Existence of Embedding
$begingroup$
I'm not sure how to solve the following task
Let $Ksubset L_{0} subset L$ be a tower field extension, $L$ is algebraic over $L_{0}$ and $Ksubset E$ is a field extension, $E$ is the algebraic closure of $K$. Let $theta_{0} : L_{0} to E$ be an embedding over K.
Show that there exists an embedding $theta : L to E$ over $theta_{0}$.
These are the definitions which we used in the lecture but it doesn't make sense to me:
A field extension of a field $K$ is a ring homomorphism $a: K to N$, $N$ is also a field.
Let $K,M$ be fields. The ring homomorphism $b: K to M$ is called embedding.
Let $b: K to M$ and $c:K to L$ be two embeddings then the ring homomorphism $d: Lto M$ with $dcirc b=c$ is called embedding over $c$.
$c$ is an embedding over $K$ if the two embeddings $b: K to M$ and $c:K to L$ satisfy $Ksubset M$ and $Ksubset L$.
Many thanks for any help in advance.
Edit: Definitions
abstract-algebra
asked Dec 6 '18 at 0:26
AnzuAnzu
146
$endgroup$
add a comment |
$begingroup$
I'm not sure how to solve the following task
Let $Ksubset L_{0} subset L$ be a tower field extension, $L$ is algebraic over $L_{0}$ and $Ksubset E$ is a field extension, $E$ is the algebraic closure of $K$. Let $theta_{0} : L_{0} to E$ be an embedding over K.
Show that there exists an embedding $theta : L to E$ over $theta_{0}$.
These are the definitions which we used in the lecture but it doesn't make sense to me:
A field extension of a field $K$ is a ring homomorphism $a: K to N$, $N$ is also a field.
Let $K,M$ be fields. The ring homomorphism $b: K to M$ is called embedding.
Let $b: K to M$ and $c:K to L$ be two embeddings then the ring homomorphism $d: Lto M$ with $dcirc b=c$ is called embedding over $c$.
$c$ is an embedding over $K$ if the two embeddings $b: K to M$ and $c:K to L$ satisfy $Ksubset M$ and $Ksubset L$.
Many thanks for any help in advance.
Edit: Definitions
abstract-algebra
asked Dec 6 '18 at 0:26
AnzuAnzu
146
$endgroup$
$begingroup$
As you've formulated it, this isn't true. For all you know, $theta_0$ may be surjective and $L/L_0$ not trivial.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:29
$begingroup$
I'll ask my professor if there's a missing condition, thank you
$endgroup$
– Anzu
Dec 6 '18 at 0:34
$begingroup$
Ah, no sorry. I had missed the fact that $E$ is not any extension, but an algebraic closure of $K$. Then, it's ok.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:36
$begingroup$
That was my bad, I just edited that. Sorry about that
$endgroup$
– Anzu
Dec 6 '18 at 0:37
add a comment |
$begingroup$
I'm not sure how to solve the following task
Let $Ksubset L_{0} subset L$ be a tower field extension, $L$ is algebraic over $L_{0}$ and $Ksubset E$ is a field extension, $E$ is the algebraic closure of $K$. Let $theta_{0} : L_{0} to E$ be an embedding over K.
Show that there exists an embedding $theta : L to E$ over $theta_{0}$.
These are the definitions which we used in the lecture but it doesn't make sense to me:
A field extension of a field $K$ is a ring homomorphism $a: K to N$, $N$ is also a field.
Let $K,M$ be fields. The ring homomorphism $b: K to M$ is called embedding.
Let $b: K to M$ and $c:K to L$ be two embeddings then the ring homomorphism $d: Lto M$ with $dcirc b=c$ is called embedding over $c$.
$c$ is an embedding over $K$ if the two embeddings $b: K to M$ and $c:K to L$ satisfy $Ksubset M$ and $Ksubset L$.
Many thanks for any help in advance.
Edit: Definitions
abstract-algebra
asked Dec 6 '18 at 0:26
AnzuAnzu
146
$endgroup$
I'm not sure how to solve the following task
Let $Ksubset L_{0} subset L$ be a tower field extension, $L$ is algebraic over $L_{0}$ and $Ksubset E$ is a field extension, $E$ is the algebraic closure of $K$. Let $theta_{0} : L_{0} to E$ be an embedding over K.
Show that there exists an embedding $theta : L to E$ over $theta_{0}$.
These are the definitions which we used in the lecture but it doesn't make sense to me:
A field extension of a field $K$ is a ring homomorphism $a: K to N$, $N$ is also a field.
Let $K,M$ be fields. The ring homomorphism $b: K to M$ is called embedding.
Let $b: K to M$ and $c:K to L$ be two embeddings then the ring homomorphism $d: Lto M$ with $dcirc b=c$ is called embedding over $c$.
$c$ is an embedding over $K$ if the two embeddings $b: K to M$ and $c:K to L$ satisfy $Ksubset M$ and $Ksubset L$.
Many thanks for any help in advance.
Edit: Definitions
abstract-algebra
abstract-algebra
asked Dec 6 '18 at 0:26
AnzuAnzu
146
asked Dec 6 '18 at 0:26
AnzuAnzu
146
edited Dec 6 '18 at 13:46
Anzu
asked Dec 6 '18 at 0:26
AnzuAnzu
146
asked Dec 6 '18 at 0:26
AnzuAnzu
146
asked Dec 6 '18 at 0:26
AnzuAnzu
146
146
$begingroup$
As you've formulated it, this isn't true. For all you know, $theta_0$ may be surjective and $L/L_0$ not trivial.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:29
$begingroup$
I'll ask my professor if there's a missing condition, thank you
$endgroup$
– Anzu
Dec 6 '18 at 0:34
$begingroup$
Ah, no sorry. I had missed the fact that $E$ is not any extension, but an algebraic closure of $K$. Then, it's ok.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:36
$begingroup$
That was my bad, I just edited that. Sorry about that
$endgroup$
– Anzu
Dec 6 '18 at 0:37
add a comment |
$begingroup$
As you've formulated it, this isn't true. For all you know, $theta_0$ may be surjective and $L/L_0$ not trivial.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:29
$begingroup$
I'll ask my professor if there's a missing condition, thank you
$endgroup$
– Anzu
Dec 6 '18 at 0:34
$begingroup$
Ah, no sorry. I had missed the fact that $E$ is not any extension, but an algebraic closure of $K$. Then, it's ok.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:36
$begingroup$
That was my bad, I just edited that. Sorry about that
$endgroup$
– Anzu
Dec 6 '18 at 0:37
$begingroup$
As you've formulated it, this isn't true. For all you know, $theta_0$ may be surjective and $L/L_0$ not trivial.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:29
$begingroup$
As you've formulated it, this isn't true. For all you know, $theta_0$ may be surjective and $L/L_0$ not trivial.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:29
$begingroup$
I'll ask my professor if there's a missing condition, thank you
$endgroup$
– Anzu
Dec 6 '18 at 0:34
$begingroup$
I'll ask my professor if there's a missing condition, thank you
$endgroup$
– Anzu
Dec 6 '18 at 0:34
$begingroup$
Ah, no sorry. I had missed the fact that $E$ is not any extension, but an algebraic closure of $K$. Then, it's ok.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:36
$begingroup$
Ah, no sorry. I had missed the fact that $E$ is not any extension, but an algebraic closure of $K$. Then, it's ok.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:36
$begingroup$
That was my bad, I just edited that. Sorry about that
$endgroup$
– Anzu
Dec 6 '18 at 0:37
$begingroup$
That was my bad, I just edited that. Sorry about that
$endgroup$
– Anzu
Dec 6 '18 at 0:37
add a comment |
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$begingroup$
As you've formulated it, this isn't true. For all you know, $theta_0$ may be surjective and $L/L_0$ not trivial.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:29
$begingroup$
I'll ask my professor if there's a missing condition, thank you
$endgroup$
– Anzu
Dec 6 '18 at 0:34
$begingroup$
Ah, no sorry. I had missed the fact that $E$ is not any extension, but an algebraic closure of $K$. Then, it's ok.
$endgroup$
– Saucy O'Path
Dec 6 '18 at 0:36
$begingroup$
That was my bad, I just edited that. Sorry about that
$endgroup$
– Anzu
Dec 6 '18 at 0:37