Transitivity of Algebraicity of Field Extensions: the reciprocal
$begingroup$
Let $ksubset K subset L$ be field extensions.
If K/k and L/K are algebraic then L/k is algebraic.
What about the reciprocal?
Thank you!
abstract-algebra
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
$endgroup$
add a comment |
$begingroup$
Let $ksubset K subset L$ be field extensions.
If K/k and L/K are algebraic then L/k is algebraic.
What about the reciprocal?
Thank you!
abstract-algebra
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
$endgroup$
2
$begingroup$
You may want to run argument at the level of individual elements. Also in the converse direction. By the way, isn't the converse rather easy? If every element of $L$ is algebraic over $k$ then surely the same holds for the elements of $K$. And being algebraic over $k$ implies algebraic over a bigger field, no?
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 16:49
$begingroup$
Right but what about L/K if L/k is algebraic? are all L elements algebaric over K?
$endgroup$
– PerelMan
Dec 18 '18 at 16:58
2
$begingroup$
Let $zin L$ be arbitrary. Because $z$ is algebraic over $k$, there exists a non-zero polynomial $m(x)in k[x]$ such that $m(z)=0$. Because $k$ is a subset of $K$, $m(x)in K[x]$. Therefore $z$ is algebraic over $K$.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 17:01
$begingroup$
Thank you! it's clear now
$endgroup$
– PerelMan
Dec 18 '18 at 17:05
1
$begingroup$
Great! You are welcome to flesh out the argument as an answer. That way you get feedback on any residual unclear points (if any).
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 19:31
add a comment |
$begingroup$
Let $ksubset K subset L$ be field extensions.
If K/k and L/K are algebraic then L/k is algebraic.
What about the reciprocal?
Thank you!
abstract-algebra
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
$endgroup$
Let $ksubset K subset L$ be field extensions.
If K/k and L/K are algebraic then L/k is algebraic.
What about the reciprocal?
Thank you!
abstract-algebra
abstract-algebra
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
edited Dec 18 '18 at 17:11
PerelMan
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
asked Dec 18 '18 at 15:43
PerelManPerelMan
649313
649313
2
$begingroup$
You may want to run argument at the level of individual elements. Also in the converse direction. By the way, isn't the converse rather easy? If every element of $L$ is algebraic over $k$ then surely the same holds for the elements of $K$. And being algebraic over $k$ implies algebraic over a bigger field, no?
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 16:49
$begingroup$
Right but what about L/K if L/k is algebraic? are all L elements algebaric over K?
$endgroup$
– PerelMan
Dec 18 '18 at 16:58
2
$begingroup$
Let $zin L$ be arbitrary. Because $z$ is algebraic over $k$, there exists a non-zero polynomial $m(x)in k[x]$ such that $m(z)=0$. Because $k$ is a subset of $K$, $m(x)in K[x]$. Therefore $z$ is algebraic over $K$.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 17:01
$begingroup$
Thank you! it's clear now
$endgroup$
– PerelMan
Dec 18 '18 at 17:05
1
$begingroup$
Great! You are welcome to flesh out the argument as an answer. That way you get feedback on any residual unclear points (if any).
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 19:31
add a comment |
2
$begingroup$
You may want to run argument at the level of individual elements. Also in the converse direction. By the way, isn't the converse rather easy? If every element of $L$ is algebraic over $k$ then surely the same holds for the elements of $K$. And being algebraic over $k$ implies algebraic over a bigger field, no?
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 16:49
$begingroup$
Right but what about L/K if L/k is algebraic? are all L elements algebaric over K?
$endgroup$
– PerelMan
Dec 18 '18 at 16:58
2
$begingroup$
Let $zin L$ be arbitrary. Because $z$ is algebraic over $k$, there exists a non-zero polynomial $m(x)in k[x]$ such that $m(z)=0$. Because $k$ is a subset of $K$, $m(x)in K[x]$. Therefore $z$ is algebraic over $K$.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 17:01
$begingroup$
Thank you! it's clear now
$endgroup$
– PerelMan
Dec 18 '18 at 17:05
1
$begingroup$
Great! You are welcome to flesh out the argument as an answer. That way you get feedback on any residual unclear points (if any).
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 19:31
2
2
$begingroup$
You may want to run argument at the level of individual elements. Also in the converse direction. By the way, isn't the converse rather easy? If every element of $L$ is algebraic over $k$ then surely the same holds for the elements of $K$. And being algebraic over $k$ implies algebraic over a bigger field, no?
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 16:49
$begingroup$
You may want to run argument at the level of individual elements. Also in the converse direction. By the way, isn't the converse rather easy? If every element of $L$ is algebraic over $k$ then surely the same holds for the elements of $K$. And being algebraic over $k$ implies algebraic over a bigger field, no?
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 16:49
$begingroup$
Right but what about L/K if L/k is algebraic? are all L elements algebaric over K?
$endgroup$
– PerelMan
Dec 18 '18 at 16:58
$begingroup$
Right but what about L/K if L/k is algebraic? are all L elements algebaric over K?
$endgroup$
– PerelMan
Dec 18 '18 at 16:58
2
2
$begingroup$
Let $zin L$ be arbitrary. Because $z$ is algebraic over $k$, there exists a non-zero polynomial $m(x)in k[x]$ such that $m(z)=0$. Because $k$ is a subset of $K$, $m(x)in K[x]$. Therefore $z$ is algebraic over $K$.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 17:01
$begingroup$
Let $zin L$ be arbitrary. Because $z$ is algebraic over $k$, there exists a non-zero polynomial $m(x)in k[x]$ such that $m(z)=0$. Because $k$ is a subset of $K$, $m(x)in K[x]$. Therefore $z$ is algebraic over $K$.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 17:01
$begingroup$
Thank you! it's clear now
$endgroup$
– PerelMan
Dec 18 '18 at 17:05
$begingroup$
Thank you! it's clear now
$endgroup$
– PerelMan
Dec 18 '18 at 17:05
1
1
$begingroup$
Great! You are welcome to flesh out the argument as an answer. That way you get feedback on any residual unclear points (if any).
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 19:31
$begingroup$
Great! You are welcome to flesh out the argument as an answer. That way you get feedback on any residual unclear points (if any).
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 19:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This answer is mainly inspired from Jyrki's explanatory Comment.
Suppose L/k is algebraic.
Let $xin L$. x is algebraic over $k$, which means : $exists Pin k[X]: P(x)=0$.
But $Pin k[X] subset K[X]$ so x is algebraic over $K$ and $L/K$ is algebraic.
Let $yin K$. $yin L$ so $y$ is algebraic over $k$ and $K/k$ is also algebraic.
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This answer is mainly inspired from Jyrki's explanatory Comment.
Suppose L/k is algebraic.
Let $xin L$. x is algebraic over $k$, which means : $exists Pin k[X]: P(x)=0$.
But $Pin k[X] subset K[X]$ so x is algebraic over $K$ and $L/K$ is algebraic.
Let $yin K$. $yin L$ so $y$ is algebraic over $k$ and $K/k$ is also algebraic.
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
$endgroup$
add a comment |
$begingroup$
This answer is mainly inspired from Jyrki's explanatory Comment.
Suppose L/k is algebraic.
Let $xin L$. x is algebraic over $k$, which means : $exists Pin k[X]: P(x)=0$.
But $Pin k[X] subset K[X]$ so x is algebraic over $K$ and $L/K$ is algebraic.
Let $yin K$. $yin L$ so $y$ is algebraic over $k$ and $K/k$ is also algebraic.
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
$endgroup$
add a comment |
$begingroup$
This answer is mainly inspired from Jyrki's explanatory Comment.
Suppose L/k is algebraic.
Let $xin L$. x is algebraic over $k$, which means : $exists Pin k[X]: P(x)=0$.
But $Pin k[X] subset K[X]$ so x is algebraic over $K$ and $L/K$ is algebraic.
Let $yin K$. $yin L$ so $y$ is algebraic over $k$ and $K/k$ is also algebraic.
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
$endgroup$
This answer is mainly inspired from Jyrki's explanatory Comment.
Suppose L/k is algebraic.
Let $xin L$. x is algebraic over $k$, which means : $exists Pin k[X]: P(x)=0$.
But $Pin k[X] subset K[X]$ so x is algebraic over $K$ and $L/K$ is algebraic.
Let $yin K$. $yin L$ so $y$ is algebraic over $k$ and $K/k$ is also algebraic.
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
answered Dec 19 '18 at 0:24
PerelManPerelMan
649313
649313
add a comment |
add a comment |
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$begingroup$
You may want to run argument at the level of individual elements. Also in the converse direction. By the way, isn't the converse rather easy? If every element of $L$ is algebraic over $k$ then surely the same holds for the elements of $K$. And being algebraic over $k$ implies algebraic over a bigger field, no?
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 16:49
$begingroup$
Right but what about L/K if L/k is algebraic? are all L elements algebaric over K?
$endgroup$
– PerelMan
Dec 18 '18 at 16:58
2
$begingroup$
Let $zin L$ be arbitrary. Because $z$ is algebraic over $k$, there exists a non-zero polynomial $m(x)in k[x]$ such that $m(z)=0$. Because $k$ is a subset of $K$, $m(x)in K[x]$. Therefore $z$ is algebraic over $K$.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 17:01
$begingroup$
Thank you! it's clear now
$endgroup$
– PerelMan
Dec 18 '18 at 17:05
1
$begingroup$
Great! You are welcome to flesh out the argument as an answer. That way you get feedback on any residual unclear points (if any).
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 19:31