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Given an algebraic extension of $F$, $K$ how do we know we can “reach†$K$
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So I am getting confused with extensions and I would appreciate some clarification.
My confusion arised when solving the problem: "Suppose that $p(x)in F[x]$ and $E$ is a finite extension of $F$. If $p(x)$ is irreducible over $F$, and deg $p(x)$ and $[E:F]$ are relatively prime, show that $p(x)$ is irreducible over $E$."
So we know that $[E:F]=p$ and so $E$ is an algebraic extension. Say $a$ is a zero of $p(x)$ then $F[x]/<p(x)>simeq F(a)$. So now do we know there is always a field that contains $E$ and $F(a)$? Or that is not always the case? Here we can just take $E(a)$ which is obviously contains both $E$ and $F(a)$. Now here is where my problem comes in, how do we know $E(a)$ is an extension of $F(a)$? I can make the argument that $E(a)$ is algebraic over $F$ and so it must be algebraic over $F(a)$. But I fail to see how we can "reach" $E(a)$ by simply using simple extensions over and over, it is not intuitive to me, perhaps it should be? I thought on it and came up with the following proof, is it correct?
Let's say we have the field $F$, and the algebraic extension of it $K$. Now by definition, we know that any element of $K$ is algebraic over $F$. So if we take every element of $Fsetminus K $ say $a_1,a_2...$ then $F(a_1,a_2...)=K$.
extension-field
edited Jul 16 at 22:18
Ethan Bolker
35.7k54199
asked Jul 16 at 22:16
Sorfosh
910616
add a comment |Â
up vote
0
down vote
favorite
So I am getting confused with extensions and I would appreciate some clarification.
My confusion arised when solving the problem: "Suppose that $p(x)in F[x]$ and $E$ is a finite extension of $F$. If $p(x)$ is irreducible over $F$, and deg $p(x)$ and $[E:F]$ are relatively prime, show that $p(x)$ is irreducible over $E$."
So we know that $[E:F]=p$ and so $E$ is an algebraic extension. Say $a$ is a zero of $p(x)$ then $F[x]/<p(x)>simeq F(a)$. So now do we know there is always a field that contains $E$ and $F(a)$? Or that is not always the case? Here we can just take $E(a)$ which is obviously contains both $E$ and $F(a)$. Now here is where my problem comes in, how do we know $E(a)$ is an extension of $F(a)$? I can make the argument that $E(a)$ is algebraic over $F$ and so it must be algebraic over $F(a)$. But I fail to see how we can "reach" $E(a)$ by simply using simple extensions over and over, it is not intuitive to me, perhaps it should be? I thought on it and came up with the following proof, is it correct?
Let's say we have the field $F$, and the algebraic extension of it $K$. Now by definition, we know that any element of $K$ is algebraic over $F$. So if we take every element of $Fsetminus K $ say $a_1,a_2...$ then $F(a_1,a_2...)=K$.
extension-field
edited Jul 16 at 22:18
Ethan Bolker
35.7k54199
asked Jul 16 at 22:16
Sorfosh
910616
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I am getting confused with extensions and I would appreciate some clarification.
My confusion arised when solving the problem: "Suppose that $p(x)in F[x]$ and $E$ is a finite extension of $F$. If $p(x)$ is irreducible over $F$, and deg $p(x)$ and $[E:F]$ are relatively prime, show that $p(x)$ is irreducible over $E$."
So we know that $[E:F]=p$ and so $E$ is an algebraic extension. Say $a$ is a zero of $p(x)$ then $F[x]/<p(x)>simeq F(a)$. So now do we know there is always a field that contains $E$ and $F(a)$? Or that is not always the case? Here we can just take $E(a)$ which is obviously contains both $E$ and $F(a)$. Now here is where my problem comes in, how do we know $E(a)$ is an extension of $F(a)$? I can make the argument that $E(a)$ is algebraic over $F$ and so it must be algebraic over $F(a)$. But I fail to see how we can "reach" $E(a)$ by simply using simple extensions over and over, it is not intuitive to me, perhaps it should be? I thought on it and came up with the following proof, is it correct?
Let's say we have the field $F$, and the algebraic extension of it $K$. Now by definition, we know that any element of $K$ is algebraic over $F$. So if we take every element of $Fsetminus K $ say $a_1,a_2...$ then $F(a_1,a_2...)=K$.
extension-field
edited Jul 16 at 22:18
Ethan Bolker
35.7k54199
asked Jul 16 at 22:16
Sorfosh
910616
So I am getting confused with extensions and I would appreciate some clarification.
My confusion arised when solving the problem: "Suppose that $p(x)in F[x]$ and $E$ is a finite extension of $F$. If $p(x)$ is irreducible over $F$, and deg $p(x)$ and $[E:F]$ are relatively prime, show that $p(x)$ is irreducible over $E$."
So we know that $[E:F]=p$ and so $E$ is an algebraic extension. Say $a$ is a zero of $p(x)$ then $F[x]/<p(x)>simeq F(a)$. So now do we know there is always a field that contains $E$ and $F(a)$? Or that is not always the case? Here we can just take $E(a)$ which is obviously contains both $E$ and $F(a)$. Now here is where my problem comes in, how do we know $E(a)$ is an extension of $F(a)$? I can make the argument that $E(a)$ is algebraic over $F$ and so it must be algebraic over $F(a)$. But I fail to see how we can "reach" $E(a)$ by simply using simple extensions over and over, it is not intuitive to me, perhaps it should be? I thought on it and came up with the following proof, is it correct?
Let's say we have the field $F$, and the algebraic extension of it $K$. Now by definition, we know that any element of $K$ is algebraic over $F$. So if we take every element of $Fsetminus K $ say $a_1,a_2...$ then $F(a_1,a_2...)=K$.
extension-field
edited Jul 16 at 22:18
Ethan Bolker
35.7k54199
asked Jul 16 at 22:16
Sorfosh
910616
edited Jul 16 at 22:18
Ethan Bolker
35.7k54199
edited Jul 16 at 22:18
Ethan Bolker
35.7k54199
edited Jul 16 at 22:18
Ethan Bolker
35.7k54199
35.7k54199
asked Jul 16 at 22:16
Sorfosh
910616
asked Jul 16 at 22:16
Sorfosh
910616
asked Jul 16 at 22:16
Sorfosh
910616
910616
add a comment |Â
add a comment |Â
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