Field extensions contradicting tower law The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Field Extensions over the RationalsArriving at a Contradiction with the Tower Law3 questions on field extensionsField Extensions and Number of IsomorphismsSimple algebraic field extensionsApplication of Tower LawNon-separable, infinite field extensions of non-zero characteristicHow can I see that the tower of extension collapse when we have the square root of discriminant is in the base field?Galois correspondence with the following field extensionsAlgebraic Field Extension of Finite Field
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Field extensions contradicting tower law
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Field Extensions over the RationalsArriving at a Contradiction with the Tower Law3 questions on field extensionsField Extensions and Number of IsomorphismsSimple algebraic field extensionsApplication of Tower LawNon-separable, infinite field extensions of non-zero characteristicHow can I see that the tower of extension collapse when we have the square root of discriminant is in the base field?Galois correspondence with the following field extensionsAlgebraic Field Extension of Finite Field
$begingroup$
$[mathbbR(x):mathbbR(x+frac1x)]=2$ since $x$ satisfies a quadratic polynomial over $mathbbR(x+frac1x)$. Similarly, $[mathbbR(x):mathbbR(x^2+frac1x^2)]=4$. But, as $(x+frac1x)^2=(x^2+frac1x^2)+2$, so
$mathbbR(x):mathbbR(x^2+frac1x^2):mathbbR(x+frac1x)$ which clearly contradicts Tower law. What is wrong here?
field-theory galois-theory extension-field
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
$endgroup$
add a comment |
$begingroup$
$[mathbbR(x):mathbbR(x+frac1x)]=2$ since $x$ satisfies a quadratic polynomial over $mathbbR(x+frac1x)$. Similarly, $[mathbbR(x):mathbbR(x^2+frac1x^2)]=4$. But, as $(x+frac1x)^2=(x^2+frac1x^2)+2$, so
$mathbbR(x):mathbbR(x^2+frac1x^2):mathbbR(x+frac1x)$ which clearly contradicts Tower law. What is wrong here?
field-theory galois-theory extension-field
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
$endgroup$
2
$begingroup$
? What actually are you asserting after "so"?
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:34
$begingroup$
I got confused. I got it. Thanks@LordSharktheUnknown
$endgroup$
– Shanghaikid
Mar 31 at 7:47
add a comment |
$begingroup$
$[mathbbR(x):mathbbR(x+frac1x)]=2$ since $x$ satisfies a quadratic polynomial over $mathbbR(x+frac1x)$. Similarly, $[mathbbR(x):mathbbR(x^2+frac1x^2)]=4$. But, as $(x+frac1x)^2=(x^2+frac1x^2)+2$, so
$mathbbR(x):mathbbR(x^2+frac1x^2):mathbbR(x+frac1x)$ which clearly contradicts Tower law. What is wrong here?
field-theory galois-theory extension-field
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
$endgroup$
$[mathbbR(x):mathbbR(x+frac1x)]=2$ since $x$ satisfies a quadratic polynomial over $mathbbR(x+frac1x)$. Similarly, $[mathbbR(x):mathbbR(x^2+frac1x^2)]=4$. But, as $(x+frac1x)^2=(x^2+frac1x^2)+2$, so
$mathbbR(x):mathbbR(x^2+frac1x^2):mathbbR(x+frac1x)$ which clearly contradicts Tower law. What is wrong here?
field-theory galois-theory extension-field
field-theory galois-theory extension-field
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
asked Mar 31 at 6:10
ShanghaikidShanghaikid
669
669
2
$begingroup$
? What actually are you asserting after "so"?
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:34
$begingroup$
I got confused. I got it. Thanks@LordSharktheUnknown
$endgroup$
– Shanghaikid
Mar 31 at 7:47
add a comment |
2
$begingroup$
? What actually are you asserting after "so"?
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:34
$begingroup$
I got confused. I got it. Thanks@LordSharktheUnknown
$endgroup$
– Shanghaikid
Mar 31 at 7:47
2
2
$begingroup$
? What actually are you asserting after "so"?
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:34
$begingroup$
? What actually are you asserting after "so"?
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:34
$begingroup$
I got confused. I got it. Thanks@LordSharktheUnknown
$endgroup$
– Shanghaikid
Mar 31 at 7:47
$begingroup$
I got confused. I got it. Thanks@LordSharktheUnknown
$endgroup$
– Shanghaikid
Mar 31 at 7:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I can't see what you are asserting, but here are some facts:
$$|k(x):k(x+1/x)|=2,$$
$$|k(x):k(x^2+1/x^2)|=4,$$
$$|k(x+1/x):k(x^2+1/x^2)|=2.$$
Here $k$ is any field, and $x$ is transcendental over $k$.
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
add a comment |
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$begingroup$
I can't see what you are asserting, but here are some facts:
$$|k(x):k(x+1/x)|=2,$$
$$|k(x):k(x^2+1/x^2)|=4,$$
$$|k(x+1/x):k(x^2+1/x^2)|=2.$$
Here $k$ is any field, and $x$ is transcendental over $k$.
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
add a comment |
$begingroup$
I can't see what you are asserting, but here are some facts:
$$|k(x):k(x+1/x)|=2,$$
$$|k(x):k(x^2+1/x^2)|=4,$$
$$|k(x+1/x):k(x^2+1/x^2)|=2.$$
Here $k$ is any field, and $x$ is transcendental over $k$.
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
add a comment |
$begingroup$
I can't see what you are asserting, but here are some facts:
$$|k(x):k(x+1/x)|=2,$$
$$|k(x):k(x^2+1/x^2)|=4,$$
$$|k(x+1/x):k(x^2+1/x^2)|=2.$$
Here $k$ is any field, and $x$ is transcendental over $k$.
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
I can't see what you are asserting, but here are some facts:
$$|k(x):k(x+1/x)|=2,$$
$$|k(x):k(x^2+1/x^2)|=4,$$
$$|k(x+1/x):k(x^2+1/x^2)|=2.$$
Here $k$ is any field, and $x$ is transcendental over $k$.
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 31 at 6:58
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
add a comment |
add a comment |
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$begingroup$
? What actually are you asserting after "so"?
$endgroup$
– Lord Shark the Unknown
Mar 31 at 6:34
$begingroup$
I got confused. I got it. Thanks@LordSharktheUnknown
$endgroup$
– Shanghaikid
Mar 31 at 7:47