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Constructing extension field $K_n$ of $mathbbQ$ such that $[K_n : mathbbQ] = n$, for any $n geq 1$.
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The following problem is from Abstract Algebra, Herstein. Problem 10 in Chapter 5.6. I'm trying to self-study abstract algebra using this book and am unsure how to think about extension fields. I assume there are many answers to this problem. Any answers would give me an idea of what type of examples to think about when thinking about finite extensions.
Construct an extension field $K_n$ of $mathbbQ$ such that $[K_n : mathbbQ] = n$, for any $n geq 1$.
abstract-algebra extension-field
asked Jul 21 at 19:53
Kam
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up vote
1
down vote
favorite
The following problem is from Abstract Algebra, Herstein. Problem 10 in Chapter 5.6. I'm trying to self-study abstract algebra using this book and am unsure how to think about extension fields. I assume there are many answers to this problem. Any answers would give me an idea of what type of examples to think about when thinking about finite extensions.
Construct an extension field $K_n$ of $mathbbQ$ such that $[K_n : mathbbQ] = n$, for any $n geq 1$.
abstract-algebra extension-field
asked Jul 21 at 19:53
Kam
61
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The following problem is from Abstract Algebra, Herstein. Problem 10 in Chapter 5.6. I'm trying to self-study abstract algebra using this book and am unsure how to think about extension fields. I assume there are many answers to this problem. Any answers would give me an idea of what type of examples to think about when thinking about finite extensions.
Construct an extension field $K_n$ of $mathbbQ$ such that $[K_n : mathbbQ] = n$, for any $n geq 1$.
abstract-algebra extension-field
asked Jul 21 at 19:53
Kam
61
The following problem is from Abstract Algebra, Herstein. Problem 10 in Chapter 5.6. I'm trying to self-study abstract algebra using this book and am unsure how to think about extension fields. I assume there are many answers to this problem. Any answers would give me an idea of what type of examples to think about when thinking about finite extensions.
Construct an extension field $K_n$ of $mathbbQ$ such that $[K_n : mathbbQ] = n$, for any $n geq 1$.
abstract-algebra extension-field
asked Jul 21 at 19:53
Kam
61
asked Jul 21 at 19:53
Kam
61
asked Jul 21 at 19:53
Kam
61
asked Jul 21 at 19:53
Kam
61
61
add a comment |Â
add a comment |Â
1 Answer
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Pick an irreducible polynomial $f$ of degree $n$. For example $f(x)=x^n+p$, for some prime $p$ and extend $mathbbQ$ by one of the roots $r$ of $f$.
Then $K_n=mathbbQ(r)$ will have $1,r,r^2,...,r^n-1$ as a basis over $mathbbQ$.
In fact, they are linearly independent, because otherwise a linear dependence between them will result in a polynomial over $mathbbQ$ of which $r$ is a root, but of degree strictly smaller than $n$. Such a polynomial would have a common factor with $f$ over $mathbbQ$, contradicting that $f$ is irreducible.
They generate $K_n$ because any polynomial $g(r)$ can be expressed as $f(r)q(r)+s(r)$ with degree of $s<n$.
answered Jul 21 at 19:56
porca
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Pick an irreducible polynomial $f$ of degree $n$. For example $f(x)=x^n+p$, for some prime $p$ and extend $mathbbQ$ by one of the roots $r$ of $f$.
Then $K_n=mathbbQ(r)$ will have $1,r,r^2,...,r^n-1$ as a basis over $mathbbQ$.
In fact, they are linearly independent, because otherwise a linear dependence between them will result in a polynomial over $mathbbQ$ of which $r$ is a root, but of degree strictly smaller than $n$. Such a polynomial would have a common factor with $f$ over $mathbbQ$, contradicting that $f$ is irreducible.
They generate $K_n$ because any polynomial $g(r)$ can be expressed as $f(r)q(r)+s(r)$ with degree of $s<n$.
answered Jul 21 at 19:56
porca
211
add a comment |Â
up vote
2
down vote
Pick an irreducible polynomial $f$ of degree $n$. For example $f(x)=x^n+p$, for some prime $p$ and extend $mathbbQ$ by one of the roots $r$ of $f$.
Then $K_n=mathbbQ(r)$ will have $1,r,r^2,...,r^n-1$ as a basis over $mathbbQ$.
In fact, they are linearly independent, because otherwise a linear dependence between them will result in a polynomial over $mathbbQ$ of which $r$ is a root, but of degree strictly smaller than $n$. Such a polynomial would have a common factor with $f$ over $mathbbQ$, contradicting that $f$ is irreducible.
They generate $K_n$ because any polynomial $g(r)$ can be expressed as $f(r)q(r)+s(r)$ with degree of $s<n$.
answered Jul 21 at 19:56
porca
211
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Pick an irreducible polynomial $f$ of degree $n$. For example $f(x)=x^n+p$, for some prime $p$ and extend $mathbbQ$ by one of the roots $r$ of $f$.
Then $K_n=mathbbQ(r)$ will have $1,r,r^2,...,r^n-1$ as a basis over $mathbbQ$.
In fact, they are linearly independent, because otherwise a linear dependence between them will result in a polynomial over $mathbbQ$ of which $r$ is a root, but of degree strictly smaller than $n$. Such a polynomial would have a common factor with $f$ over $mathbbQ$, contradicting that $f$ is irreducible.
They generate $K_n$ because any polynomial $g(r)$ can be expressed as $f(r)q(r)+s(r)$ with degree of $s<n$.
answered Jul 21 at 19:56
porca
211
Pick an irreducible polynomial $f$ of degree $n$. For example $f(x)=x^n+p$, for some prime $p$ and extend $mathbbQ$ by one of the roots $r$ of $f$.
Then $K_n=mathbbQ(r)$ will have $1,r,r^2,...,r^n-1$ as a basis over $mathbbQ$.
In fact, they are linearly independent, because otherwise a linear dependence between them will result in a polynomial over $mathbbQ$ of which $r$ is a root, but of degree strictly smaller than $n$. Such a polynomial would have a common factor with $f$ over $mathbbQ$, contradicting that $f$ is irreducible.
They generate $K_n$ because any polynomial $g(r)$ can be expressed as $f(r)q(r)+s(r)$ with degree of $s<n$.
answered Jul 21 at 19:56
porca
211
answered Jul 21 at 19:56
porca
211
answered Jul 21 at 19:56
porca
211
answered Jul 21 at 19:56
porca
211
211
add a comment |Â
add a comment |Â
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