The algebraic elements of $K$ extension of $F$ are a subfield of $K$
$begingroup$
the question I wanna try to answer is this:
The algebraic elements of $K$ extension of $F$ are a subfield of $K$
I know that:
An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.
My approach is that I have two algebraic elements $a,b in K$.
$a+b in K$ and also $ab in K$, and of course they belong to $S$ too.
Maybe I need to write a polynomial with the algebraic elements and show that is finite?
I've some difficult to glue together those pieces for a satisfying proof. Any hints?
Thanks
abstract-algebra
edited Dec 13 '18 at 10:09
Bernard
121k740116
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
$endgroup$
add a comment |
$begingroup$
the question I wanna try to answer is this:
The algebraic elements of $K$ extension of $F$ are a subfield of $K$
I know that:
An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.
My approach is that I have two algebraic elements $a,b in K$.
$a+b in K$ and also $ab in K$, and of course they belong to $S$ too.
Maybe I need to write a polynomial with the algebraic elements and show that is finite?
I've some difficult to glue together those pieces for a satisfying proof. Any hints?
Thanks
abstract-algebra
edited Dec 13 '18 at 10:09
Bernard
121k740116
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
$endgroup$
add a comment |
$begingroup$
the question I wanna try to answer is this:
The algebraic elements of $K$ extension of $F$ are a subfield of $K$
I know that:
An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.
My approach is that I have two algebraic elements $a,b in K$.
$a+b in K$ and also $ab in K$, and of course they belong to $S$ too.
Maybe I need to write a polynomial with the algebraic elements and show that is finite?
I've some difficult to glue together those pieces for a satisfying proof. Any hints?
Thanks
abstract-algebra
edited Dec 13 '18 at 10:09
Bernard
121k740116
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
$endgroup$
the question I wanna try to answer is this:
The algebraic elements of $K$ extension of $F$ are a subfield of $K$
I know that:
An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.
My approach is that I have two algebraic elements $a,b in K$.
$a+b in K$ and also $ab in K$, and of course they belong to $S$ too.
Maybe I need to write a polynomial with the algebraic elements and show that is finite?
I've some difficult to glue together those pieces for a satisfying proof. Any hints?
Thanks
abstract-algebra
abstract-algebra
edited Dec 13 '18 at 10:09
Bernard
121k740116
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
edited Dec 13 '18 at 10:09
Bernard
121k740116
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
edited Dec 13 '18 at 10:09
Bernard
121k740116
edited Dec 13 '18 at 10:09
Bernard
121k740116
edited Dec 13 '18 at 10:09
Bernard
121k740116
121k740116
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
asked Dec 13 '18 at 9:56
AlessarAlessar
313115
313115
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
$endgroup$
$begingroup$
To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
$endgroup$
– Alessar
Dec 13 '18 at 10:22
$begingroup$
I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
$endgroup$
– Alessar
Dec 13 '18 at 10:27
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
$endgroup$
$begingroup$
To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
$endgroup$
– Alessar
Dec 13 '18 at 10:22
$begingroup$
I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
$endgroup$
– Alessar
Dec 13 '18 at 10:27
add a comment |
$begingroup$
Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
$endgroup$
$begingroup$
To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
$endgroup$
– Alessar
Dec 13 '18 at 10:22
$begingroup$
I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
$endgroup$
– Alessar
Dec 13 '18 at 10:27
add a comment |
$begingroup$
Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
$endgroup$
Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
answered Dec 13 '18 at 10:03
lhflhf
165k10171396
165k10171396
$begingroup$
To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
$endgroup$
– Alessar
Dec 13 '18 at 10:22
$begingroup$
I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
$endgroup$
– Alessar
Dec 13 '18 at 10:27
add a comment |
$begingroup$
To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
$endgroup$
– Alessar
Dec 13 '18 at 10:22
$begingroup$
I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
$endgroup$
– Alessar
Dec 13 '18 at 10:27
$begingroup$
To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
$endgroup$
– Alessar
Dec 13 '18 at 10:22
$begingroup$
To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
$endgroup$
– Alessar
Dec 13 '18 at 10:22
$begingroup$
I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
$endgroup$
– Alessar
Dec 13 '18 at 10:27
$begingroup$
I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
$endgroup$
– Alessar
Dec 13 '18 at 10:27
add a comment |
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