Mixing
When is $F[X]/(f)$ over $F$ an integral extension?
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I was looking at the concept of integral extension. So $K/F$ is integral means that for every $a in K$ is the root of a monic polynomial with coefficient in $F$. For example $mathbb{Q}/mathbb{Z}$ is not integral since $1/3$ is not the root of a monic polynomial with integral coefficients. I was wondering when the extension $F[X]/(f)$ over $F$ was an integral extension. I was suggested it is integral if and only if $f$ is a monic polynomial? Why? If I take for instance $[p]_f in F[X]/(f)$ it is not a root of $f$ I think.
commutative-algebra integral-extensions
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
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show 4 more comments
$begingroup$
I was looking at the concept of integral extension. So $K/F$ is integral means that for every $a in K$ is the root of a monic polynomial with coefficient in $F$. For example $mathbb{Q}/mathbb{Z}$ is not integral since $1/3$ is not the root of a monic polynomial with integral coefficients. I was wondering when the extension $F[X]/(f)$ over $F$ was an integral extension. I was suggested it is integral if and only if $f$ is a monic polynomial? Why? If I take for instance $[p]_f in F[X]/(f)$ it is not a root of $f$ I think.
commutative-algebra integral-extensions
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
$endgroup$
$begingroup$
Or if its leading coefficient is a unit. That being said, to answer your question, $xbmod f$ is a root of $f$.
$endgroup$
– Bernard
Jan 27 at 13:44
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@Bernard I also thought I should use $[x]_f$ is a root of $f$ but if I consider $mathbb{Q}[x]/(x^2-2)$ for instance, and $[p]_f=[x+3]_f$, then $f([p]_f)=[x]_f^2+6[x]_f+[7]_f neq 0$ no?
$endgroup$
– roi_saumon
Jan 27 at 13:54
$begingroup$
Why should $[x+3]$ be a root of $x^2-2$? Also, you can simplify further: $[x^2]_f+6[x]_f+[7]_f=6[x]_f+[9]_f$.
$endgroup$
– Bernard
Jan 27 at 14:32
$begingroup$
Oh, but so I didn't understand how "$[x]_f$ if a root of $f$" should help me.
$endgroup$
– roi_saumon
Jan 27 at 14:41
1
$begingroup$
You can simply say that the set of integral elements is a subring, hence any polynomial in the integral element $[x]$ is itself an integral element.
$endgroup$
– Bernard
Jan 27 at 15:15
|
show 4 more comments
$begingroup$
I was looking at the concept of integral extension. So $K/F$ is integral means that for every $a in K$ is the root of a monic polynomial with coefficient in $F$. For example $mathbb{Q}/mathbb{Z}$ is not integral since $1/3$ is not the root of a monic polynomial with integral coefficients. I was wondering when the extension $F[X]/(f)$ over $F$ was an integral extension. I was suggested it is integral if and only if $f$ is a monic polynomial? Why? If I take for instance $[p]_f in F[X]/(f)$ it is not a root of $f$ I think.
commutative-algebra integral-extensions
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
$endgroup$
I was looking at the concept of integral extension. So $K/F$ is integral means that for every $a in K$ is the root of a monic polynomial with coefficient in $F$. For example $mathbb{Q}/mathbb{Z}$ is not integral since $1/3$ is not the root of a monic polynomial with integral coefficients. I was wondering when the extension $F[X]/(f)$ over $F$ was an integral extension. I was suggested it is integral if and only if $f$ is a monic polynomial? Why? If I take for instance $[p]_f in F[X]/(f)$ it is not a root of $f$ I think.
commutative-algebra integral-extensions
commutative-algebra integral-extensions
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
asked Jan 27 at 13:37
roi_saumonroi_saumon
63438
63438
$begingroup$
Or if its leading coefficient is a unit. That being said, to answer your question, $xbmod f$ is a root of $f$.
$endgroup$
– Bernard
Jan 27 at 13:44
$begingroup$
@Bernard I also thought I should use $[x]_f$ is a root of $f$ but if I consider $mathbb{Q}[x]/(x^2-2)$ for instance, and $[p]_f=[x+3]_f$, then $f([p]_f)=[x]_f^2+6[x]_f+[7]_f neq 0$ no?
$endgroup$
– roi_saumon
Jan 27 at 13:54
$begingroup$
Why should $[x+3]$ be a root of $x^2-2$? Also, you can simplify further: $[x^2]_f+6[x]_f+[7]_f=6[x]_f+[9]_f$.
$endgroup$
– Bernard
Jan 27 at 14:32
$begingroup$
Oh, but so I didn't understand how "$[x]_f$ if a root of $f$" should help me.
$endgroup$
– roi_saumon
Jan 27 at 14:41
1
$begingroup$
You can simply say that the set of integral elements is a subring, hence any polynomial in the integral element $[x]$ is itself an integral element.
$endgroup$
– Bernard
Jan 27 at 15:15
|
show 4 more comments
$begingroup$
Or if its leading coefficient is a unit. That being said, to answer your question, $xbmod f$ is a root of $f$.
$endgroup$
– Bernard
Jan 27 at 13:44
$begingroup$
@Bernard I also thought I should use $[x]_f$ is a root of $f$ but if I consider $mathbb{Q}[x]/(x^2-2)$ for instance, and $[p]_f=[x+3]_f$, then $f([p]_f)=[x]_f^2+6[x]_f+[7]_f neq 0$ no?
$endgroup$
– roi_saumon
Jan 27 at 13:54
$begingroup$
Why should $[x+3]$ be a root of $x^2-2$? Also, you can simplify further: $[x^2]_f+6[x]_f+[7]_f=6[x]_f+[9]_f$.
$endgroup$
– Bernard
Jan 27 at 14:32
$begingroup$
Oh, but so I didn't understand how "$[x]_f$ if a root of $f$" should help me.
$endgroup$
– roi_saumon
Jan 27 at 14:41
1
$begingroup$
You can simply say that the set of integral elements is a subring, hence any polynomial in the integral element $[x]$ is itself an integral element.
$endgroup$
– Bernard
Jan 27 at 15:15
$begingroup$
Or if its leading coefficient is a unit. That being said, to answer your question, $xbmod f$ is a root of $f$.
$endgroup$
– Bernard
Jan 27 at 13:44
$begingroup$
Or if its leading coefficient is a unit. That being said, to answer your question, $xbmod f$ is a root of $f$.
$endgroup$
– Bernard
Jan 27 at 13:44
$begingroup$
@Bernard I also thought I should use $[x]_f$ is a root of $f$ but if I consider $mathbb{Q}[x]/(x^2-2)$ for instance, and $[p]_f=[x+3]_f$, then $f([p]_f)=[x]_f^2+6[x]_f+[7]_f neq 0$ no?
$endgroup$
– roi_saumon
Jan 27 at 13:54
$begingroup$
@Bernard I also thought I should use $[x]_f$ is a root of $f$ but if I consider $mathbb{Q}[x]/(x^2-2)$ for instance, and $[p]_f=[x+3]_f$, then $f([p]_f)=[x]_f^2+6[x]_f+[7]_f neq 0$ no?
$endgroup$
– roi_saumon
Jan 27 at 13:54
$begingroup$
Why should $[x+3]$ be a root of $x^2-2$? Also, you can simplify further: $[x^2]_f+6[x]_f+[7]_f=6[x]_f+[9]_f$.
$endgroup$
– Bernard
Jan 27 at 14:32
$begingroup$
Why should $[x+3]$ be a root of $x^2-2$? Also, you can simplify further: $[x^2]_f+6[x]_f+[7]_f=6[x]_f+[9]_f$.
$endgroup$
– Bernard
Jan 27 at 14:32
$begingroup$
Oh, but so I didn't understand how "$[x]_f$ if a root of $f$" should help me.
$endgroup$
– roi_saumon
Jan 27 at 14:41
$begingroup$
Oh, but so I didn't understand how "$[x]_f$ if a root of $f$" should help me.
$endgroup$
– roi_saumon
Jan 27 at 14:41
1
1
$begingroup$
You can simply say that the set of integral elements is a subring, hence any polynomial in the integral element $[x]$ is itself an integral element.
$endgroup$
– Bernard
Jan 27 at 15:15
$begingroup$
You can simply say that the set of integral elements is a subring, hence any polynomial in the integral element $[x]$ is itself an integral element.
$endgroup$
– Bernard
Jan 27 at 15:15
|
show 4 more comments
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$begingroup$
Or if its leading coefficient is a unit. That being said, to answer your question, $xbmod f$ is a root of $f$.
$endgroup$
– Bernard
Jan 27 at 13:44
$begingroup$
@Bernard I also thought I should use $[x]_f$ is a root of $f$ but if I consider $mathbb{Q}[x]/(x^2-2)$ for instance, and $[p]_f=[x+3]_f$, then $f([p]_f)=[x]_f^2+6[x]_f+[7]_f neq 0$ no?
$endgroup$
– roi_saumon
Jan 27 at 13:54
$begingroup$
Why should $[x+3]$ be a root of $x^2-2$? Also, you can simplify further: $[x^2]_f+6[x]_f+[7]_f=6[x]_f+[9]_f$.
$endgroup$
– Bernard
Jan 27 at 14:32
$begingroup$
Oh, but so I didn't understand how "$[x]_f$ if a root of $f$" should help me.
$endgroup$
– roi_saumon
Jan 27 at 14:41
1
$begingroup$
You can simply say that the set of integral elements is a subring, hence any polynomial in the integral element $[x]$ is itself an integral element.
$endgroup$
– Bernard
Jan 27 at 15:15