If $x$ is algebraic over a quotient field $K$ of $A$, then there exists an integral element $cx$ for some $A...
$begingroup$
Let $A$ be a commutative ring, $K$ its quotient field and $x$ algebraic over $K$. This means that there exists a polynomial $f(X)$ with coefficients in $K$ such that $f(x) = 0$. In other words, if we write the coefficients of $f$ as $dfrac{a_i}{b_i}, i = 0,.., n$, then we can multiply both sides of $f(x) = 0$ by $b = prod_{i=0}^n b_i$ to achieve $f'(x) = 0$ where the coefficients of $f'$ can be viewed as elements of $A$. How do we get this to be $f(cx)$ though?
abstract-algebra algebraic-number-theory fractions integral-extensions algebraic-equations
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
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add a comment |
$begingroup$
Let $A$ be a commutative ring, $K$ its quotient field and $x$ algebraic over $K$. This means that there exists a polynomial $f(X)$ with coefficients in $K$ such that $f(x) = 0$. In other words, if we write the coefficients of $f$ as $dfrac{a_i}{b_i}, i = 0,.., n$, then we can multiply both sides of $f(x) = 0$ by $b = prod_{i=0}^n b_i$ to achieve $f'(x) = 0$ where the coefficients of $f'$ can be viewed as elements of $A$. How do we get this to be $f(cx)$ though?
abstract-algebra algebraic-number-theory fractions integral-extensions algebraic-equations
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
$endgroup$
2
$begingroup$
$b f(X) in A[X]$, its leading coefficient is $c$, then $c^{n-1} b f( X/c) $ is monic. What do you expect with $f'$ ?
$endgroup$
– reuns
Jan 9 at 23:52
1
$begingroup$
Just write $x$ in the form $x=p/q$ with $p$ and $q$ coprime and you'll see what polynomial that vanishes at $qx$.
$endgroup$
– mouthetics
Jan 10 at 0:07
1
$begingroup$
Thanks @reuns you're right. So that makes sense. $(b f)_n = a_n prod_{i neq n} b_i = c$ is the leading coefficient but doesn't that mean $(X/c)^n = X^n / c^n$ so we would indeed need $c^{n-1}$ to make it monic. Thanks !
$endgroup$
– Roll up and smoke Adjoint
Jan 10 at 0:08
$begingroup$
@mouthetics: but $x$ is algebraic over $K$, so how do we know $x = p/q$?
$endgroup$
– Robert Lewis
Jan 10 at 0:10
1
$begingroup$
@RobertLewis Sorry! I thought $x$ is in the quotient field of $A$.
$endgroup$
– mouthetics
Jan 10 at 0:15
add a comment |
$begingroup$
Let $A$ be a commutative ring, $K$ its quotient field and $x$ algebraic over $K$. This means that there exists a polynomial $f(X)$ with coefficients in $K$ such that $f(x) = 0$. In other words, if we write the coefficients of $f$ as $dfrac{a_i}{b_i}, i = 0,.., n$, then we can multiply both sides of $f(x) = 0$ by $b = prod_{i=0}^n b_i$ to achieve $f'(x) = 0$ where the coefficients of $f'$ can be viewed as elements of $A$. How do we get this to be $f(cx)$ though?
abstract-algebra algebraic-number-theory fractions integral-extensions algebraic-equations
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
$endgroup$
Let $A$ be a commutative ring, $K$ its quotient field and $x$ algebraic over $K$. This means that there exists a polynomial $f(X)$ with coefficients in $K$ such that $f(x) = 0$. In other words, if we write the coefficients of $f$ as $dfrac{a_i}{b_i}, i = 0,.., n$, then we can multiply both sides of $f(x) = 0$ by $b = prod_{i=0}^n b_i$ to achieve $f'(x) = 0$ where the coefficients of $f'$ can be viewed as elements of $A$. How do we get this to be $f(cx)$ though?
abstract-algebra algebraic-number-theory fractions integral-extensions algebraic-equations
abstract-algebra algebraic-number-theory fractions integral-extensions algebraic-equations
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
asked Jan 9 at 23:47
Roll up and smoke AdjointRoll up and smoke Adjoint
9,11052458
9,11052458
2
$begingroup$
$b f(X) in A[X]$, its leading coefficient is $c$, then $c^{n-1} b f( X/c) $ is monic. What do you expect with $f'$ ?
$endgroup$
– reuns
Jan 9 at 23:52
1
$begingroup$
Just write $x$ in the form $x=p/q$ with $p$ and $q$ coprime and you'll see what polynomial that vanishes at $qx$.
$endgroup$
– mouthetics
Jan 10 at 0:07
1
$begingroup$
Thanks @reuns you're right. So that makes sense. $(b f)_n = a_n prod_{i neq n} b_i = c$ is the leading coefficient but doesn't that mean $(X/c)^n = X^n / c^n$ so we would indeed need $c^{n-1}$ to make it monic. Thanks !
$endgroup$
– Roll up and smoke Adjoint
Jan 10 at 0:08
$begingroup$
@mouthetics: but $x$ is algebraic over $K$, so how do we know $x = p/q$?
$endgroup$
– Robert Lewis
Jan 10 at 0:10
1
$begingroup$
@RobertLewis Sorry! I thought $x$ is in the quotient field of $A$.
$endgroup$
– mouthetics
Jan 10 at 0:15
add a comment |
2
$begingroup$
$b f(X) in A[X]$, its leading coefficient is $c$, then $c^{n-1} b f( X/c) $ is monic. What do you expect with $f'$ ?
$endgroup$
– reuns
Jan 9 at 23:52
1
$begingroup$
Just write $x$ in the form $x=p/q$ with $p$ and $q$ coprime and you'll see what polynomial that vanishes at $qx$.
$endgroup$
– mouthetics
Jan 10 at 0:07
1
$begingroup$
Thanks @reuns you're right. So that makes sense. $(b f)_n = a_n prod_{i neq n} b_i = c$ is the leading coefficient but doesn't that mean $(X/c)^n = X^n / c^n$ so we would indeed need $c^{n-1}$ to make it monic. Thanks !
$endgroup$
– Roll up and smoke Adjoint
Jan 10 at 0:08
$begingroup$
@mouthetics: but $x$ is algebraic over $K$, so how do we know $x = p/q$?
$endgroup$
– Robert Lewis
Jan 10 at 0:10
1
$begingroup$
@RobertLewis Sorry! I thought $x$ is in the quotient field of $A$.
$endgroup$
– mouthetics
Jan 10 at 0:15
2
2
$begingroup$
$b f(X) in A[X]$, its leading coefficient is $c$, then $c^{n-1} b f( X/c) $ is monic. What do you expect with $f'$ ?
$endgroup$
– reuns
Jan 9 at 23:52
$begingroup$
$b f(X) in A[X]$, its leading coefficient is $c$, then $c^{n-1} b f( X/c) $ is monic. What do you expect with $f'$ ?
$endgroup$
– reuns
Jan 9 at 23:52
1
1
$begingroup$
Just write $x$ in the form $x=p/q$ with $p$ and $q$ coprime and you'll see what polynomial that vanishes at $qx$.
$endgroup$
– mouthetics
Jan 10 at 0:07
$begingroup$
Just write $x$ in the form $x=p/q$ with $p$ and $q$ coprime and you'll see what polynomial that vanishes at $qx$.
$endgroup$
– mouthetics
Jan 10 at 0:07
1
1
$begingroup$
Thanks @reuns you're right. So that makes sense. $(b f)_n = a_n prod_{i neq n} b_i = c$ is the leading coefficient but doesn't that mean $(X/c)^n = X^n / c^n$ so we would indeed need $c^{n-1}$ to make it monic. Thanks !
$endgroup$
– Roll up and smoke Adjoint
Jan 10 at 0:08
$begingroup$
Thanks @reuns you're right. So that makes sense. $(b f)_n = a_n prod_{i neq n} b_i = c$ is the leading coefficient but doesn't that mean $(X/c)^n = X^n / c^n$ so we would indeed need $c^{n-1}$ to make it monic. Thanks !
$endgroup$
– Roll up and smoke Adjoint
Jan 10 at 0:08
$begingroup$
@mouthetics: but $x$ is algebraic over $K$, so how do we know $x = p/q$?
$endgroup$
– Robert Lewis
Jan 10 at 0:10
$begingroup$
@mouthetics: but $x$ is algebraic over $K$, so how do we know $x = p/q$?
$endgroup$
– Robert Lewis
Jan 10 at 0:10
1
1
$begingroup$
@RobertLewis Sorry! I thought $x$ is in the quotient field of $A$.
$endgroup$
– mouthetics
Jan 10 at 0:15
$begingroup$
@RobertLewis Sorry! I thought $x$ is in the quotient field of $A$.
$endgroup$
– mouthetics
Jan 10 at 0:15
add a comment |
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$begingroup$
$b f(X) in A[X]$, its leading coefficient is $c$, then $c^{n-1} b f( X/c) $ is monic. What do you expect with $f'$ ?
$endgroup$
– reuns
Jan 9 at 23:52
1
$begingroup$
Just write $x$ in the form $x=p/q$ with $p$ and $q$ coprime and you'll see what polynomial that vanishes at $qx$.
$endgroup$
– mouthetics
Jan 10 at 0:07
1
$begingroup$
Thanks @reuns you're right. So that makes sense. $(b f)_n = a_n prod_{i neq n} b_i = c$ is the leading coefficient but doesn't that mean $(X/c)^n = X^n / c^n$ so we would indeed need $c^{n-1}$ to make it monic. Thanks !
$endgroup$
– Roll up and smoke Adjoint
Jan 10 at 0:08
$begingroup$
@mouthetics: but $x$ is algebraic over $K$, so how do we know $x = p/q$?
$endgroup$
– Robert Lewis
Jan 10 at 0:10
1
$begingroup$
@RobertLewis Sorry! I thought $x$ is in the quotient field of $A$.
$endgroup$
– mouthetics
Jan 10 at 0:15