Connected Smooth Projective curve $C$ is rational if unirational












1












$begingroup$


Following question: Why and how to see that a connected, smooth, projective curve $C$ (so a so a $1$-dimensional, proper $k$-scheme) is rational if it is unirational.



Remark: unirational means that there exist a dense morphism $phi: mathbb{P}^n to C$



Especially that would mean that every unirational normal curve have a rational point.



Remark #2: This question arises from the answer given https://mathoverflow.net/questions/319483/extend-group-action-of-mathbba1-g-to-projective-line



but it seems to me that my question above about the step in the given answer would be too low for a further MO-thread.










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$endgroup$












  • $begingroup$
    Your mention a curve several times but also say "$2$-dimensional, proper $k$-scheme" in your post while clarifying your definition of a curve. Is this a typo? If not, could you clarify your meaning?
    $endgroup$
    – KReiser
    Dec 28 '18 at 23:41










  • $begingroup$
    of course a typo :)
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:05
















1












$begingroup$


Following question: Why and how to see that a connected, smooth, projective curve $C$ (so a so a $1$-dimensional, proper $k$-scheme) is rational if it is unirational.



Remark: unirational means that there exist a dense morphism $phi: mathbb{P}^n to C$



Especially that would mean that every unirational normal curve have a rational point.



Remark #2: This question arises from the answer given https://mathoverflow.net/questions/319483/extend-group-action-of-mathbba1-g-to-projective-line



but it seems to me that my question above about the step in the given answer would be too low for a further MO-thread.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Your mention a curve several times but also say "$2$-dimensional, proper $k$-scheme" in your post while clarifying your definition of a curve. Is this a typo? If not, could you clarify your meaning?
    $endgroup$
    – KReiser
    Dec 28 '18 at 23:41










  • $begingroup$
    of course a typo :)
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:05














1












1








1


1



$begingroup$


Following question: Why and how to see that a connected, smooth, projective curve $C$ (so a so a $1$-dimensional, proper $k$-scheme) is rational if it is unirational.



Remark: unirational means that there exist a dense morphism $phi: mathbb{P}^n to C$



Especially that would mean that every unirational normal curve have a rational point.



Remark #2: This question arises from the answer given https://mathoverflow.net/questions/319483/extend-group-action-of-mathbba1-g-to-projective-line



but it seems to me that my question above about the step in the given answer would be too low for a further MO-thread.










share|cite|improve this question











$endgroup$




Following question: Why and how to see that a connected, smooth, projective curve $C$ (so a so a $1$-dimensional, proper $k$-scheme) is rational if it is unirational.



Remark: unirational means that there exist a dense morphism $phi: mathbb{P}^n to C$



Especially that would mean that every unirational normal curve have a rational point.



Remark #2: This question arises from the answer given https://mathoverflow.net/questions/319483/extend-group-action-of-mathbba1-g-to-projective-line



but it seems to me that my question above about the step in the given answer would be too low for a further MO-thread.







algebraic-geometry algebraic-curves projective-schemes






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share|cite|improve this question













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share|cite|improve this question








edited Dec 29 '18 at 3:06







Tim Grosskreutz

















asked Dec 28 '18 at 5:07









Tim GrosskreutzTim Grosskreutz

1576




1576












  • $begingroup$
    Your mention a curve several times but also say "$2$-dimensional, proper $k$-scheme" in your post while clarifying your definition of a curve. Is this a typo? If not, could you clarify your meaning?
    $endgroup$
    – KReiser
    Dec 28 '18 at 23:41










  • $begingroup$
    of course a typo :)
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:05


















  • $begingroup$
    Your mention a curve several times but also say "$2$-dimensional, proper $k$-scheme" in your post while clarifying your definition of a curve. Is this a typo? If not, could you clarify your meaning?
    $endgroup$
    – KReiser
    Dec 28 '18 at 23:41










  • $begingroup$
    of course a typo :)
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:05
















$begingroup$
Your mention a curve several times but also say "$2$-dimensional, proper $k$-scheme" in your post while clarifying your definition of a curve. Is this a typo? If not, could you clarify your meaning?
$endgroup$
– KReiser
Dec 28 '18 at 23:41




$begingroup$
Your mention a curve several times but also say "$2$-dimensional, proper $k$-scheme" in your post while clarifying your definition of a curve. Is this a typo? If not, could you clarify your meaning?
$endgroup$
– KReiser
Dec 28 '18 at 23:41












$begingroup$
of course a typo :)
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 3:05




$begingroup$
of course a typo :)
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 3:05










1 Answer
1






active

oldest

votes


















1












$begingroup$

This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.



For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.



Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.



Proof.
We show that if $k subseteq K subseteq k(t)$ is a sequence of field
extensions such that $operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K simeq k(X)$ for a transcendental element $X$ over $k$. Since
$operatorname{trdeg}_k K = 1$, it is enough to find some $a in K$ such that
$K = k(a)$.



First, the second extension $K subseteq k(t)$ must be algebraic, so
$t$ is algebraic over $K$. Let
$$
f(X) = X^n + a_1X^{n-1} + cdots + a_n in K[X]
$$

be the minimal polynomial of $t$ over $K$. Since $t$ is
transcendental over $k$, we cannot have all $a_i in k$, and so there is some
$i$ such that $a_i in K smallsetminus k$. We will show that in this case, $K
= k(a_i)$
. We know that the $a_i in K subseteq k(t)$,
and so we may write
$$
a_i = frac{u(t)}{v(t)},
$$

where $u,v in k[t]$ are relatively prime, and where at least one of them of
positive degree by the assumption that $a_i notin k$. Now consider the following polynomial:
$$
F(X) = u(X) - a_iv(X) in k(a_i)[X] subseteq K[X].
$$

Since $F(t) = 0$, we have that $f(X) mid F(X)$ in $K[X]$ by minimality of
$f(X)$, and so
begin{equation}
u(X) - a_iv(X) = f(X)g(X).tag{1}label{eq:0105star}
end{equation}

where $g in K[X]$. We then claim the following:



Claim. $g in K$.



Showing the Claim would conclude the proof, for then we have a sequence of
extensions
$$
k hookrightarrow k(a_i) hookrightarrow K hookrightarrow k(t)
$$

where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence
$[k(t) : K] ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.



To prove the Claim, the first step is to get rid of all denominators: by
multiplying eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we
get a relation
$$
c(t)bigl(u(X)v(t) - v(X)u(t)bigr) = f_1(t,X)g_1(t,X),
$$

where $c(t) in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) in k[t,X]$ are obtained from $f$ resp. $g$ by
multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$.
Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$
to get a relation
begin{equation}
u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),tag{2}label{eq:0105rel3}
end{equation}

where now $f_2(t,X),g_2(t,X) in k[t,X]$ are obtained from $f,g$ by
multiplication by a nonzero element in $k(t)$. The trick is now to look at the
degrees in $t$ on both sides. First,
$$
deg_tbigl(u(X)v(t) - v(X)u(t)bigr) le maxbigl{deg u(t),deg
v(t)bigr}.
$$

Letting $f_2(t,X) = gamma_0(t)X^n + cdots + gamma_n(t)$, we see that
$$
frac{gamma_i(t)}{gamma_0(t)} = a_i(t) = frac{u(t)}{v(t)},
$$

where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact,
$$
deg_tbigl(f_2(t,X)bigr) ge maxbigl{deg u(t),deg v(t)bigr}.
$$

By looking at degrees in $t$ on both sides of the relation
eqref{eq:0105rel3}, we have that $deg_t(g_2(t,X)) = 0$, hence $g_2 in k[X]$.



Now we show that $g_2 in k$. For sake of contradiction, suppose that
$g_2 notin k$. Then, there is a root $gamma in overline{k}$ such that
$g_2(gamma) = 0$, which implies that $u(gamma)v(t) = v(gamma)u(t)$ by eqref{eq:0105rel3}. But
since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we
must have $u(gamma) = 0 = v(gamma)$. This contradicts that $u,v$ are
relatively prime, and so $g_2 in k$.
Finally, since $g in K[X]$ and $g = g_2 cdot (text{element of
$k(t)$})$
, we have that $g in K$. $blacksquare$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    hi, thank you a lot for the detailed explanation. Essentially Lüroth theorem proves $mathbb{P}^1 /G cong mathbb{P}^1$ taking into account famous category equivalence between the categories of smooth projective algebraic curves and function fields of trdeg $1$ showing that both have isomorphic function fields, right? the aspect that I unfortunately can't find in your answer ...please correct me if I have ovenseen this point ...is that the thread primary not refers to the statement from my previous MO question about $mathbb{A}^1 /G cong mathbb{A}^1$
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54












  • $begingroup$
    but to two intermediate results occured in the answer concerning unirational curves in general. namely why normal unirational curve has a rational point (or the more stronger result that which appropriate extra conditions it is already ration. I'm not sure how this can be also answered with Lüroth.
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54










  • $begingroup$
    @TimGrosskreutz I am a bit confused, since I thought you were asking about why unirational implies rational for curves in your original question, so that is the question that I answered. In any case, a unirational curve always has rational points: Suppose $fcolon mathbf{P}^1 dashrightarrow C$ is a dominant rational map to a complete curve. Since $mathbf{P}^1$ is regular, $f$ is in fact a morphism. Now $mathbf{P}^1$ has multiple $k$-rational points, and these map to $k$-rational points on $C$.
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:52










  • $begingroup$
    @TimGrosskreutz I also notice you said "two intermediate results". Do you have other questions in addition to why unirational implies rational for curves, and why unirational curves have rational points? I apologize for missing them, if you have already mentioned them on this page!
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:55












  • $begingroup$
    no sorry I have just overseen that Lüroth's statement for intermediate fields from wiki - namely that every intermedadiate extension between $K$ and $K(X)$ is generated by single element - has a geometric version: and this is exactly that unirational implies rational so what I was asking for. this I exactly what I meant by "intermediate result".
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 21:35











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.



For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.



Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.



Proof.
We show that if $k subseteq K subseteq k(t)$ is a sequence of field
extensions such that $operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K simeq k(X)$ for a transcendental element $X$ over $k$. Since
$operatorname{trdeg}_k K = 1$, it is enough to find some $a in K$ such that
$K = k(a)$.



First, the second extension $K subseteq k(t)$ must be algebraic, so
$t$ is algebraic over $K$. Let
$$
f(X) = X^n + a_1X^{n-1} + cdots + a_n in K[X]
$$

be the minimal polynomial of $t$ over $K$. Since $t$ is
transcendental over $k$, we cannot have all $a_i in k$, and so there is some
$i$ such that $a_i in K smallsetminus k$. We will show that in this case, $K
= k(a_i)$
. We know that the $a_i in K subseteq k(t)$,
and so we may write
$$
a_i = frac{u(t)}{v(t)},
$$

where $u,v in k[t]$ are relatively prime, and where at least one of them of
positive degree by the assumption that $a_i notin k$. Now consider the following polynomial:
$$
F(X) = u(X) - a_iv(X) in k(a_i)[X] subseteq K[X].
$$

Since $F(t) = 0$, we have that $f(X) mid F(X)$ in $K[X]$ by minimality of
$f(X)$, and so
begin{equation}
u(X) - a_iv(X) = f(X)g(X).tag{1}label{eq:0105star}
end{equation}

where $g in K[X]$. We then claim the following:



Claim. $g in K$.



Showing the Claim would conclude the proof, for then we have a sequence of
extensions
$$
k hookrightarrow k(a_i) hookrightarrow K hookrightarrow k(t)
$$

where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence
$[k(t) : K] ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.



To prove the Claim, the first step is to get rid of all denominators: by
multiplying eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we
get a relation
$$
c(t)bigl(u(X)v(t) - v(X)u(t)bigr) = f_1(t,X)g_1(t,X),
$$

where $c(t) in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) in k[t,X]$ are obtained from $f$ resp. $g$ by
multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$.
Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$
to get a relation
begin{equation}
u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),tag{2}label{eq:0105rel3}
end{equation}

where now $f_2(t,X),g_2(t,X) in k[t,X]$ are obtained from $f,g$ by
multiplication by a nonzero element in $k(t)$. The trick is now to look at the
degrees in $t$ on both sides. First,
$$
deg_tbigl(u(X)v(t) - v(X)u(t)bigr) le maxbigl{deg u(t),deg
v(t)bigr}.
$$

Letting $f_2(t,X) = gamma_0(t)X^n + cdots + gamma_n(t)$, we see that
$$
frac{gamma_i(t)}{gamma_0(t)} = a_i(t) = frac{u(t)}{v(t)},
$$

where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact,
$$
deg_tbigl(f_2(t,X)bigr) ge maxbigl{deg u(t),deg v(t)bigr}.
$$

By looking at degrees in $t$ on both sides of the relation
eqref{eq:0105rel3}, we have that $deg_t(g_2(t,X)) = 0$, hence $g_2 in k[X]$.



Now we show that $g_2 in k$. For sake of contradiction, suppose that
$g_2 notin k$. Then, there is a root $gamma in overline{k}$ such that
$g_2(gamma) = 0$, which implies that $u(gamma)v(t) = v(gamma)u(t)$ by eqref{eq:0105rel3}. But
since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we
must have $u(gamma) = 0 = v(gamma)$. This contradicts that $u,v$ are
relatively prime, and so $g_2 in k$.
Finally, since $g in K[X]$ and $g = g_2 cdot (text{element of
$k(t)$})$
, we have that $g in K$. $blacksquare$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    hi, thank you a lot for the detailed explanation. Essentially Lüroth theorem proves $mathbb{P}^1 /G cong mathbb{P}^1$ taking into account famous category equivalence between the categories of smooth projective algebraic curves and function fields of trdeg $1$ showing that both have isomorphic function fields, right? the aspect that I unfortunately can't find in your answer ...please correct me if I have ovenseen this point ...is that the thread primary not refers to the statement from my previous MO question about $mathbb{A}^1 /G cong mathbb{A}^1$
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54












  • $begingroup$
    but to two intermediate results occured in the answer concerning unirational curves in general. namely why normal unirational curve has a rational point (or the more stronger result that which appropriate extra conditions it is already ration. I'm not sure how this can be also answered with Lüroth.
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54










  • $begingroup$
    @TimGrosskreutz I am a bit confused, since I thought you were asking about why unirational implies rational for curves in your original question, so that is the question that I answered. In any case, a unirational curve always has rational points: Suppose $fcolon mathbf{P}^1 dashrightarrow C$ is a dominant rational map to a complete curve. Since $mathbf{P}^1$ is regular, $f$ is in fact a morphism. Now $mathbf{P}^1$ has multiple $k$-rational points, and these map to $k$-rational points on $C$.
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:52










  • $begingroup$
    @TimGrosskreutz I also notice you said "two intermediate results". Do you have other questions in addition to why unirational implies rational for curves, and why unirational curves have rational points? I apologize for missing them, if you have already mentioned them on this page!
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:55












  • $begingroup$
    no sorry I have just overseen that Lüroth's statement for intermediate fields from wiki - namely that every intermedadiate extension between $K$ and $K(X)$ is generated by single element - has a geometric version: and this is exactly that unirational implies rational so what I was asking for. this I exactly what I meant by "intermediate result".
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 21:35
















1












$begingroup$

This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.



For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.



Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.



Proof.
We show that if $k subseteq K subseteq k(t)$ is a sequence of field
extensions such that $operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K simeq k(X)$ for a transcendental element $X$ over $k$. Since
$operatorname{trdeg}_k K = 1$, it is enough to find some $a in K$ such that
$K = k(a)$.



First, the second extension $K subseteq k(t)$ must be algebraic, so
$t$ is algebraic over $K$. Let
$$
f(X) = X^n + a_1X^{n-1} + cdots + a_n in K[X]
$$

be the minimal polynomial of $t$ over $K$. Since $t$ is
transcendental over $k$, we cannot have all $a_i in k$, and so there is some
$i$ such that $a_i in K smallsetminus k$. We will show that in this case, $K
= k(a_i)$
. We know that the $a_i in K subseteq k(t)$,
and so we may write
$$
a_i = frac{u(t)}{v(t)},
$$

where $u,v in k[t]$ are relatively prime, and where at least one of them of
positive degree by the assumption that $a_i notin k$. Now consider the following polynomial:
$$
F(X) = u(X) - a_iv(X) in k(a_i)[X] subseteq K[X].
$$

Since $F(t) = 0$, we have that $f(X) mid F(X)$ in $K[X]$ by minimality of
$f(X)$, and so
begin{equation}
u(X) - a_iv(X) = f(X)g(X).tag{1}label{eq:0105star}
end{equation}

where $g in K[X]$. We then claim the following:



Claim. $g in K$.



Showing the Claim would conclude the proof, for then we have a sequence of
extensions
$$
k hookrightarrow k(a_i) hookrightarrow K hookrightarrow k(t)
$$

where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence
$[k(t) : K] ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.



To prove the Claim, the first step is to get rid of all denominators: by
multiplying eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we
get a relation
$$
c(t)bigl(u(X)v(t) - v(X)u(t)bigr) = f_1(t,X)g_1(t,X),
$$

where $c(t) in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) in k[t,X]$ are obtained from $f$ resp. $g$ by
multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$.
Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$
to get a relation
begin{equation}
u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),tag{2}label{eq:0105rel3}
end{equation}

where now $f_2(t,X),g_2(t,X) in k[t,X]$ are obtained from $f,g$ by
multiplication by a nonzero element in $k(t)$. The trick is now to look at the
degrees in $t$ on both sides. First,
$$
deg_tbigl(u(X)v(t) - v(X)u(t)bigr) le maxbigl{deg u(t),deg
v(t)bigr}.
$$

Letting $f_2(t,X) = gamma_0(t)X^n + cdots + gamma_n(t)$, we see that
$$
frac{gamma_i(t)}{gamma_0(t)} = a_i(t) = frac{u(t)}{v(t)},
$$

where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact,
$$
deg_tbigl(f_2(t,X)bigr) ge maxbigl{deg u(t),deg v(t)bigr}.
$$

By looking at degrees in $t$ on both sides of the relation
eqref{eq:0105rel3}, we have that $deg_t(g_2(t,X)) = 0$, hence $g_2 in k[X]$.



Now we show that $g_2 in k$. For sake of contradiction, suppose that
$g_2 notin k$. Then, there is a root $gamma in overline{k}$ such that
$g_2(gamma) = 0$, which implies that $u(gamma)v(t) = v(gamma)u(t)$ by eqref{eq:0105rel3}. But
since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we
must have $u(gamma) = 0 = v(gamma)$. This contradicts that $u,v$ are
relatively prime, and so $g_2 in k$.
Finally, since $g in K[X]$ and $g = g_2 cdot (text{element of
$k(t)$})$
, we have that $g in K$. $blacksquare$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    hi, thank you a lot for the detailed explanation. Essentially Lüroth theorem proves $mathbb{P}^1 /G cong mathbb{P}^1$ taking into account famous category equivalence between the categories of smooth projective algebraic curves and function fields of trdeg $1$ showing that both have isomorphic function fields, right? the aspect that I unfortunately can't find in your answer ...please correct me if I have ovenseen this point ...is that the thread primary not refers to the statement from my previous MO question about $mathbb{A}^1 /G cong mathbb{A}^1$
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54












  • $begingroup$
    but to two intermediate results occured in the answer concerning unirational curves in general. namely why normal unirational curve has a rational point (or the more stronger result that which appropriate extra conditions it is already ration. I'm not sure how this can be also answered with Lüroth.
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54










  • $begingroup$
    @TimGrosskreutz I am a bit confused, since I thought you were asking about why unirational implies rational for curves in your original question, so that is the question that I answered. In any case, a unirational curve always has rational points: Suppose $fcolon mathbf{P}^1 dashrightarrow C$ is a dominant rational map to a complete curve. Since $mathbf{P}^1$ is regular, $f$ is in fact a morphism. Now $mathbf{P}^1$ has multiple $k$-rational points, and these map to $k$-rational points on $C$.
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:52










  • $begingroup$
    @TimGrosskreutz I also notice you said "two intermediate results". Do you have other questions in addition to why unirational implies rational for curves, and why unirational curves have rational points? I apologize for missing them, if you have already mentioned them on this page!
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:55












  • $begingroup$
    no sorry I have just overseen that Lüroth's statement for intermediate fields from wiki - namely that every intermedadiate extension between $K$ and $K(X)$ is generated by single element - has a geometric version: and this is exactly that unirational implies rational so what I was asking for. this I exactly what I meant by "intermediate result".
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 21:35














1












1








1





$begingroup$

This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.



For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.



Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.



Proof.
We show that if $k subseteq K subseteq k(t)$ is a sequence of field
extensions such that $operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K simeq k(X)$ for a transcendental element $X$ over $k$. Since
$operatorname{trdeg}_k K = 1$, it is enough to find some $a in K$ such that
$K = k(a)$.



First, the second extension $K subseteq k(t)$ must be algebraic, so
$t$ is algebraic over $K$. Let
$$
f(X) = X^n + a_1X^{n-1} + cdots + a_n in K[X]
$$

be the minimal polynomial of $t$ over $K$. Since $t$ is
transcendental over $k$, we cannot have all $a_i in k$, and so there is some
$i$ such that $a_i in K smallsetminus k$. We will show that in this case, $K
= k(a_i)$
. We know that the $a_i in K subseteq k(t)$,
and so we may write
$$
a_i = frac{u(t)}{v(t)},
$$

where $u,v in k[t]$ are relatively prime, and where at least one of them of
positive degree by the assumption that $a_i notin k$. Now consider the following polynomial:
$$
F(X) = u(X) - a_iv(X) in k(a_i)[X] subseteq K[X].
$$

Since $F(t) = 0$, we have that $f(X) mid F(X)$ in $K[X]$ by minimality of
$f(X)$, and so
begin{equation}
u(X) - a_iv(X) = f(X)g(X).tag{1}label{eq:0105star}
end{equation}

where $g in K[X]$. We then claim the following:



Claim. $g in K$.



Showing the Claim would conclude the proof, for then we have a sequence of
extensions
$$
k hookrightarrow k(a_i) hookrightarrow K hookrightarrow k(t)
$$

where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence
$[k(t) : K] ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.



To prove the Claim, the first step is to get rid of all denominators: by
multiplying eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we
get a relation
$$
c(t)bigl(u(X)v(t) - v(X)u(t)bigr) = f_1(t,X)g_1(t,X),
$$

where $c(t) in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) in k[t,X]$ are obtained from $f$ resp. $g$ by
multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$.
Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$
to get a relation
begin{equation}
u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),tag{2}label{eq:0105rel3}
end{equation}

where now $f_2(t,X),g_2(t,X) in k[t,X]$ are obtained from $f,g$ by
multiplication by a nonzero element in $k(t)$. The trick is now to look at the
degrees in $t$ on both sides. First,
$$
deg_tbigl(u(X)v(t) - v(X)u(t)bigr) le maxbigl{deg u(t),deg
v(t)bigr}.
$$

Letting $f_2(t,X) = gamma_0(t)X^n + cdots + gamma_n(t)$, we see that
$$
frac{gamma_i(t)}{gamma_0(t)} = a_i(t) = frac{u(t)}{v(t)},
$$

where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact,
$$
deg_tbigl(f_2(t,X)bigr) ge maxbigl{deg u(t),deg v(t)bigr}.
$$

By looking at degrees in $t$ on both sides of the relation
eqref{eq:0105rel3}, we have that $deg_t(g_2(t,X)) = 0$, hence $g_2 in k[X]$.



Now we show that $g_2 in k$. For sake of contradiction, suppose that
$g_2 notin k$. Then, there is a root $gamma in overline{k}$ such that
$g_2(gamma) = 0$, which implies that $u(gamma)v(t) = v(gamma)u(t)$ by eqref{eq:0105rel3}. But
since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we
must have $u(gamma) = 0 = v(gamma)$. This contradicts that $u,v$ are
relatively prime, and so $g_2 in k$.
Finally, since $g in K[X]$ and $g = g_2 cdot (text{element of
$k(t)$})$
, we have that $g in K$. $blacksquare$






share|cite|improve this answer









$endgroup$



This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.



For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.



Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.



Proof.
We show that if $k subseteq K subseteq k(t)$ is a sequence of field
extensions such that $operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K simeq k(X)$ for a transcendental element $X$ over $k$. Since
$operatorname{trdeg}_k K = 1$, it is enough to find some $a in K$ such that
$K = k(a)$.



First, the second extension $K subseteq k(t)$ must be algebraic, so
$t$ is algebraic over $K$. Let
$$
f(X) = X^n + a_1X^{n-1} + cdots + a_n in K[X]
$$

be the minimal polynomial of $t$ over $K$. Since $t$ is
transcendental over $k$, we cannot have all $a_i in k$, and so there is some
$i$ such that $a_i in K smallsetminus k$. We will show that in this case, $K
= k(a_i)$
. We know that the $a_i in K subseteq k(t)$,
and so we may write
$$
a_i = frac{u(t)}{v(t)},
$$

where $u,v in k[t]$ are relatively prime, and where at least one of them of
positive degree by the assumption that $a_i notin k$. Now consider the following polynomial:
$$
F(X) = u(X) - a_iv(X) in k(a_i)[X] subseteq K[X].
$$

Since $F(t) = 0$, we have that $f(X) mid F(X)$ in $K[X]$ by minimality of
$f(X)$, and so
begin{equation}
u(X) - a_iv(X) = f(X)g(X).tag{1}label{eq:0105star}
end{equation}

where $g in K[X]$. We then claim the following:



Claim. $g in K$.



Showing the Claim would conclude the proof, for then we have a sequence of
extensions
$$
k hookrightarrow k(a_i) hookrightarrow K hookrightarrow k(t)
$$

where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence
$[k(t) : K] ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.



To prove the Claim, the first step is to get rid of all denominators: by
multiplying eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we
get a relation
$$
c(t)bigl(u(X)v(t) - v(X)u(t)bigr) = f_1(t,X)g_1(t,X),
$$

where $c(t) in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) in k[t,X]$ are obtained from $f$ resp. $g$ by
multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$.
Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$
to get a relation
begin{equation}
u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),tag{2}label{eq:0105rel3}
end{equation}

where now $f_2(t,X),g_2(t,X) in k[t,X]$ are obtained from $f,g$ by
multiplication by a nonzero element in $k(t)$. The trick is now to look at the
degrees in $t$ on both sides. First,
$$
deg_tbigl(u(X)v(t) - v(X)u(t)bigr) le maxbigl{deg u(t),deg
v(t)bigr}.
$$

Letting $f_2(t,X) = gamma_0(t)X^n + cdots + gamma_n(t)$, we see that
$$
frac{gamma_i(t)}{gamma_0(t)} = a_i(t) = frac{u(t)}{v(t)},
$$

where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact,
$$
deg_tbigl(f_2(t,X)bigr) ge maxbigl{deg u(t),deg v(t)bigr}.
$$

By looking at degrees in $t$ on both sides of the relation
eqref{eq:0105rel3}, we have that $deg_t(g_2(t,X)) = 0$, hence $g_2 in k[X]$.



Now we show that $g_2 in k$. For sake of contradiction, suppose that
$g_2 notin k$. Then, there is a root $gamma in overline{k}$ such that
$g_2(gamma) = 0$, which implies that $u(gamma)v(t) = v(gamma)u(t)$ by eqref{eq:0105rel3}. But
since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we
must have $u(gamma) = 0 = v(gamma)$. This contradicts that $u,v$ are
relatively prime, and so $g_2 in k$.
Finally, since $g in K[X]$ and $g = g_2 cdot (text{element of
$k(t)$})$
, we have that $g in K$. $blacksquare$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 1:20









Takumi MurayamaTakumi Murayama

6,39611646




6,39611646












  • $begingroup$
    hi, thank you a lot for the detailed explanation. Essentially Lüroth theorem proves $mathbb{P}^1 /G cong mathbb{P}^1$ taking into account famous category equivalence between the categories of smooth projective algebraic curves and function fields of trdeg $1$ showing that both have isomorphic function fields, right? the aspect that I unfortunately can't find in your answer ...please correct me if I have ovenseen this point ...is that the thread primary not refers to the statement from my previous MO question about $mathbb{A}^1 /G cong mathbb{A}^1$
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54












  • $begingroup$
    but to two intermediate results occured in the answer concerning unirational curves in general. namely why normal unirational curve has a rational point (or the more stronger result that which appropriate extra conditions it is already ration. I'm not sure how this can be also answered with Lüroth.
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54










  • $begingroup$
    @TimGrosskreutz I am a bit confused, since I thought you were asking about why unirational implies rational for curves in your original question, so that is the question that I answered. In any case, a unirational curve always has rational points: Suppose $fcolon mathbf{P}^1 dashrightarrow C$ is a dominant rational map to a complete curve. Since $mathbf{P}^1$ is regular, $f$ is in fact a morphism. Now $mathbf{P}^1$ has multiple $k$-rational points, and these map to $k$-rational points on $C$.
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:52










  • $begingroup$
    @TimGrosskreutz I also notice you said "two intermediate results". Do you have other questions in addition to why unirational implies rational for curves, and why unirational curves have rational points? I apologize for missing them, if you have already mentioned them on this page!
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:55












  • $begingroup$
    no sorry I have just overseen that Lüroth's statement for intermediate fields from wiki - namely that every intermedadiate extension between $K$ and $K(X)$ is generated by single element - has a geometric version: and this is exactly that unirational implies rational so what I was asking for. this I exactly what I meant by "intermediate result".
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 21:35


















  • $begingroup$
    hi, thank you a lot for the detailed explanation. Essentially Lüroth theorem proves $mathbb{P}^1 /G cong mathbb{P}^1$ taking into account famous category equivalence between the categories of smooth projective algebraic curves and function fields of trdeg $1$ showing that both have isomorphic function fields, right? the aspect that I unfortunately can't find in your answer ...please correct me if I have ovenseen this point ...is that the thread primary not refers to the statement from my previous MO question about $mathbb{A}^1 /G cong mathbb{A}^1$
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54












  • $begingroup$
    but to two intermediate results occured in the answer concerning unirational curves in general. namely why normal unirational curve has a rational point (or the more stronger result that which appropriate extra conditions it is already ration. I'm not sure how this can be also answered with Lüroth.
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 3:54










  • $begingroup$
    @TimGrosskreutz I am a bit confused, since I thought you were asking about why unirational implies rational for curves in your original question, so that is the question that I answered. In any case, a unirational curve always has rational points: Suppose $fcolon mathbf{P}^1 dashrightarrow C$ is a dominant rational map to a complete curve. Since $mathbf{P}^1$ is regular, $f$ is in fact a morphism. Now $mathbf{P}^1$ has multiple $k$-rational points, and these map to $k$-rational points on $C$.
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:52










  • $begingroup$
    @TimGrosskreutz I also notice you said "two intermediate results". Do you have other questions in addition to why unirational implies rational for curves, and why unirational curves have rational points? I apologize for missing them, if you have already mentioned them on this page!
    $endgroup$
    – Takumi Murayama
    Dec 29 '18 at 6:55












  • $begingroup$
    no sorry I have just overseen that Lüroth's statement for intermediate fields from wiki - namely that every intermedadiate extension between $K$ and $K(X)$ is generated by single element - has a geometric version: and this is exactly that unirational implies rational so what I was asking for. this I exactly what I meant by "intermediate result".
    $endgroup$
    – Tim Grosskreutz
    Dec 29 '18 at 21:35
















$begingroup$
hi, thank you a lot for the detailed explanation. Essentially Lüroth theorem proves $mathbb{P}^1 /G cong mathbb{P}^1$ taking into account famous category equivalence between the categories of smooth projective algebraic curves and function fields of trdeg $1$ showing that both have isomorphic function fields, right? the aspect that I unfortunately can't find in your answer ...please correct me if I have ovenseen this point ...is that the thread primary not refers to the statement from my previous MO question about $mathbb{A}^1 /G cong mathbb{A}^1$
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 3:54






$begingroup$
hi, thank you a lot for the detailed explanation. Essentially Lüroth theorem proves $mathbb{P}^1 /G cong mathbb{P}^1$ taking into account famous category equivalence between the categories of smooth projective algebraic curves and function fields of trdeg $1$ showing that both have isomorphic function fields, right? the aspect that I unfortunately can't find in your answer ...please correct me if I have ovenseen this point ...is that the thread primary not refers to the statement from my previous MO question about $mathbb{A}^1 /G cong mathbb{A}^1$
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 3:54














$begingroup$
but to two intermediate results occured in the answer concerning unirational curves in general. namely why normal unirational curve has a rational point (or the more stronger result that which appropriate extra conditions it is already ration. I'm not sure how this can be also answered with Lüroth.
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 3:54




$begingroup$
but to two intermediate results occured in the answer concerning unirational curves in general. namely why normal unirational curve has a rational point (or the more stronger result that which appropriate extra conditions it is already ration. I'm not sure how this can be also answered with Lüroth.
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 3:54












$begingroup$
@TimGrosskreutz I am a bit confused, since I thought you were asking about why unirational implies rational for curves in your original question, so that is the question that I answered. In any case, a unirational curve always has rational points: Suppose $fcolon mathbf{P}^1 dashrightarrow C$ is a dominant rational map to a complete curve. Since $mathbf{P}^1$ is regular, $f$ is in fact a morphism. Now $mathbf{P}^1$ has multiple $k$-rational points, and these map to $k$-rational points on $C$.
$endgroup$
– Takumi Murayama
Dec 29 '18 at 6:52




$begingroup$
@TimGrosskreutz I am a bit confused, since I thought you were asking about why unirational implies rational for curves in your original question, so that is the question that I answered. In any case, a unirational curve always has rational points: Suppose $fcolon mathbf{P}^1 dashrightarrow C$ is a dominant rational map to a complete curve. Since $mathbf{P}^1$ is regular, $f$ is in fact a morphism. Now $mathbf{P}^1$ has multiple $k$-rational points, and these map to $k$-rational points on $C$.
$endgroup$
– Takumi Murayama
Dec 29 '18 at 6:52












$begingroup$
@TimGrosskreutz I also notice you said "two intermediate results". Do you have other questions in addition to why unirational implies rational for curves, and why unirational curves have rational points? I apologize for missing them, if you have already mentioned them on this page!
$endgroup$
– Takumi Murayama
Dec 29 '18 at 6:55






$begingroup$
@TimGrosskreutz I also notice you said "two intermediate results". Do you have other questions in addition to why unirational implies rational for curves, and why unirational curves have rational points? I apologize for missing them, if you have already mentioned them on this page!
$endgroup$
– Takumi Murayama
Dec 29 '18 at 6:55














$begingroup$
no sorry I have just overseen that Lüroth's statement for intermediate fields from wiki - namely that every intermedadiate extension between $K$ and $K(X)$ is generated by single element - has a geometric version: and this is exactly that unirational implies rational so what I was asking for. this I exactly what I meant by "intermediate result".
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 21:35




$begingroup$
no sorry I have just overseen that Lüroth's statement for intermediate fields from wiki - namely that every intermedadiate extension between $K$ and $K(X)$ is generated by single element - has a geometric version: and this is exactly that unirational implies rational so what I was asking for. this I exactly what I meant by "intermediate result".
$endgroup$
– Tim Grosskreutz
Dec 29 '18 at 21:35


















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