Mixing
Simple field extension where primitive element is algebraic
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I know that when $E$ is a field extension of field $F,$ $t in E$ $F,$ $t$ algebraic over $F,$ than the smallest field, containing $F cup t$ is the same as set of polynomials, with coefficients in $F$ and a variable $t.$
Do you know how to prove it? Or maybe you've seen a proof of that statement?
field-theory extension-field
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
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add a comment |
$begingroup$
I know that when $E$ is a field extension of field $F,$ $t in E$ $F,$ $t$ algebraic over $F,$ than the smallest field, containing $F cup t$ is the same as set of polynomials, with coefficients in $F$ and a variable $t.$
Do you know how to prove it? Or maybe you've seen a proof of that statement?
field-theory extension-field
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
$endgroup$
$begingroup$
That's what I wrote, if t is algebraic over F, then F[t] is field. In case it is transcendent, it isn't, I know that.
$endgroup$
– benjamin1996
Jan 18 at 11:56
add a comment |
$begingroup$
I know that when $E$ is a field extension of field $F,$ $t in E$ $F,$ $t$ algebraic over $F,$ than the smallest field, containing $F cup t$ is the same as set of polynomials, with coefficients in $F$ and a variable $t.$
Do you know how to prove it? Or maybe you've seen a proof of that statement?
field-theory extension-field
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
$endgroup$
I know that when $E$ is a field extension of field $F,$ $t in E$ $F,$ $t$ algebraic over $F,$ than the smallest field, containing $F cup t$ is the same as set of polynomials, with coefficients in $F$ and a variable $t.$
Do you know how to prove it? Or maybe you've seen a proof of that statement?
field-theory extension-field
field-theory extension-field
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
asked Jan 18 at 11:12
benjamin1996benjamin1996
62
62
$begingroup$
That's what I wrote, if t is algebraic over F, then F[t] is field. In case it is transcendent, it isn't, I know that.
$endgroup$
– benjamin1996
Jan 18 at 11:56
add a comment |
$begingroup$
That's what I wrote, if t is algebraic over F, then F[t] is field. In case it is transcendent, it isn't, I know that.
$endgroup$
– benjamin1996
Jan 18 at 11:56
$begingroup$
That's what I wrote, if t is algebraic over F, then F[t] is field. In case it is transcendent, it isn't, I know that.
$endgroup$
– benjamin1996
Jan 18 at 11:56
$begingroup$
That's what I wrote, if t is algebraic over F, then F[t] is field. In case it is transcendent, it isn't, I know that.
$endgroup$
– benjamin1996
Jan 18 at 11:56
add a comment |
1 Answer
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Hint: If $E$ is an extension of $F$ and $tin E$, then the ring extension $F[t]$ is also a field extension.
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
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Hint: If $E$ is an extension of $F$ and $tin E$, then the ring extension $F[t]$ is also a field extension.
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
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Hint: If $E$ is an extension of $F$ and $tin E$, then the ring extension $F[t]$ is also a field extension.
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
$endgroup$
add a comment |
$begingroup$
Hint: If $E$ is an extension of $F$ and $tin E$, then the ring extension $F[t]$ is also a field extension.
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
$endgroup$
Hint: If $E$ is an extension of $F$ and $tin E$, then the ring extension $F[t]$ is also a field extension.
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
answered Jan 18 at 11:43
WuestenfuxWuestenfux
4,7291513
4,7291513
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$begingroup$
That's what I wrote, if t is algebraic over F, then F[t] is field. In case it is transcendent, it isn't, I know that.
$endgroup$
– benjamin1996
Jan 18 at 11:56