Mixing
$x^2-3$ is separable over $mathbb Q$ but not separable over $F_2$
$begingroup$
$x^2-3=(x-sqrt{3})(x+sqrt{3})$ over $mathbb Q$, so that part makes sense.
Now, when it says $x^2-3$ is a polynomial over $F_2$, I imagine it means all the coefficients are calculated mod $3$, so $p(x)=x^2-3=x^2$ in $F_2$.
$x^2$ doesn't have a second solution other than $0$ in $F_2={0,1,2}$, but how do I know there is not some algebraic extension that it will have two roots in?
Also, is this discussion so far accurate?
polynomials field-theory galois-theory finite-fields separable-extension
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
asked Jan 30 at 0:41
MikeMike
783415
$endgroup$
add a comment |
$begingroup$
$x^2-3=(x-sqrt{3})(x+sqrt{3})$ over $mathbb Q$, so that part makes sense.
Now, when it says $x^2-3$ is a polynomial over $F_2$, I imagine it means all the coefficients are calculated mod $3$, so $p(x)=x^2-3=x^2$ in $F_2$.
$x^2$ doesn't have a second solution other than $0$ in $F_2={0,1,2}$, but how do I know there is not some algebraic extension that it will have two roots in?
Also, is this discussion so far accurate?
polynomials field-theory galois-theory finite-fields separable-extension
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
asked Jan 30 at 0:41
MikeMike
783415
$endgroup$
1
$begingroup$
Because a polynomial of degree $n$ has at most $n$ Distinct roots counting multaplicities. Moreover, any field is integral domain, so $x^n = 0$ if and only if $x=0$ in any field extension of any ground field.
$endgroup$
– Adam Higgins
Jan 30 at 0:54
2
$begingroup$
Also, $F_2 = {0,1}$, not ${0,1,2}$.
$endgroup$
– Adam Higgins
Jan 30 at 0:55
$begingroup$
Thus, if you have a polynomial $f$ of degree $n$ over a field $K$ that has $n$-roots in $K$ counting multaplicities, then if $K’$ is any exntesion of $K$, then if an element of $kin K’$ is a root of $f$, then $kin K$.
$endgroup$
– Adam Higgins
Jan 30 at 0:57
add a comment |
$begingroup$
$x^2-3=(x-sqrt{3})(x+sqrt{3})$ over $mathbb Q$, so that part makes sense.
Now, when it says $x^2-3$ is a polynomial over $F_2$, I imagine it means all the coefficients are calculated mod $3$, so $p(x)=x^2-3=x^2$ in $F_2$.
$x^2$ doesn't have a second solution other than $0$ in $F_2={0,1,2}$, but how do I know there is not some algebraic extension that it will have two roots in?
Also, is this discussion so far accurate?
polynomials field-theory galois-theory finite-fields separable-extension
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
asked Jan 30 at 0:41
MikeMike
783415
$endgroup$
$x^2-3=(x-sqrt{3})(x+sqrt{3})$ over $mathbb Q$, so that part makes sense.
Now, when it says $x^2-3$ is a polynomial over $F_2$, I imagine it means all the coefficients are calculated mod $3$, so $p(x)=x^2-3=x^2$ in $F_2$.
$x^2$ doesn't have a second solution other than $0$ in $F_2={0,1,2}$, but how do I know there is not some algebraic extension that it will have two roots in?
Also, is this discussion so far accurate?
polynomials field-theory galois-theory finite-fields separable-extension
polynomials field-theory galois-theory finite-fields separable-extension
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
asked Jan 30 at 0:41
MikeMike
783415
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
asked Jan 30 at 0:41
MikeMike
783415
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
edited Feb 3 at 15:07
J. W. Tanner
4,2561320
4,2561320
asked Jan 30 at 0:41
MikeMike
783415
asked Jan 30 at 0:41
MikeMike
783415
asked Jan 30 at 0:41
MikeMike
783415
783415
1
$begingroup$
Because a polynomial of degree $n$ has at most $n$ Distinct roots counting multaplicities. Moreover, any field is integral domain, so $x^n = 0$ if and only if $x=0$ in any field extension of any ground field.
$endgroup$
– Adam Higgins
Jan 30 at 0:54
2
$begingroup$
Also, $F_2 = {0,1}$, not ${0,1,2}$.
$endgroup$
– Adam Higgins
Jan 30 at 0:55
$begingroup$
Thus, if you have a polynomial $f$ of degree $n$ over a field $K$ that has $n$-roots in $K$ counting multaplicities, then if $K’$ is any exntesion of $K$, then if an element of $kin K’$ is a root of $f$, then $kin K$.
$endgroup$
– Adam Higgins
Jan 30 at 0:57
add a comment |
1
$begingroup$
Because a polynomial of degree $n$ has at most $n$ Distinct roots counting multaplicities. Moreover, any field is integral domain, so $x^n = 0$ if and only if $x=0$ in any field extension of any ground field.
$endgroup$
– Adam Higgins
Jan 30 at 0:54
2
$begingroup$
Also, $F_2 = {0,1}$, not ${0,1,2}$.
$endgroup$
– Adam Higgins
Jan 30 at 0:55
$begingroup$
Thus, if you have a polynomial $f$ of degree $n$ over a field $K$ that has $n$-roots in $K$ counting multaplicities, then if $K’$ is any exntesion of $K$, then if an element of $kin K’$ is a root of $f$, then $kin K$.
$endgroup$
– Adam Higgins
Jan 30 at 0:57
1
1
$begingroup$
Because a polynomial of degree $n$ has at most $n$ Distinct roots counting multaplicities. Moreover, any field is integral domain, so $x^n = 0$ if and only if $x=0$ in any field extension of any ground field.
$endgroup$
– Adam Higgins
Jan 30 at 0:54
$begingroup$
Because a polynomial of degree $n$ has at most $n$ Distinct roots counting multaplicities. Moreover, any field is integral domain, so $x^n = 0$ if and only if $x=0$ in any field extension of any ground field.
$endgroup$
– Adam Higgins
Jan 30 at 0:54
2
2
$begingroup$
Also, $F_2 = {0,1}$, not ${0,1,2}$.
$endgroup$
– Adam Higgins
Jan 30 at 0:55
$begingroup$
Also, $F_2 = {0,1}$, not ${0,1,2}$.
$endgroup$
– Adam Higgins
Jan 30 at 0:55
$begingroup$
Thus, if you have a polynomial $f$ of degree $n$ over a field $K$ that has $n$-roots in $K$ counting multaplicities, then if $K’$ is any exntesion of $K$, then if an element of $kin K’$ is a root of $f$, then $kin K$.
$endgroup$
– Adam Higgins
Jan 30 at 0:57
$begingroup$
Thus, if you have a polynomial $f$ of degree $n$ over a field $K$ that has $n$-roots in $K$ counting multaplicities, then if $K’$ is any exntesion of $K$, then if an element of $kin K’$ is a root of $f$, then $kin K$.
$endgroup$
– Adam Higgins
Jan 30 at 0:57
add a comment |
1 Answer
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$begingroup$
$X^2-3$ in $F_2$ means that the coefficients are calculated $mod$ $2$ and not $mod$ $3$.
So $x^2-3=x^2+1=(x+1)^2$ mod $2$. You cannot have an extension with two different roots since the factorization over a field is unique. If you assume that the coefficient of higher degree of factors of degree $>1$ is $1$.
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
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$begingroup$
$X^2-3$ in $F_2$ means that the coefficients are calculated $mod$ $2$ and not $mod$ $3$.
So $x^2-3=x^2+1=(x+1)^2$ mod $2$. You cannot have an extension with two different roots since the factorization over a field is unique. If you assume that the coefficient of higher degree of factors of degree $>1$ is $1$.
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
$X^2-3$ in $F_2$ means that the coefficients are calculated $mod$ $2$ and not $mod$ $3$.
So $x^2-3=x^2+1=(x+1)^2$ mod $2$. You cannot have an extension with two different roots since the factorization over a field is unique. If you assume that the coefficient of higher degree of factors of degree $>1$ is $1$.
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
$X^2-3$ in $F_2$ means that the coefficients are calculated $mod$ $2$ and not $mod$ $3$.
So $x^2-3=x^2+1=(x+1)^2$ mod $2$. You cannot have an extension with two different roots since the factorization over a field is unique. If you assume that the coefficient of higher degree of factors of degree $>1$ is $1$.
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$X^2-3$ in $F_2$ means that the coefficients are calculated $mod$ $2$ and not $mod$ $3$.
So $x^2-3=x^2+1=(x+1)^2$ mod $2$. You cannot have an extension with two different roots since the factorization over a field is unique. If you assume that the coefficient of higher degree of factors of degree $>1$ is $1$.
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 0:54
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
add a comment |
add a comment |
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1
$begingroup$
Because a polynomial of degree $n$ has at most $n$ Distinct roots counting multaplicities. Moreover, any field is integral domain, so $x^n = 0$ if and only if $x=0$ in any field extension of any ground field.
$endgroup$
– Adam Higgins
Jan 30 at 0:54
2
$begingroup$
Also, $F_2 = {0,1}$, not ${0,1,2}$.
$endgroup$
– Adam Higgins
Jan 30 at 0:55
$begingroup$
Thus, if you have a polynomial $f$ of degree $n$ over a field $K$ that has $n$-roots in $K$ counting multaplicities, then if $K’$ is any exntesion of $K$, then if an element of $kin K’$ is a root of $f$, then $kin K$.
$endgroup$
– Adam Higgins
Jan 30 at 0:57