a point exists since we have a conic and $C(K)$ is empty. Thus we have $deg(P)=2$ and $deg(Q)=m\geq 3$. We way thus write $m=2a+1$ for some positive integer $a$. Now let $$ D:=[Q]-a[P]\in Div_K(C) $$
We have $deg(D)=1$. Now let us consider the line bundle $L_D$ on $C$ where $$ L_D=\{f\in K(C):div(f)\geq -D\} $$ By Riemann-Roch, we have that $dim_K(L_D)=2$ and thus there exists a non-constant function $f\in L_D$. Note that the map $$ C(\overline{K})\rightarrow P^1(\overline{K}) $$ given by $x\mapsto [f(x),1]$ has degree $deg(div(f)_{\infty})$. So in general, it is not an embedding. Now let us work over $\overline{K}$ so that $[Q]=[Q_1]+[Q_2]\ldots+[Q_m]$ and $[P]=[P_1]+[P_2]$. Since $$ div(f)\geq -D, $$ $f\in K(C)$ (so $deg(div(f))=0$ and $div(f)$ is $G_K$-invariant) we must have that over $\overline{K}$ $$ div(f)=a[P_1]+a[P_2]+[P_3]-[Q_1]-[Q_2]-\ldots -[Q_m] $$ where $deg([P_3])=1$. In other words we have that $P_3$ is defined over $K$. This contradicts the fact that $C(K)$ was empty.

So if we use Riemann-Roch for smooth projective curves over $K$ the problem becomes easy. So without lost of generality we may assume that $C/K$ is smooth and projective since a conic admits a point over $K$ iff $C(K)$ is infinite (this is because the existence of a parametrization over $K$).

If $C(K)$ is not empty we are done. So now assume that $C(K)$ is empty. We will try to reach a contradiction. Let $Q$ be a point in $C(L)$ with minimal field of definition $L$ and let $[L:K]=m\equiv 1\pmod{2}$. Choose a point $P_1$ in $C(\overline{K})$ that lives in a quadratic extension of $K$. Such a point exists since we have a conic and $C(K)$ is empty. Let $P:=P_1+P_1^{\sigma}\in Div_K(C)$. Thus we have $deg(P)=2$ and $deg(Q)=m\geq 3$. We way thus write $m=2a+1$ for some positive integer $a$. Now let $$ D:=[Q]-a[P]\in Div_K(C) $$
We have $deg(D)=1$. Now let us consider the line bundle $L_D$ on $C$ where $$ L_D=\{f\in K(C):div(f)\geq -D\} $$ By Riemann-Roch, we have that $dim_K(L_D)=2$ and thus there exists a non-constant function $f\in L_D$. Note that the map $$ C(\overline{K})\rightarrow P^1(\overline{K}) $$ given by $x\mapsto [f(x),1]$ has degree $deg(div(f)_{\infty})$. So in general, it is not an embedding. Now let us work over $\overline{K}$ so that $[Q]=[Q_1]+[Q_2]\ldots+[Q_m]$ and $[P]=[P_1]+[P_2]$. Since $$ div(f)\geq -D, $$ $f\in K(C)$ (so $deg(div(f))=0$ and $div(f)$ is $G_K$-invariant) we must have that over $\overline{K}$ $$ div(f)=a[P_1]+a[P_2]+[P_3]-[Q_1]-[Q_2]-\ldots -[Q_m] $$ where $deg([P_3])=1$. This forces $P_3$ to be defined over $K$. This contradicts the fact that $C(K)$ was empty.

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Hugo Chapdelaine
  • 7.6k
  • 2
  • 28
  • 70

So if we use Riemann-Roch for smooth projective curves over $K$ the problem becomes easy. So without lost of generality we may assume that $C/K$ is smooth and projective since a conic admits a point over $K$ iff $C(K)$ is infinite (this is because the existence of a parametrization over $K$).

If $C(K)$ is not empty we are done. So now assume that $C(K)$ is empty. We will try to reach a contradiction. Let $Q$ be a point in $C(L)$ with minimal field of definition $L$ and let $[L:K]=m\equiv 1\pmod{2}$. Choose a point $P$ in $C(\overline{K})$ that lives in a quadratic extension of $K$. Such a point exists since we have a conic and $C(K)$ is empty. Thus we have $deg(P)=2$ and $deg(Q)=m\geq 3$. We way thus write $m=2a+1$ for some positive integer $a$. Now let $$ D:=[Q]-a[P]\in Div_K(C) $$
We have $deg(D)=1$. Now let us consider the line bundle $L_D$ on $C$ where $$ L_D=\{f\in K(C):div(f)\geq -D\} $$ By Riemann-Roch, we have that $dim_K(L_D)=2$ and thus there exists a non-constant function $f\in L_D$. Note that the map $$ C(\overline{K})\rightarrow P^1(\overline{K}) $$ given by $x\mapsto [f(x),1]$ has degree $deg(div(f)_{\infty})$. So in general, it is not an embedding. Now let us work over $\overline{K}$ so that $[Q]=[Q_1]+[Q_2]\ldots+[Q_m]$ and $[P]=[P_1]+[P_2]$. Since $$ div(f)\geq -D, $$ $f\in K(C)$ (so $deg(div(f))=0$ and $div(f)$ is $G_K$-invariant) we must have that over $\overline{K}$ $$ div(f)=a[P_1]+a[P_2]+[P_3]-[Q_1]-[Q_2]-\ldots -[Q_m] $$ where $deg([P_3])=1$. In other words we have that $P_3$ is defined over $K$. This contradicts the fact that $C(K)$ was empty.