If a field $k$ is not algebraically closed, $k$ is not a $C_R$ field for any $0le R<1$.
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I'm reading Dr. Pete Clark's paper Around Chevalley-Warning Theorem and I have a question about the proof of the following proposition on page 14:
Proposition: For a field $k$, the following are equivalent:
(i) $k$ is algebraically closed.
(ii)$k$ is $C_0.$
(iii) There is some $0le R<1$, such that $k$ is $C_R$.
Recall the definition of $C_R$ property of a field $k$: $Rgeq 0$, for $forall$ $d,n in mathbb Z^{+}$, with $d^{R}<n$, $forall fin k[t_1,cdots,t_n]$ homogeneous, $deg f=d$ , $exists$ $0neq x in k^n$ such that $f(x)=0$.
Specifically, I don't quite understand the part of the proof from $(iii)implies (i)$. The proof goes as following: we prove it by contrapositive, so we assume first k is not algebraically closed. Let $l/k$ be a field extension of degree d. Choose a basis $alpha_1, cdots, alpha_d$ of $l/k$, and use this to identify $l$ with $k^{d}$, and view $alpha in l$ as giving a k-linear endomorphism of $k^{d}$ by multiplication. For indeterminates $t_1,cdots,t_d$, let $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$. Thus P is a homogeneous polynomial of degree d with coefficients in k. Moreover, every nonzero $xin k^{n}$ corresponds to a nonzero element of $l$, and since $l$ is a field, multiplication by $x$ has nonzero determinant. It follows that $P(x_1,cdots,x_d)=0$ if and only if $x=0$. So whenever we have a degree d field extension , we have a degree d homogeneous polynomial in d variables which has only trivial solution, which means $k$ is not $C_R,_d$ for any $R<1$.
I'm not very familiar the concept of field extension but I have seen a few examples. I have the following questions regarding the proof:
I don't think I fully understand the meaning of the sentence: "view $alpha in l$ as giving a $k$-linear endomorphism of $k^{d}$ by multiplication." Does it mean we could multiply $alpha$ with any element in $k^{d}$ and get a new element in $k^d$? Is it true that this endomorphism is an automorphism? Could anyone elaborate what this means?
How do you evaluate the determinant appeared in this equation: $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$? Isn't $alpha_1 t_1+cdots+ alpha_d t_d$ a polynomial? How to calculate the determinant of a polynomial? Why $P$ is a homogeneous polynomial of degree $d$?
Thank you in advance for any feedback and advice!
field-theory extension-field
asked Nov 21 at 16:14
Jiguang Li
63
add a comment |
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1
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I'm reading Dr. Pete Clark's paper Around Chevalley-Warning Theorem and I have a question about the proof of the following proposition on page 14:
Proposition: For a field $k$, the following are equivalent:
(i) $k$ is algebraically closed.
(ii)$k$ is $C_0.$
(iii) There is some $0le R<1$, such that $k$ is $C_R$.
Recall the definition of $C_R$ property of a field $k$: $Rgeq 0$, for $forall$ $d,n in mathbb Z^{+}$, with $d^{R}<n$, $forall fin k[t_1,cdots,t_n]$ homogeneous, $deg f=d$ , $exists$ $0neq x in k^n$ such that $f(x)=0$.
Specifically, I don't quite understand the part of the proof from $(iii)implies (i)$. The proof goes as following: we prove it by contrapositive, so we assume first k is not algebraically closed. Let $l/k$ be a field extension of degree d. Choose a basis $alpha_1, cdots, alpha_d$ of $l/k$, and use this to identify $l$ with $k^{d}$, and view $alpha in l$ as giving a k-linear endomorphism of $k^{d}$ by multiplication. For indeterminates $t_1,cdots,t_d$, let $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$. Thus P is a homogeneous polynomial of degree d with coefficients in k. Moreover, every nonzero $xin k^{n}$ corresponds to a nonzero element of $l$, and since $l$ is a field, multiplication by $x$ has nonzero determinant. It follows that $P(x_1,cdots,x_d)=0$ if and only if $x=0$. So whenever we have a degree d field extension , we have a degree d homogeneous polynomial in d variables which has only trivial solution, which means $k$ is not $C_R,_d$ for any $R<1$.
I'm not very familiar the concept of field extension but I have seen a few examples. I have the following questions regarding the proof:
I don't think I fully understand the meaning of the sentence: "view $alpha in l$ as giving a $k$-linear endomorphism of $k^{d}$ by multiplication." Does it mean we could multiply $alpha$ with any element in $k^{d}$ and get a new element in $k^d$? Is it true that this endomorphism is an automorphism? Could anyone elaborate what this means?
How do you evaluate the determinant appeared in this equation: $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$? Isn't $alpha_1 t_1+cdots+ alpha_d t_d$ a polynomial? How to calculate the determinant of a polynomial? Why $P$ is a homogeneous polynomial of degree $d$?
Thank you in advance for any feedback and advice!
field-theory extension-field
asked Nov 21 at 16:14
Jiguang Li
63
For the first question, we're just talking about the map $k^d rightarrow k^d$, $beta mapsto beta alpha$. It's clearly an endomorphism; the inverse is the map coming from multiplication by $alpha^{-1}$. Also, thanks for asking this question! I've been thinking about this sort of thing for a while and it's hard to find references. Hope you get a good answer!
– Sam Streeter
Nov 21 at 16:19
For the second question, I'm thinking we're maybe really talking about the norm.
– Sam Streeter
Nov 21 at 16:24
Thank you for your reply to question 1: it's more clear now. But I'm still confused by the second question. Did you treat $a_1t_1+cdots+ a_dt_d$ as a vector? If it refers to the norm, which norm do we use here?
– Jiguang Li
Nov 22 at 16:41
I'm struggling to come up with a concise and easy explanation. I think taking a look at Example 1.2.3 of Poonen's book "Rational Points on Varieties" might be enlightening though. It is also a good reference for material on $C_r$ fields. You can find it here: www-math.mit.edu/~poonen/papers/Qpoints.pdf
– Sam Streeter
Nov 22 at 16:52
1
@ Sam Streeter: thank you for the reference ! I believe the idea that $det$ refers to the norm really makes sense now.
– Jiguang Li
Nov 25 at 22:20
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading Dr. Pete Clark's paper Around Chevalley-Warning Theorem and I have a question about the proof of the following proposition on page 14:
Proposition: For a field $k$, the following are equivalent:
(i) $k$ is algebraically closed.
(ii)$k$ is $C_0.$
(iii) There is some $0le R<1$, such that $k$ is $C_R$.
Recall the definition of $C_R$ property of a field $k$: $Rgeq 0$, for $forall$ $d,n in mathbb Z^{+}$, with $d^{R}<n$, $forall fin k[t_1,cdots,t_n]$ homogeneous, $deg f=d$ , $exists$ $0neq x in k^n$ such that $f(x)=0$.
Specifically, I don't quite understand the part of the proof from $(iii)implies (i)$. The proof goes as following: we prove it by contrapositive, so we assume first k is not algebraically closed. Let $l/k$ be a field extension of degree d. Choose a basis $alpha_1, cdots, alpha_d$ of $l/k$, and use this to identify $l$ with $k^{d}$, and view $alpha in l$ as giving a k-linear endomorphism of $k^{d}$ by multiplication. For indeterminates $t_1,cdots,t_d$, let $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$. Thus P is a homogeneous polynomial of degree d with coefficients in k. Moreover, every nonzero $xin k^{n}$ corresponds to a nonzero element of $l$, and since $l$ is a field, multiplication by $x$ has nonzero determinant. It follows that $P(x_1,cdots,x_d)=0$ if and only if $x=0$. So whenever we have a degree d field extension , we have a degree d homogeneous polynomial in d variables which has only trivial solution, which means $k$ is not $C_R,_d$ for any $R<1$.
I'm not very familiar the concept of field extension but I have seen a few examples. I have the following questions regarding the proof:
I don't think I fully understand the meaning of the sentence: "view $alpha in l$ as giving a $k$-linear endomorphism of $k^{d}$ by multiplication." Does it mean we could multiply $alpha$ with any element in $k^{d}$ and get a new element in $k^d$? Is it true that this endomorphism is an automorphism? Could anyone elaborate what this means?
How do you evaluate the determinant appeared in this equation: $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$? Isn't $alpha_1 t_1+cdots+ alpha_d t_d$ a polynomial? How to calculate the determinant of a polynomial? Why $P$ is a homogeneous polynomial of degree $d$?
Thank you in advance for any feedback and advice!
field-theory extension-field
asked Nov 21 at 16:14
Jiguang Li
63
I'm reading Dr. Pete Clark's paper Around Chevalley-Warning Theorem and I have a question about the proof of the following proposition on page 14:
Proposition: For a field $k$, the following are equivalent:
(i) $k$ is algebraically closed.
(ii)$k$ is $C_0.$
(iii) There is some $0le R<1$, such that $k$ is $C_R$.
Recall the definition of $C_R$ property of a field $k$: $Rgeq 0$, for $forall$ $d,n in mathbb Z^{+}$, with $d^{R}<n$, $forall fin k[t_1,cdots,t_n]$ homogeneous, $deg f=d$ , $exists$ $0neq x in k^n$ such that $f(x)=0$.
Specifically, I don't quite understand the part of the proof from $(iii)implies (i)$. The proof goes as following: we prove it by contrapositive, so we assume first k is not algebraically closed. Let $l/k$ be a field extension of degree d. Choose a basis $alpha_1, cdots, alpha_d$ of $l/k$, and use this to identify $l$ with $k^{d}$, and view $alpha in l$ as giving a k-linear endomorphism of $k^{d}$ by multiplication. For indeterminates $t_1,cdots,t_d$, let $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$. Thus P is a homogeneous polynomial of degree d with coefficients in k. Moreover, every nonzero $xin k^{n}$ corresponds to a nonzero element of $l$, and since $l$ is a field, multiplication by $x$ has nonzero determinant. It follows that $P(x_1,cdots,x_d)=0$ if and only if $x=0$. So whenever we have a degree d field extension , we have a degree d homogeneous polynomial in d variables which has only trivial solution, which means $k$ is not $C_R,_d$ for any $R<1$.
I'm not very familiar the concept of field extension but I have seen a few examples. I have the following questions regarding the proof:
I don't think I fully understand the meaning of the sentence: "view $alpha in l$ as giving a $k$-linear endomorphism of $k^{d}$ by multiplication." Does it mean we could multiply $alpha$ with any element in $k^{d}$ and get a new element in $k^d$? Is it true that this endomorphism is an automorphism? Could anyone elaborate what this means?
How do you evaluate the determinant appeared in this equation: $p(t_1,cdots,t_d)=det(alpha_1 t_1+cdots+ alpha_d t_d)$? Isn't $alpha_1 t_1+cdots+ alpha_d t_d$ a polynomial? How to calculate the determinant of a polynomial? Why $P$ is a homogeneous polynomial of degree $d$?
Thank you in advance for any feedback and advice!
field-theory extension-field
field-theory extension-field
asked Nov 21 at 16:14
Jiguang Li
63
asked Nov 21 at 16:14
Jiguang Li
63
edited Nov 21 at 16:21
asked Nov 21 at 16:14
Jiguang Li
63
asked Nov 21 at 16:14
Jiguang Li
63
asked Nov 21 at 16:14
Jiguang Li
63
63
For the first question, we're just talking about the map $k^d rightarrow k^d$, $beta mapsto beta alpha$. It's clearly an endomorphism; the inverse is the map coming from multiplication by $alpha^{-1}$. Also, thanks for asking this question! I've been thinking about this sort of thing for a while and it's hard to find references. Hope you get a good answer!
– Sam Streeter
Nov 21 at 16:19
For the second question, I'm thinking we're maybe really talking about the norm.
– Sam Streeter
Nov 21 at 16:24
Thank you for your reply to question 1: it's more clear now. But I'm still confused by the second question. Did you treat $a_1t_1+cdots+ a_dt_d$ as a vector? If it refers to the norm, which norm do we use here?
– Jiguang Li
Nov 22 at 16:41
I'm struggling to come up with a concise and easy explanation. I think taking a look at Example 1.2.3 of Poonen's book "Rational Points on Varieties" might be enlightening though. It is also a good reference for material on $C_r$ fields. You can find it here: www-math.mit.edu/~poonen/papers/Qpoints.pdf
– Sam Streeter
Nov 22 at 16:52
1
@ Sam Streeter: thank you for the reference ! I believe the idea that $det$ refers to the norm really makes sense now.
– Jiguang Li
Nov 25 at 22:20
add a comment |
For the first question, we're just talking about the map $k^d rightarrow k^d$, $beta mapsto beta alpha$. It's clearly an endomorphism; the inverse is the map coming from multiplication by $alpha^{-1}$. Also, thanks for asking this question! I've been thinking about this sort of thing for a while and it's hard to find references. Hope you get a good answer!
– Sam Streeter
Nov 21 at 16:19
For the second question, I'm thinking we're maybe really talking about the norm.
– Sam Streeter
Nov 21 at 16:24
Thank you for your reply to question 1: it's more clear now. But I'm still confused by the second question. Did you treat $a_1t_1+cdots+ a_dt_d$ as a vector? If it refers to the norm, which norm do we use here?
– Jiguang Li
Nov 22 at 16:41
I'm struggling to come up with a concise and easy explanation. I think taking a look at Example 1.2.3 of Poonen's book "Rational Points on Varieties" might be enlightening though. It is also a good reference for material on $C_r$ fields. You can find it here: www-math.mit.edu/~poonen/papers/Qpoints.pdf
– Sam Streeter
Nov 22 at 16:52
1
@ Sam Streeter: thank you for the reference ! I believe the idea that $det$ refers to the norm really makes sense now.
– Jiguang Li
Nov 25 at 22:20
For the first question, we're just talking about the map $k^d rightarrow k^d$, $beta mapsto beta alpha$. It's clearly an endomorphism; the inverse is the map coming from multiplication by $alpha^{-1}$. Also, thanks for asking this question! I've been thinking about this sort of thing for a while and it's hard to find references. Hope you get a good answer!
– Sam Streeter
Nov 21 at 16:19
For the first question, we're just talking about the map $k^d rightarrow k^d$, $beta mapsto beta alpha$. It's clearly an endomorphism; the inverse is the map coming from multiplication by $alpha^{-1}$. Also, thanks for asking this question! I've been thinking about this sort of thing for a while and it's hard to find references. Hope you get a good answer!
– Sam Streeter
Nov 21 at 16:19
For the second question, I'm thinking we're maybe really talking about the norm.
– Sam Streeter
Nov 21 at 16:24
For the second question, I'm thinking we're maybe really talking about the norm.
– Sam Streeter
Nov 21 at 16:24
Thank you for your reply to question 1: it's more clear now. But I'm still confused by the second question. Did you treat $a_1t_1+cdots+ a_dt_d$ as a vector? If it refers to the norm, which norm do we use here?
– Jiguang Li
Nov 22 at 16:41
Thank you for your reply to question 1: it's more clear now. But I'm still confused by the second question. Did you treat $a_1t_1+cdots+ a_dt_d$ as a vector? If it refers to the norm, which norm do we use here?
– Jiguang Li
Nov 22 at 16:41
I'm struggling to come up with a concise and easy explanation. I think taking a look at Example 1.2.3 of Poonen's book "Rational Points on Varieties" might be enlightening though. It is also a good reference for material on $C_r$ fields. You can find it here: www-math.mit.edu/~poonen/papers/Qpoints.pdf
– Sam Streeter
Nov 22 at 16:52
I'm struggling to come up with a concise and easy explanation. I think taking a look at Example 1.2.3 of Poonen's book "Rational Points on Varieties" might be enlightening though. It is also a good reference for material on $C_r$ fields. You can find it here: www-math.mit.edu/~poonen/papers/Qpoints.pdf
– Sam Streeter
Nov 22 at 16:52
1
1
@ Sam Streeter: thank you for the reference ! I believe the idea that $det$ refers to the norm really makes sense now.
– Jiguang Li
Nov 25 at 22:20
@ Sam Streeter: thank you for the reference ! I believe the idea that $det$ refers to the norm really makes sense now.
– Jiguang Li
Nov 25 at 22:20
add a comment |
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For the first question, we're just talking about the map $k^d rightarrow k^d$, $beta mapsto beta alpha$. It's clearly an endomorphism; the inverse is the map coming from multiplication by $alpha^{-1}$. Also, thanks for asking this question! I've been thinking about this sort of thing for a while and it's hard to find references. Hope you get a good answer!
– Sam Streeter
Nov 21 at 16:19
For the second question, I'm thinking we're maybe really talking about the norm.
– Sam Streeter
Nov 21 at 16:24
Thank you for your reply to question 1: it's more clear now. But I'm still confused by the second question. Did you treat $a_1t_1+cdots+ a_dt_d$ as a vector? If it refers to the norm, which norm do we use here?
– Jiguang Li
Nov 22 at 16:41
I'm struggling to come up with a concise and easy explanation. I think taking a look at Example 1.2.3 of Poonen's book "Rational Points on Varieties" might be enlightening though. It is also a good reference for material on $C_r$ fields. You can find it here: www-math.mit.edu/~poonen/papers/Qpoints.pdf
– Sam Streeter
Nov 22 at 16:52
1
@ Sam Streeter: thank you for the reference ! I believe the idea that $det$ refers to the norm really makes sense now.
– Jiguang Li
Nov 25 at 22:20