Mixing
Uniqueness of representation of elements of quotient by minimal polynomial in a polynomial ring
$begingroup$
Let $F$ be a subfield of $K$. It is easy to see that if $alpha $ is algebraic over $F$, then its minimal polynomial $p(x)$ is (monic) unique and irreducible.
Taking the quotient, $F[x]/p(x)F[x]$ is a field (as $p(x)$ is irreducible), and any element $gin F$ can be represented uniquely as $p(x)F(x)$+ a polynomial of $degree<n$, where $n=deg p(x).$
I can immediately see that if instead of taking the minimal polynomial $p(x)$, we take any polynomial $f$ and quotient by its ideal $(f)$, we get a ring (not necessarly a field).
In this case, the representation does not need to be unique, right?
abstract-algebra polynomials ring-theory field-theory
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
$endgroup$
add a comment |
$begingroup$
Let $F$ be a subfield of $K$. It is easy to see that if $alpha $ is algebraic over $F$, then its minimal polynomial $p(x)$ is (monic) unique and irreducible.
Taking the quotient, $F[x]/p(x)F[x]$ is a field (as $p(x)$ is irreducible), and any element $gin F$ can be represented uniquely as $p(x)F(x)$+ a polynomial of $degree<n$, where $n=deg p(x).$
I can immediately see that if instead of taking the minimal polynomial $p(x)$, we take any polynomial $f$ and quotient by its ideal $(f)$, we get a ring (not necessarly a field).
In this case, the representation does not need to be unique, right?
abstract-algebra polynomials ring-theory field-theory
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
$endgroup$
$begingroup$
yes, thank you.
$endgroup$
– Dan Leonte
Jul 6 '17 at 17:44
$begingroup$
The representation of what is not unique ? $ mathbb{Q}[x]/(x^2+1) cong mathbb{Q}(i) = mathbb{Q}(di+e) cong mathbb{Q}[x]/(ax^2+bx+c)$ whenever $frac{b^2-4ac}{4a^2} = -d^2$
$endgroup$
– reuns
Jul 6 '17 at 17:47
$begingroup$
any element $g in F[x] $ can be repesented ,by quotinent, $F[x]/p(x)F[x]$ as ${a_0+a_1*x+...+a_{n-1}*x^{n-1}}$. Is this representations unique if $f$ is a random poly?
$endgroup$
– Dan Leonte
Jul 6 '17 at 21:52
$begingroup$
Yes of course it is unique : the elements of the quotient are of the form ${ f(x) + (p(x)), deg(f) < deg(p) }$
$endgroup$
– reuns
Jul 6 '17 at 23:10
add a comment |
$begingroup$
Let $F$ be a subfield of $K$. It is easy to see that if $alpha $ is algebraic over $F$, then its minimal polynomial $p(x)$ is (monic) unique and irreducible.
Taking the quotient, $F[x]/p(x)F[x]$ is a field (as $p(x)$ is irreducible), and any element $gin F$ can be represented uniquely as $p(x)F(x)$+ a polynomial of $degree<n$, where $n=deg p(x).$
I can immediately see that if instead of taking the minimal polynomial $p(x)$, we take any polynomial $f$ and quotient by its ideal $(f)$, we get a ring (not necessarly a field).
In this case, the representation does not need to be unique, right?
abstract-algebra polynomials ring-theory field-theory
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
$endgroup$
Let $F$ be a subfield of $K$. It is easy to see that if $alpha $ is algebraic over $F$, then its minimal polynomial $p(x)$ is (monic) unique and irreducible.
Taking the quotient, $F[x]/p(x)F[x]$ is a field (as $p(x)$ is irreducible), and any element $gin F$ can be represented uniquely as $p(x)F(x)$+ a polynomial of $degree<n$, where $n=deg p(x).$
I can immediately see that if instead of taking the minimal polynomial $p(x)$, we take any polynomial $f$ and quotient by its ideal $(f)$, we get a ring (not necessarly a field).
In this case, the representation does not need to be unique, right?
abstract-algebra polynomials ring-theory field-theory
abstract-algebra polynomials ring-theory field-theory
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
edited Jan 25 at 6:41
Eric Wofsey
190k14216347
190k14216347
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
asked Jul 6 '17 at 17:40
Dan LeonteDan Leonte
1338
1338
$begingroup$
yes, thank you.
$endgroup$
– Dan Leonte
Jul 6 '17 at 17:44
$begingroup$
The representation of what is not unique ? $ mathbb{Q}[x]/(x^2+1) cong mathbb{Q}(i) = mathbb{Q}(di+e) cong mathbb{Q}[x]/(ax^2+bx+c)$ whenever $frac{b^2-4ac}{4a^2} = -d^2$
$endgroup$
– reuns
Jul 6 '17 at 17:47
$begingroup$
any element $g in F[x] $ can be repesented ,by quotinent, $F[x]/p(x)F[x]$ as ${a_0+a_1*x+...+a_{n-1}*x^{n-1}}$. Is this representations unique if $f$ is a random poly?
$endgroup$
– Dan Leonte
Jul 6 '17 at 21:52
$begingroup$
Yes of course it is unique : the elements of the quotient are of the form ${ f(x) + (p(x)), deg(f) < deg(p) }$
$endgroup$
– reuns
Jul 6 '17 at 23:10
add a comment |
$begingroup$
yes, thank you.
$endgroup$
– Dan Leonte
Jul 6 '17 at 17:44
$begingroup$
The representation of what is not unique ? $ mathbb{Q}[x]/(x^2+1) cong mathbb{Q}(i) = mathbb{Q}(di+e) cong mathbb{Q}[x]/(ax^2+bx+c)$ whenever $frac{b^2-4ac}{4a^2} = -d^2$
$endgroup$
– reuns
Jul 6 '17 at 17:47
$begingroup$
any element $g in F[x] $ can be repesented ,by quotinent, $F[x]/p(x)F[x]$ as ${a_0+a_1*x+...+a_{n-1}*x^{n-1}}$. Is this representations unique if $f$ is a random poly?
$endgroup$
– Dan Leonte
Jul 6 '17 at 21:52
$begingroup$
Yes of course it is unique : the elements of the quotient are of the form ${ f(x) + (p(x)), deg(f) < deg(p) }$
$endgroup$
– reuns
Jul 6 '17 at 23:10
$begingroup$
yes, thank you.
$endgroup$
– Dan Leonte
Jul 6 '17 at 17:44
$begingroup$
yes, thank you.
$endgroup$
– Dan Leonte
Jul 6 '17 at 17:44
$begingroup$
The representation of what is not unique ? $ mathbb{Q}[x]/(x^2+1) cong mathbb{Q}(i) = mathbb{Q}(di+e) cong mathbb{Q}[x]/(ax^2+bx+c)$ whenever $frac{b^2-4ac}{4a^2} = -d^2$
$endgroup$
– reuns
Jul 6 '17 at 17:47
$begingroup$
The representation of what is not unique ? $ mathbb{Q}[x]/(x^2+1) cong mathbb{Q}(i) = mathbb{Q}(di+e) cong mathbb{Q}[x]/(ax^2+bx+c)$ whenever $frac{b^2-4ac}{4a^2} = -d^2$
$endgroup$
– reuns
Jul 6 '17 at 17:47
$begingroup$
any element $g in F[x] $ can be repesented ,by quotinent, $F[x]/p(x)F[x]$ as ${a_0+a_1*x+...+a_{n-1}*x^{n-1}}$. Is this representations unique if $f$ is a random poly?
$endgroup$
– Dan Leonte
Jul 6 '17 at 21:52
$begingroup$
any element $g in F[x] $ can be repesented ,by quotinent, $F[x]/p(x)F[x]$ as ${a_0+a_1*x+...+a_{n-1}*x^{n-1}}$. Is this representations unique if $f$ is a random poly?
$endgroup$
– Dan Leonte
Jul 6 '17 at 21:52
$begingroup$
Yes of course it is unique : the elements of the quotient are of the form ${ f(x) + (p(x)), deg(f) < deg(p) }$
$endgroup$
– reuns
Jul 6 '17 at 23:10
$begingroup$
Yes of course it is unique : the elements of the quotient are of the form ${ f(x) + (p(x)), deg(f) < deg(p) }$
$endgroup$
– reuns
Jul 6 '17 at 23:10
add a comment |
1 Answer
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$begingroup$
No, the representation is unique regardless of what (nonzero) polynomial $f$ is. For any polynomials $f$ and $g$ with $f$ nonzero, there exist unique polynomials $q$ and $r$ with $deg r<deg f$ such that $$g=qf+r.$$ This is just the statement of polynomial division with remainder, which has nothing to do with whether the polynomials are irreducible.
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
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1 Answer
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$begingroup$
No, the representation is unique regardless of what (nonzero) polynomial $f$ is. For any polynomials $f$ and $g$ with $f$ nonzero, there exist unique polynomials $q$ and $r$ with $deg r<deg f$ such that $$g=qf+r.$$ This is just the statement of polynomial division with remainder, which has nothing to do with whether the polynomials are irreducible.
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
$begingroup$
No, the representation is unique regardless of what (nonzero) polynomial $f$ is. For any polynomials $f$ and $g$ with $f$ nonzero, there exist unique polynomials $q$ and $r$ with $deg r<deg f$ such that $$g=qf+r.$$ This is just the statement of polynomial division with remainder, which has nothing to do with whether the polynomials are irreducible.
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
$begingroup$
No, the representation is unique regardless of what (nonzero) polynomial $f$ is. For any polynomials $f$ and $g$ with $f$ nonzero, there exist unique polynomials $q$ and $r$ with $deg r<deg f$ such that $$g=qf+r.$$ This is just the statement of polynomial division with remainder, which has nothing to do with whether the polynomials are irreducible.
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
No, the representation is unique regardless of what (nonzero) polynomial $f$ is. For any polynomials $f$ and $g$ with $f$ nonzero, there exist unique polynomials $q$ and $r$ with $deg r<deg f$ such that $$g=qf+r.$$ This is just the statement of polynomial division with remainder, which has nothing to do with whether the polynomials are irreducible.
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
answered Jul 6 '17 at 21:57
Eric WofseyEric Wofsey
190k14216347
190k14216347
add a comment |
add a comment |
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$begingroup$
yes, thank you.
$endgroup$
– Dan Leonte
Jul 6 '17 at 17:44
$begingroup$
The representation of what is not unique ? $ mathbb{Q}[x]/(x^2+1) cong mathbb{Q}(i) = mathbb{Q}(di+e) cong mathbb{Q}[x]/(ax^2+bx+c)$ whenever $frac{b^2-4ac}{4a^2} = -d^2$
$endgroup$
– reuns
Jul 6 '17 at 17:47
$begingroup$
any element $g in F[x] $ can be repesented ,by quotinent, $F[x]/p(x)F[x]$ as ${a_0+a_1*x+...+a_{n-1}*x^{n-1}}$. Is this representations unique if $f$ is a random poly?
$endgroup$
– Dan Leonte
Jul 6 '17 at 21:52
$begingroup$
Yes of course it is unique : the elements of the quotient are of the form ${ f(x) + (p(x)), deg(f) < deg(p) }$
$endgroup$
– reuns
Jul 6 '17 at 23:10