Form of an element in an algebraic field extension

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Let $e_1,...,e_n$ be algebraic elements over a prime field $F$. Since $F$ is a prime field, we know that the automorphism group of $F$ is trivial. Let $K = F(e_1,...,e_n)$ and consider the extension $F subset K$. Is it true that every $F$-automorphism of $K$ is determined by the images of the $e_i$? I can also reformulate the question in the following sense: Can every element of an $F$-basis of $K$ be expressed in terms of the $e_i$?



I personally believe it is true due to all the examples i tried out, but cannot seem to find a good formal argument.






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  • Welcome to Maths SX! It is obviously true, but these images are not arbitrary since the elements are not algebraically independent.
    – Bernard
    Sep 3 at 11:40














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Let $e_1,...,e_n$ be algebraic elements over a prime field $F$. Since $F$ is a prime field, we know that the automorphism group of $F$ is trivial. Let $K = F(e_1,...,e_n)$ and consider the extension $F subset K$. Is it true that every $F$-automorphism of $K$ is determined by the images of the $e_i$? I can also reformulate the question in the following sense: Can every element of an $F$-basis of $K$ be expressed in terms of the $e_i$?



I personally believe it is true due to all the examples i tried out, but cannot seem to find a good formal argument.






abstract-algebra extension-field







share|cite|improve this question





















  • Welcome to Maths SX! It is obviously true, but these images are not arbitrary since the elements are not algebraically independent.
    – Bernard
    Sep 3 at 11:40












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $e_1,...,e_n$ be algebraic elements over a prime field $F$. Since $F$ is a prime field, we know that the automorphism group of $F$ is trivial. Let $K = F(e_1,...,e_n)$ and consider the extension $F subset K$. Is it true that every $F$-automorphism of $K$ is determined by the images of the $e_i$? I can also reformulate the question in the following sense: Can every element of an $F$-basis of $K$ be expressed in terms of the $e_i$?



I personally believe it is true due to all the examples i tried out, but cannot seem to find a good formal argument.






abstract-algebra extension-field







share|cite|improve this question













Let $e_1,...,e_n$ be algebraic elements over a prime field $F$. Since $F$ is a prime field, we know that the automorphism group of $F$ is trivial. Let $K = F(e_1,...,e_n)$ and consider the extension $F subset K$. Is it true that every $F$-automorphism of $K$ is determined by the images of the $e_i$? I can also reformulate the question in the following sense: Can every element of an $F$-basis of $K$ be expressed in terms of the $e_i$?



I personally believe it is true due to all the examples i tried out, but cannot seem to find a good formal argument.






abstract-algebra extension-field




abstract-algebra extension-field






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asked Sep 3 at 11:36









Peter Blane

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  • Welcome to Maths SX! It is obviously true, but these images are not arbitrary since the elements are not algebraically independent.
    – Bernard
    Sep 3 at 11:40
















  • Welcome to Maths SX! It is obviously true, but these images are not arbitrary since the elements are not algebraically independent.
    – Bernard
    Sep 3 at 11:40















Welcome to Maths SX! It is obviously true, but these images are not arbitrary since the elements are not algebraically independent.
– Bernard
Sep 3 at 11:40




Welcome to Maths SX! It is obviously true, but these images are not arbitrary since the elements are not algebraically independent.
– Bernard
Sep 3 at 11:40










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Using the fact that the $e_i$ are algebraic over $F$, it is easy to show that $K=F[e_1,dots,e_n]$, i.e. $K$ coincides with the smallest sub-$F$-algebra of $K$ containing the $e_i$. Now this shows that we have $Kcong F[X_1,dots,X_n]/mathfrak m$ for some maximal ideal $mathfrak m$. In particular we see that any automorphism of $K$ is determined by the image of the $e_i$ since any automorphism of the above quotient comes from a map $F[X_1,dots,X_n]to K$ subject to the condition that $mathfrak m$ is the kernel of that map.






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    Using the fact that the $e_i$ are algebraic over $F$, it is easy to show that $K=F[e_1,dots,e_n]$, i.e. $K$ coincides with the smallest sub-$F$-algebra of $K$ containing the $e_i$. Now this shows that we have $Kcong F[X_1,dots,X_n]/mathfrak m$ for some maximal ideal $mathfrak m$. In particular we see that any automorphism of $K$ is determined by the image of the $e_i$ since any automorphism of the above quotient comes from a map $F[X_1,dots,X_n]to K$ subject to the condition that $mathfrak m$ is the kernel of that map.






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      Using the fact that the $e_i$ are algebraic over $F$, it is easy to show that $K=F[e_1,dots,e_n]$, i.e. $K$ coincides with the smallest sub-$F$-algebra of $K$ containing the $e_i$. Now this shows that we have $Kcong F[X_1,dots,X_n]/mathfrak m$ for some maximal ideal $mathfrak m$. In particular we see that any automorphism of $K$ is determined by the image of the $e_i$ since any automorphism of the above quotient comes from a map $F[X_1,dots,X_n]to K$ subject to the condition that $mathfrak m$ is the kernel of that map.






      share|cite|improve this answer






















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        Using the fact that the $e_i$ are algebraic over $F$, it is easy to show that $K=F[e_1,dots,e_n]$, i.e. $K$ coincides with the smallest sub-$F$-algebra of $K$ containing the $e_i$. Now this shows that we have $Kcong F[X_1,dots,X_n]/mathfrak m$ for some maximal ideal $mathfrak m$. In particular we see that any automorphism of $K$ is determined by the image of the $e_i$ since any automorphism of the above quotient comes from a map $F[X_1,dots,X_n]to K$ subject to the condition that $mathfrak m$ is the kernel of that map.






        share|cite|improve this answer












        Using the fact that the $e_i$ are algebraic over $F$, it is easy to show that $K=F[e_1,dots,e_n]$, i.e. $K$ coincides with the smallest sub-$F$-algebra of $K$ containing the $e_i$. Now this shows that we have $Kcong F[X_1,dots,X_n]/mathfrak m$ for some maximal ideal $mathfrak m$. In particular we see that any automorphism of $K$ is determined by the image of the $e_i$ since any automorphism of the above quotient comes from a map $F[X_1,dots,X_n]to K$ subject to the condition that $mathfrak m$ is the kernel of that map.







        share|cite|improve this answer












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        answered Sep 3 at 12:03









        asdq

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