A field extension of prime degree
$begingroup$
Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$
Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.
Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$
$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$
Have I inferred it correctly? How do I move ahead?
Thank you for your help.
abstract-algebra ring-theory field-theory extension-field
edited Aug 14 '14 at 15:41
user26857
39.5k124284
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
$endgroup$
|
show 1 more comment
$begingroup$
Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$
Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.
Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$
$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$
Have I inferred it correctly? How do I move ahead?
Thank you for your help.
abstract-algebra ring-theory field-theory extension-field
edited Aug 14 '14 at 15:41
user26857
39.5k124284
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
$endgroup$
2
$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44
$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50
1
$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52
$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45
|
show 1 more comment
$begingroup$
Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$
Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.
Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$
$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$
Have I inferred it correctly? How do I move ahead?
Thank you for your help.
abstract-algebra ring-theory field-theory extension-field
edited Aug 14 '14 at 15:41
user26857
39.5k124284
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
$endgroup$
Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$
Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.
Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$
$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$
Have I inferred it correctly? How do I move ahead?
Thank you for your help.
abstract-algebra ring-theory field-theory extension-field
abstract-algebra ring-theory field-theory extension-field
edited Aug 14 '14 at 15:41
user26857
39.5k124284
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
edited Aug 14 '14 at 15:41
user26857
39.5k124284
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
edited Aug 14 '14 at 15:41
user26857
39.5k124284
edited Aug 14 '14 at 15:41
user26857
39.5k124284
edited Aug 14 '14 at 15:41
user26857
39.5k124284
39.5k124284
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
asked Aug 14 '14 at 13:39
MathManMathMan
3,64642072
3,64642072
2
$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44
$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50
1
$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52
$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45
|
show 1 more comment
2
$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44
$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50
1
$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52
$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45
2
2
$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44
$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44
$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50
$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50
1
1
$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52
$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52
$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45
$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.
Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.
The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.
You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
1
$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.
Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.
The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.
You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
1
$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56
add a comment |
$begingroup$
Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.
Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.
The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.
You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
1
$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56
add a comment |
$begingroup$
Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.
Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.
The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.
You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
$endgroup$
Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.
Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.
The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.
You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
answered Aug 14 '14 at 13:47
rschwiebrschwieb
108k12104253
108k12104253
$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
1
$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56
add a comment |
$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
1
$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56
$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
1
1
$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56
$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56
add a comment |
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2
$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44
$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50
1
$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52
$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55
$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45