Some questions regarding the proof of Luroth theorem given by G. Bergman
G. Bergman gave a series of exercises leading to the proof of Luroth's Theorem: https://math.berkeley.edu/~gbergman/grad.hndts/Luroth.ps
My concern lays in problem (7) and (8).
$L$ is any intermediate field: $k subsetneq L subset k(t)$, where $t$ is transcendental over $k$. $u=P(t)/Q(t)$ is an element of $L-k$.
Please refer to the pdf file for other details.
Show that if $P(x)-uQ(x)$ is divisible in $L[x]$ by an element of $k[x]$, then this element must divide both $P(x)$ and $Q(x)$.
how to prove it bothers me; my thought is like this: $ {1,u} subset B$, where $B$ is a basis of $L$ over $k$. Then $L[x] = k[u,..][x]$. After we embed $k[x]$ into $L[x]$, $u$ clearly is not divisible by any element from $k[x]$ then $P(x)$ and $Q(x)$ has to be divided by the element respectively. Am I correct?
In addition, is there any theorem leading to such result that if $t$ has an identical minimal polynomial over $k(u)$ and $L$ while $k(u) subset L$, then $L=k(u)$?
abstract-algebra field-theory divisibility extension-field
asked Nov 20 at 0:16
Alex Gong
11
add a comment |
G. Bergman gave a series of exercises leading to the proof of Luroth's Theorem: https://math.berkeley.edu/~gbergman/grad.hndts/Luroth.ps
My concern lays in problem (7) and (8).
$L$ is any intermediate field: $k subsetneq L subset k(t)$, where $t$ is transcendental over $k$. $u=P(t)/Q(t)$ is an element of $L-k$.
Please refer to the pdf file for other details.
Show that if $P(x)-uQ(x)$ is divisible in $L[x]$ by an element of $k[x]$, then this element must divide both $P(x)$ and $Q(x)$.
how to prove it bothers me; my thought is like this: $ {1,u} subset B$, where $B$ is a basis of $L$ over $k$. Then $L[x] = k[u,..][x]$. After we embed $k[x]$ into $L[x]$, $u$ clearly is not divisible by any element from $k[x]$ then $P(x)$ and $Q(x)$ has to be divided by the element respectively. Am I correct?
In addition, is there any theorem leading to such result that if $t$ has an identical minimal polynomial over $k(u)$ and $L$ while $k(u) subset L$, then $L=k(u)$?
abstract-algebra field-theory divisibility extension-field
asked Nov 20 at 0:16
Alex Gong
11
What are $P$ and $Q$? Polynomials in $kleft[xright]$ or in $Lleft[xright]$ ?
– darij grinberg
Nov 20 at 4:23
Also your last question sounds weird to me. Could it be the $subset$ sign should be $supset$?
– darij grinberg
Nov 20 at 4:23
i suggest that if possible, you shall read the pdf file. The sign is correct, since $u$ is in $L-k$.
– Alex Gong
Nov 20 at 4:48
add a comment |
G. Bergman gave a series of exercises leading to the proof of Luroth's Theorem: https://math.berkeley.edu/~gbergman/grad.hndts/Luroth.ps
My concern lays in problem (7) and (8).
$L$ is any intermediate field: $k subsetneq L subset k(t)$, where $t$ is transcendental over $k$. $u=P(t)/Q(t)$ is an element of $L-k$.
Please refer to the pdf file for other details.
Show that if $P(x)-uQ(x)$ is divisible in $L[x]$ by an element of $k[x]$, then this element must divide both $P(x)$ and $Q(x)$.
how to prove it bothers me; my thought is like this: $ {1,u} subset B$, where $B$ is a basis of $L$ over $k$. Then $L[x] = k[u,..][x]$. After we embed $k[x]$ into $L[x]$, $u$ clearly is not divisible by any element from $k[x]$ then $P(x)$ and $Q(x)$ has to be divided by the element respectively. Am I correct?
In addition, is there any theorem leading to such result that if $t$ has an identical minimal polynomial over $k(u)$ and $L$ while $k(u) subset L$, then $L=k(u)$?
abstract-algebra field-theory divisibility extension-field
asked Nov 20 at 0:16
Alex Gong
11
G. Bergman gave a series of exercises leading to the proof of Luroth's Theorem: https://math.berkeley.edu/~gbergman/grad.hndts/Luroth.ps
My concern lays in problem (7) and (8).
$L$ is any intermediate field: $k subsetneq L subset k(t)$, where $t$ is transcendental over $k$. $u=P(t)/Q(t)$ is an element of $L-k$.
Please refer to the pdf file for other details.
Show that if $P(x)-uQ(x)$ is divisible in $L[x]$ by an element of $k[x]$, then this element must divide both $P(x)$ and $Q(x)$.
how to prove it bothers me; my thought is like this: $ {1,u} subset B$, where $B$ is a basis of $L$ over $k$. Then $L[x] = k[u,..][x]$. After we embed $k[x]$ into $L[x]$, $u$ clearly is not divisible by any element from $k[x]$ then $P(x)$ and $Q(x)$ has to be divided by the element respectively. Am I correct?
In addition, is there any theorem leading to such result that if $t$ has an identical minimal polynomial over $k(u)$ and $L$ while $k(u) subset L$, then $L=k(u)$?
abstract-algebra field-theory divisibility extension-field
abstract-algebra field-theory divisibility extension-field
asked Nov 20 at 0:16
Alex Gong
11
asked Nov 20 at 0:16
Alex Gong
11
edited Nov 20 at 6:17
asked Nov 20 at 0:16
Alex Gong
11
asked Nov 20 at 0:16
Alex Gong
11
asked Nov 20 at 0:16
Alex Gong
11
11
What are $P$ and $Q$? Polynomials in $kleft[xright]$ or in $Lleft[xright]$ ?
– darij grinberg
Nov 20 at 4:23
Also your last question sounds weird to me. Could it be the $subset$ sign should be $supset$?
– darij grinberg
Nov 20 at 4:23
i suggest that if possible, you shall read the pdf file. The sign is correct, since $u$ is in $L-k$.
– Alex Gong
Nov 20 at 4:48
add a comment |
What are $P$ and $Q$? Polynomials in $kleft[xright]$ or in $Lleft[xright]$ ?
– darij grinberg
Nov 20 at 4:23
Also your last question sounds weird to me. Could it be the $subset$ sign should be $supset$?
– darij grinberg
Nov 20 at 4:23
i suggest that if possible, you shall read the pdf file. The sign is correct, since $u$ is in $L-k$.
– Alex Gong
Nov 20 at 4:48
What are $P$ and $Q$? Polynomials in $kleft[xright]$ or in $Lleft[xright]$ ?
– darij grinberg
Nov 20 at 4:23
What are $P$ and $Q$? Polynomials in $kleft[xright]$ or in $Lleft[xright]$ ?
– darij grinberg
Nov 20 at 4:23
Also your last question sounds weird to me. Could it be the $subset$ sign should be $supset$?
– darij grinberg
Nov 20 at 4:23
Also your last question sounds weird to me. Could it be the $subset$ sign should be $supset$?
– darij grinberg
Nov 20 at 4:23
i suggest that if possible, you shall read the pdf file. The sign is correct, since $u$ is in $L-k$.
– Alex Gong
Nov 20 at 4:48
i suggest that if possible, you shall read the pdf file. The sign is correct, since $u$ is in $L-k$.
– Alex Gong
Nov 20 at 4:48
add a comment |
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What are $P$ and $Q$? Polynomials in $kleft[xright]$ or in $Lleft[xright]$ ?
– darij grinberg
Nov 20 at 4:23
Also your last question sounds weird to me. Could it be the $subset$ sign should be $supset$?
– darij grinberg
Nov 20 at 4:23
i suggest that if possible, you shall read the pdf file. The sign is correct, since $u$ is in $L-k$.
– Alex Gong
Nov 20 at 4:48