Field always equipped with arbitrary degree extension?
$begingroup$
Let $K$ be a field. Over finite field $F_p$ or rational $Q$, it is clear there are extensions of any degree. The former is obtained by taking the muliplicative group adjoining units in $overline{F}_p$ and the latter is obtained by considering Eisenstein applied to any $Z[x]$ irreducible polynomial. The Eisenstein argument applies to function field case as well. Hence the statement holds for function fields.
$textbf{Q:}$ Is it true that in general $K$ will always have arbitrary degree of extensions? If so, how do I see this? If not, what is the counter example?
abstract-algebra
asked Jan 1 at 14:22
user45765user45765
2,6322724
$endgroup$
|
show 2 more comments
$begingroup$
Let $K$ be a field. Over finite field $F_p$ or rational $Q$, it is clear there are extensions of any degree. The former is obtained by taking the muliplicative group adjoining units in $overline{F}_p$ and the latter is obtained by considering Eisenstein applied to any $Z[x]$ irreducible polynomial. The Eisenstein argument applies to function field case as well. Hence the statement holds for function fields.
$textbf{Q:}$ Is it true that in general $K$ will always have arbitrary degree of extensions? If so, how do I see this? If not, what is the counter example?
abstract-algebra
asked Jan 1 at 14:22
user45765user45765
2,6322724
$endgroup$
2
$begingroup$
Try $K=mathbb C$ for a counterexample. or, $overline {mathbb F_p}$ since you mentioned it.
$endgroup$
– lulu
Jan 1 at 14:29
$begingroup$
@lulu OK. So for algebraically closed, it is not going to work. Suppose $K$ is a proper subfield of algebraically closed field. Does this still hold?
$endgroup$
– user45765
Jan 1 at 14:34
2
$begingroup$
Try $mathbb R$.
$endgroup$
– lulu
Jan 1 at 14:35
1
$begingroup$
@lulu It seems this post answers the question. math.stackexchange.com/q/304516
$endgroup$
– user45765
Jan 1 at 14:46
3
$begingroup$
More interesting example: consider $K=bigcup_{k=1}^inftyBbb F_{p^{2^k}}$. Then $K$ has extensions of every odd degree, but none of any even degree.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:15
|
show 2 more comments
$begingroup$
Let $K$ be a field. Over finite field $F_p$ or rational $Q$, it is clear there are extensions of any degree. The former is obtained by taking the muliplicative group adjoining units in $overline{F}_p$ and the latter is obtained by considering Eisenstein applied to any $Z[x]$ irreducible polynomial. The Eisenstein argument applies to function field case as well. Hence the statement holds for function fields.
$textbf{Q:}$ Is it true that in general $K$ will always have arbitrary degree of extensions? If so, how do I see this? If not, what is the counter example?
abstract-algebra
asked Jan 1 at 14:22
user45765user45765
2,6322724
$endgroup$
Let $K$ be a field. Over finite field $F_p$ or rational $Q$, it is clear there are extensions of any degree. The former is obtained by taking the muliplicative group adjoining units in $overline{F}_p$ and the latter is obtained by considering Eisenstein applied to any $Z[x]$ irreducible polynomial. The Eisenstein argument applies to function field case as well. Hence the statement holds for function fields.
$textbf{Q:}$ Is it true that in general $K$ will always have arbitrary degree of extensions? If so, how do I see this? If not, what is the counter example?
abstract-algebra
abstract-algebra
asked Jan 1 at 14:22
user45765user45765
2,6322724
asked Jan 1 at 14:22
user45765user45765
2,6322724
asked Jan 1 at 14:22
user45765user45765
2,6322724
asked Jan 1 at 14:22
user45765user45765
2,6322724
asked Jan 1 at 14:22
user45765user45765
2,6322724
2,6322724
2
$begingroup$
Try $K=mathbb C$ for a counterexample. or, $overline {mathbb F_p}$ since you mentioned it.
$endgroup$
– lulu
Jan 1 at 14:29
$begingroup$
@lulu OK. So for algebraically closed, it is not going to work. Suppose $K$ is a proper subfield of algebraically closed field. Does this still hold?
$endgroup$
– user45765
Jan 1 at 14:34
2
$begingroup$
Try $mathbb R$.
$endgroup$
– lulu
Jan 1 at 14:35
1
$begingroup$
@lulu It seems this post answers the question. math.stackexchange.com/q/304516
$endgroup$
– user45765
Jan 1 at 14:46
3
$begingroup$
More interesting example: consider $K=bigcup_{k=1}^inftyBbb F_{p^{2^k}}$. Then $K$ has extensions of every odd degree, but none of any even degree.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:15
|
show 2 more comments
2
$begingroup$
Try $K=mathbb C$ for a counterexample. or, $overline {mathbb F_p}$ since you mentioned it.
$endgroup$
– lulu
Jan 1 at 14:29
$begingroup$
@lulu OK. So for algebraically closed, it is not going to work. Suppose $K$ is a proper subfield of algebraically closed field. Does this still hold?
$endgroup$
– user45765
Jan 1 at 14:34
2
$begingroup$
Try $mathbb R$.
$endgroup$
– lulu
Jan 1 at 14:35
1
$begingroup$
@lulu It seems this post answers the question. math.stackexchange.com/q/304516
$endgroup$
– user45765
Jan 1 at 14:46
3
$begingroup$
More interesting example: consider $K=bigcup_{k=1}^inftyBbb F_{p^{2^k}}$. Then $K$ has extensions of every odd degree, but none of any even degree.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:15
2
2
$begingroup$
Try $K=mathbb C$ for a counterexample. or, $overline {mathbb F_p}$ since you mentioned it.
$endgroup$
– lulu
Jan 1 at 14:29
$begingroup$
Try $K=mathbb C$ for a counterexample. or, $overline {mathbb F_p}$ since you mentioned it.
$endgroup$
– lulu
Jan 1 at 14:29
$begingroup$
@lulu OK. So for algebraically closed, it is not going to work. Suppose $K$ is a proper subfield of algebraically closed field. Does this still hold?
$endgroup$
– user45765
Jan 1 at 14:34
$begingroup$
@lulu OK. So for algebraically closed, it is not going to work. Suppose $K$ is a proper subfield of algebraically closed field. Does this still hold?
$endgroup$
– user45765
Jan 1 at 14:34
2
2
$begingroup$
Try $mathbb R$.
$endgroup$
– lulu
Jan 1 at 14:35
$begingroup$
Try $mathbb R$.
$endgroup$
– lulu
Jan 1 at 14:35
1
1
$begingroup$
@lulu It seems this post answers the question. math.stackexchange.com/q/304516
$endgroup$
– user45765
Jan 1 at 14:46
$begingroup$
@lulu It seems this post answers the question. math.stackexchange.com/q/304516
$endgroup$
– user45765
Jan 1 at 14:46
3
3
$begingroup$
More interesting example: consider $K=bigcup_{k=1}^inftyBbb F_{p^{2^k}}$. Then $K$ has extensions of every odd degree, but none of any even degree.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:15
$begingroup$
More interesting example: consider $K=bigcup_{k=1}^inftyBbb F_{p^{2^k}}$. Then $K$ has extensions of every odd degree, but none of any even degree.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:15
|
show 2 more comments
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$begingroup$
Try $K=mathbb C$ for a counterexample. or, $overline {mathbb F_p}$ since you mentioned it.
$endgroup$
– lulu
Jan 1 at 14:29
$begingroup$
@lulu OK. So for algebraically closed, it is not going to work. Suppose $K$ is a proper subfield of algebraically closed field. Does this still hold?
$endgroup$
– user45765
Jan 1 at 14:34
2
$begingroup$
Try $mathbb R$.
$endgroup$
– lulu
Jan 1 at 14:35
1
$begingroup$
@lulu It seems this post answers the question. math.stackexchange.com/q/304516
$endgroup$
– user45765
Jan 1 at 14:46
3
$begingroup$
More interesting example: consider $K=bigcup_{k=1}^inftyBbb F_{p^{2^k}}$. Then $K$ has extensions of every odd degree, but none of any even degree.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:15