Mixing
Doubt about a paragraph in the book “Algebraic Number Theory by Neukirch”.
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Please refer to the question in the given link.
The question already has an answer here
An element is integral iff its minimal polynomial has integral coefficients.
My question is ---
Here, $A$ is integrally closed in $K$. And $beta_i$'s $in bar{K}$. Don't we have to show that $beta_i$'s $in K$ to show they belong to $A$? Because $A$ is integrally closed in $K$ not in $bar{K}$.
Thank you.
algebraic-number-theory
asked Jan 29 at 16:01
SaikatSaikat
366216
$endgroup$
add a comment |
$begingroup$
Please refer to the question in the given link.
The question already has an answer here
An element is integral iff its minimal polynomial has integral coefficients.
My question is ---
Here, $A$ is integrally closed in $K$. And $beta_i$'s $in bar{K}$. Don't we have to show that $beta_i$'s $in K$ to show they belong to $A$? Because $A$ is integrally closed in $K$ not in $bar{K}$.
Thank you.
algebraic-number-theory
asked Jan 29 at 16:01
SaikatSaikat
366216
$endgroup$
add a comment |
$begingroup$
Please refer to the question in the given link.
The question already has an answer here
An element is integral iff its minimal polynomial has integral coefficients.
My question is ---
Here, $A$ is integrally closed in $K$. And $beta_i$'s $in bar{K}$. Don't we have to show that $beta_i$'s $in K$ to show they belong to $A$? Because $A$ is integrally closed in $K$ not in $bar{K}$.
Thank you.
algebraic-number-theory
asked Jan 29 at 16:01
SaikatSaikat
366216
$endgroup$
Please refer to the question in the given link.
The question already has an answer here
An element is integral iff its minimal polynomial has integral coefficients.
My question is ---
Here, $A$ is integrally closed in $K$. And $beta_i$'s $in bar{K}$. Don't we have to show that $beta_i$'s $in K$ to show they belong to $A$? Because $A$ is integrally closed in $K$ not in $bar{K}$.
Thank you.
algebraic-number-theory
algebraic-number-theory
asked Jan 29 at 16:01
SaikatSaikat
366216
asked Jan 29 at 16:01
SaikatSaikat
366216
edited Jan 29 at 16:44
Saikat
asked Jan 29 at 16:01
SaikatSaikat
366216
asked Jan 29 at 16:01
SaikatSaikat
366216
asked Jan 29 at 16:01
SaikatSaikat
366216
366216
add a comment |
add a comment |
1 Answer
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$begingroup$
Here, $beta_i$'s belong to $bar{K}$ which are integral over $A$. So, coefficients of $p(x)$ are integral over $A$. But the coefficients of $p(x)$ belong to $K$. So, coefficients of $p(x)$ belong to $A$ as $A$ is integrally closed in $K$. Hence, $p(x)in A[x]$.
answered Jan 29 at 17:42
SaikatSaikat
366216
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here, $beta_i$'s belong to $bar{K}$ which are integral over $A$. So, coefficients of $p(x)$ are integral over $A$. But the coefficients of $p(x)$ belong to $K$. So, coefficients of $p(x)$ belong to $A$ as $A$ is integrally closed in $K$. Hence, $p(x)in A[x]$.
answered Jan 29 at 17:42
SaikatSaikat
366216
$endgroup$
add a comment |
$begingroup$
Here, $beta_i$'s belong to $bar{K}$ which are integral over $A$. So, coefficients of $p(x)$ are integral over $A$. But the coefficients of $p(x)$ belong to $K$. So, coefficients of $p(x)$ belong to $A$ as $A$ is integrally closed in $K$. Hence, $p(x)in A[x]$.
answered Jan 29 at 17:42
SaikatSaikat
366216
$endgroup$
add a comment |
$begingroup$
Here, $beta_i$'s belong to $bar{K}$ which are integral over $A$. So, coefficients of $p(x)$ are integral over $A$. But the coefficients of $p(x)$ belong to $K$. So, coefficients of $p(x)$ belong to $A$ as $A$ is integrally closed in $K$. Hence, $p(x)in A[x]$.
answered Jan 29 at 17:42
SaikatSaikat
366216
$endgroup$
Here, $beta_i$'s belong to $bar{K}$ which are integral over $A$. So, coefficients of $p(x)$ are integral over $A$. But the coefficients of $p(x)$ belong to $K$. So, coefficients of $p(x)$ belong to $A$ as $A$ is integrally closed in $K$. Hence, $p(x)in A[x]$.
answered Jan 29 at 17:42
SaikatSaikat
366216
answered Jan 29 at 17:42
SaikatSaikat
366216
answered Jan 29 at 17:42
SaikatSaikat
366216
answered Jan 29 at 17:42
SaikatSaikat
366216
366216
add a comment |
add a comment |
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