Degree of $a+b$ over a field $k$, where $a$ and $b$ are distinct roots of the same polynomial
$begingroup$
Let $K/k$ be an algebraic extension of fields with $a$ and $b$ distinct roots in $K$ of the same irreducible polynomial $f(x) in k[x]$ of degree $n$. Show that the degree of $k(a+b)/k$ is less than or equal to $frac{n(n-1)}{2}$.
Also, how does one construct fields $k$ and $K$ together with roots $a,bin K$ so that the preceding inequality is actually an equality?
I'm pretty sure I can get that $k(a+b)/k$ has degree less than or equal to $n(n-1)$ since the minimal polynomial of $b$ over $k(a)$ has degree less than or equal to $n-1$, but I'm not sure how to reduce this by a factor of $1/2$. I've also seen that there are computational techniques for computing the minimal polynomial of a sum, but a proof that avoids things such as resolvents would be ideal.
abstract-algebra field-theory galois-theory
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
$endgroup$
add a comment |
$begingroup$
Let $K/k$ be an algebraic extension of fields with $a$ and $b$ distinct roots in $K$ of the same irreducible polynomial $f(x) in k[x]$ of degree $n$. Show that the degree of $k(a+b)/k$ is less than or equal to $frac{n(n-1)}{2}$.
Also, how does one construct fields $k$ and $K$ together with roots $a,bin K$ so that the preceding inequality is actually an equality?
I'm pretty sure I can get that $k(a+b)/k$ has degree less than or equal to $n(n-1)$ since the minimal polynomial of $b$ over $k(a)$ has degree less than or equal to $n-1$, but I'm not sure how to reduce this by a factor of $1/2$. I've also seen that there are computational techniques for computing the minimal polynomial of a sum, but a proof that avoids things such as resolvents would be ideal.
abstract-algebra field-theory galois-theory
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
$endgroup$
add a comment |
$begingroup$
Let $K/k$ be an algebraic extension of fields with $a$ and $b$ distinct roots in $K$ of the same irreducible polynomial $f(x) in k[x]$ of degree $n$. Show that the degree of $k(a+b)/k$ is less than or equal to $frac{n(n-1)}{2}$.
Also, how does one construct fields $k$ and $K$ together with roots $a,bin K$ so that the preceding inequality is actually an equality?
I'm pretty sure I can get that $k(a+b)/k$ has degree less than or equal to $n(n-1)$ since the minimal polynomial of $b$ over $k(a)$ has degree less than or equal to $n-1$, but I'm not sure how to reduce this by a factor of $1/2$. I've also seen that there are computational techniques for computing the minimal polynomial of a sum, but a proof that avoids things such as resolvents would be ideal.
abstract-algebra field-theory galois-theory
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
$endgroup$
Let $K/k$ be an algebraic extension of fields with $a$ and $b$ distinct roots in $K$ of the same irreducible polynomial $f(x) in k[x]$ of degree $n$. Show that the degree of $k(a+b)/k$ is less than or equal to $frac{n(n-1)}{2}$.
Also, how does one construct fields $k$ and $K$ together with roots $a,bin K$ so that the preceding inequality is actually an equality?
I'm pretty sure I can get that $k(a+b)/k$ has degree less than or equal to $n(n-1)$ since the minimal polynomial of $b$ over $k(a)$ has degree less than or equal to $n-1$, but I'm not sure how to reduce this by a factor of $1/2$. I've also seen that there are computational techniques for computing the minimal polynomial of a sum, but a proof that avoids things such as resolvents would be ideal.
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
edited Dec 25 '18 at 19:56
leibnewtz
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
asked Dec 25 '18 at 19:43
leibnewtzleibnewtz
2,6111717
2,6111717
add a comment |
add a comment |
1 Answer
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If $f$ is inseparable, then $k$ has prime characteristic $p$ and $f(X) = X^{p^l} - t$ for some $tin k$ and $lgeq 0$. Then $a+b$ satisfies the polynomial $X^{p^l} - 2t$, which has degree $p^l=n$, which is $leq {n choose 2}$ when $ngeq 3$. $n=2$ is a special case, since $p=2$ and so $a+b=0$, while $n=1$ is impossible.
On the other hand, suppose that $f$ is separable. Then every conjugate of $a+b$ is a sum of two distinct roots of $f$, and there are ${nchoose 2}$ of these. Since $prod_{gamma sim a+b} (X-gamma)$ is fixed by the Galois group, it has coefficients in $k$, and it follows that $a+b$ satisfies a polynomial of degree $leq {nchoose 2}$.
To see that the inequality is tight, let $f$ be any polynomial with doubly-transitive Galois group, e.g. $S_n$, such that the pairwise sums of roots are distinct. (These should be common, but is there a good way to construct them?) Then $a+b$ has exactly ${nchoose 2}$ conjugates.
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
$endgroup$
$begingroup$
Very nice! I'm refraining from accepting for the moment just in case anyone is able to offer an explicit example
$endgroup$
– leibnewtz
Dec 25 '18 at 20:58
$begingroup$
@leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $sin mathbb{Q}$. See here for a discussion of why this has full Galois group.
$endgroup$
– Slade
Dec 25 '18 at 21:52
1
$begingroup$
Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-)
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 9:27
add a comment |
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1 Answer
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$begingroup$
If $f$ is inseparable, then $k$ has prime characteristic $p$ and $f(X) = X^{p^l} - t$ for some $tin k$ and $lgeq 0$. Then $a+b$ satisfies the polynomial $X^{p^l} - 2t$, which has degree $p^l=n$, which is $leq {n choose 2}$ when $ngeq 3$. $n=2$ is a special case, since $p=2$ and so $a+b=0$, while $n=1$ is impossible.
On the other hand, suppose that $f$ is separable. Then every conjugate of $a+b$ is a sum of two distinct roots of $f$, and there are ${nchoose 2}$ of these. Since $prod_{gamma sim a+b} (X-gamma)$ is fixed by the Galois group, it has coefficients in $k$, and it follows that $a+b$ satisfies a polynomial of degree $leq {nchoose 2}$.
To see that the inequality is tight, let $f$ be any polynomial with doubly-transitive Galois group, e.g. $S_n$, such that the pairwise sums of roots are distinct. (These should be common, but is there a good way to construct them?) Then $a+b$ has exactly ${nchoose 2}$ conjugates.
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
$endgroup$
$begingroup$
Very nice! I'm refraining from accepting for the moment just in case anyone is able to offer an explicit example
$endgroup$
– leibnewtz
Dec 25 '18 at 20:58
$begingroup$
@leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $sin mathbb{Q}$. See here for a discussion of why this has full Galois group.
$endgroup$
– Slade
Dec 25 '18 at 21:52
1
$begingroup$
Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-)
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 9:27
add a comment |
$begingroup$
If $f$ is inseparable, then $k$ has prime characteristic $p$ and $f(X) = X^{p^l} - t$ for some $tin k$ and $lgeq 0$. Then $a+b$ satisfies the polynomial $X^{p^l} - 2t$, which has degree $p^l=n$, which is $leq {n choose 2}$ when $ngeq 3$. $n=2$ is a special case, since $p=2$ and so $a+b=0$, while $n=1$ is impossible.
On the other hand, suppose that $f$ is separable. Then every conjugate of $a+b$ is a sum of two distinct roots of $f$, and there are ${nchoose 2}$ of these. Since $prod_{gamma sim a+b} (X-gamma)$ is fixed by the Galois group, it has coefficients in $k$, and it follows that $a+b$ satisfies a polynomial of degree $leq {nchoose 2}$.
To see that the inequality is tight, let $f$ be any polynomial with doubly-transitive Galois group, e.g. $S_n$, such that the pairwise sums of roots are distinct. (These should be common, but is there a good way to construct them?) Then $a+b$ has exactly ${nchoose 2}$ conjugates.
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
$endgroup$
$begingroup$
Very nice! I'm refraining from accepting for the moment just in case anyone is able to offer an explicit example
$endgroup$
– leibnewtz
Dec 25 '18 at 20:58
$begingroup$
@leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $sin mathbb{Q}$. See here for a discussion of why this has full Galois group.
$endgroup$
– Slade
Dec 25 '18 at 21:52
1
$begingroup$
Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-)
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 9:27
add a comment |
$begingroup$
If $f$ is inseparable, then $k$ has prime characteristic $p$ and $f(X) = X^{p^l} - t$ for some $tin k$ and $lgeq 0$. Then $a+b$ satisfies the polynomial $X^{p^l} - 2t$, which has degree $p^l=n$, which is $leq {n choose 2}$ when $ngeq 3$. $n=2$ is a special case, since $p=2$ and so $a+b=0$, while $n=1$ is impossible.
On the other hand, suppose that $f$ is separable. Then every conjugate of $a+b$ is a sum of two distinct roots of $f$, and there are ${nchoose 2}$ of these. Since $prod_{gamma sim a+b} (X-gamma)$ is fixed by the Galois group, it has coefficients in $k$, and it follows that $a+b$ satisfies a polynomial of degree $leq {nchoose 2}$.
To see that the inequality is tight, let $f$ be any polynomial with doubly-transitive Galois group, e.g. $S_n$, such that the pairwise sums of roots are distinct. (These should be common, but is there a good way to construct them?) Then $a+b$ has exactly ${nchoose 2}$ conjugates.
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
$endgroup$
If $f$ is inseparable, then $k$ has prime characteristic $p$ and $f(X) = X^{p^l} - t$ for some $tin k$ and $lgeq 0$. Then $a+b$ satisfies the polynomial $X^{p^l} - 2t$, which has degree $p^l=n$, which is $leq {n choose 2}$ when $ngeq 3$. $n=2$ is a special case, since $p=2$ and so $a+b=0$, while $n=1$ is impossible.
On the other hand, suppose that $f$ is separable. Then every conjugate of $a+b$ is a sum of two distinct roots of $f$, and there are ${nchoose 2}$ of these. Since $prod_{gamma sim a+b} (X-gamma)$ is fixed by the Galois group, it has coefficients in $k$, and it follows that $a+b$ satisfies a polynomial of degree $leq {nchoose 2}$.
To see that the inequality is tight, let $f$ be any polynomial with doubly-transitive Galois group, e.g. $S_n$, such that the pairwise sums of roots are distinct. (These should be common, but is there a good way to construct them?) Then $a+b$ has exactly ${nchoose 2}$ conjugates.
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
edited Dec 25 '18 at 20:24
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
answered Dec 25 '18 at 20:13
SladeSlade
25.2k12665
25.2k12665
$begingroup$
Very nice! I'm refraining from accepting for the moment just in case anyone is able to offer an explicit example
$endgroup$
– leibnewtz
Dec 25 '18 at 20:58
$begingroup$
@leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $sin mathbb{Q}$. See here for a discussion of why this has full Galois group.
$endgroup$
– Slade
Dec 25 '18 at 21:52
1
$begingroup$
Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-)
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 9:27
add a comment |
$begingroup$
Very nice! I'm refraining from accepting for the moment just in case anyone is able to offer an explicit example
$endgroup$
– leibnewtz
Dec 25 '18 at 20:58
$begingroup$
@leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $sin mathbb{Q}$. See here for a discussion of why this has full Galois group.
$endgroup$
– Slade
Dec 25 '18 at 21:52
1
$begingroup$
Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-)
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 9:27
$begingroup$
Very nice! I'm refraining from accepting for the moment just in case anyone is able to offer an explicit example
$endgroup$
– leibnewtz
Dec 25 '18 at 20:58
$begingroup$
Very nice! I'm refraining from accepting for the moment just in case anyone is able to offer an explicit example
$endgroup$
– leibnewtz
Dec 25 '18 at 20:58
$begingroup$
@leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $sin mathbb{Q}$. See here for a discussion of why this has full Galois group.
$endgroup$
– Slade
Dec 25 '18 at 21:52
$begingroup$
@leibnewtz For starters, I'd guess that $x^n - sx - s$ is an example for most $sin mathbb{Q}$. See here for a discussion of why this has full Galois group.
$endgroup$
– Slade
Dec 25 '18 at 21:52
1
1
$begingroup$
Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-)
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 9:27
$begingroup$
Good job! Commenting on the last paragraph. If we let $K$ be the field of fractions of the $n$ variable polynomial ring $F[x_1,x_2,ldots,x_n]$, $F$ some field, and let $k$ be the subfield of symmetric rational functions, then $f(T)=prod_i(T-x_i)$ has coefficients in $k$ and works for this purpose. IOW, the same old :-)
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 9:27
add a comment |
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