Mixing
degree of a sum of two algebraic numbers
$begingroup$
Let $a=sqrt{2}$ and $b=sqrt{3}$, then $a$ and $b$ are algebraic numbers of degree 2, while the degree of $a+b$ is not 2, actually it is 4, by the standard argument: if $x=a+b$, then we can rationalize it completely and get $x^4-10x^2+1=0$, so $a+b$ is of degree 4, since the polynomial is irreducible.
Now take $a=sqrt[2019]{2}$ and $b=sqrt[2019]{3}$. How to prove that the degree of $a+b$ is:
greater than 2019? (if it can be done with less advanced methods then the second part)
equal $2019^2$? (if this value is correct)
algebraic-number-theory minimal-polynomials
asked Jan 26 at 23:19
larry01larry01
960416
$endgroup$
add a comment |
$begingroup$
Let $a=sqrt{2}$ and $b=sqrt{3}$, then $a$ and $b$ are algebraic numbers of degree 2, while the degree of $a+b$ is not 2, actually it is 4, by the standard argument: if $x=a+b$, then we can rationalize it completely and get $x^4-10x^2+1=0$, so $a+b$ is of degree 4, since the polynomial is irreducible.
Now take $a=sqrt[2019]{2}$ and $b=sqrt[2019]{3}$. How to prove that the degree of $a+b$ is:
greater than 2019? (if it can be done with less advanced methods then the second part)
equal $2019^2$? (if this value is correct)
algebraic-number-theory minimal-polynomials
asked Jan 26 at 23:19
larry01larry01
960416
$endgroup$
3
$begingroup$
You have a simple algorithm for finding the minimal polynomial of such a number.
$endgroup$
– ÍgjøgnumMeg
Jan 26 at 23:25
2
$begingroup$
@ÍgjøgnumMeg: really? We have a simple algorithm for finding a polynomial (with degree $2019^2$) in $mathbb{Z}[x]$ which vanishes at $a+b$, but what is ensuring the irreducibility of such polynomial?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:34
add a comment |
$begingroup$
Let $a=sqrt{2}$ and $b=sqrt{3}$, then $a$ and $b$ are algebraic numbers of degree 2, while the degree of $a+b$ is not 2, actually it is 4, by the standard argument: if $x=a+b$, then we can rationalize it completely and get $x^4-10x^2+1=0$, so $a+b$ is of degree 4, since the polynomial is irreducible.
Now take $a=sqrt[2019]{2}$ and $b=sqrt[2019]{3}$. How to prove that the degree of $a+b$ is:
greater than 2019? (if it can be done with less advanced methods then the second part)
equal $2019^2$? (if this value is correct)
algebraic-number-theory minimal-polynomials
asked Jan 26 at 23:19
larry01larry01
960416
$endgroup$
Let $a=sqrt{2}$ and $b=sqrt{3}$, then $a$ and $b$ are algebraic numbers of degree 2, while the degree of $a+b$ is not 2, actually it is 4, by the standard argument: if $x=a+b$, then we can rationalize it completely and get $x^4-10x^2+1=0$, so $a+b$ is of degree 4, since the polynomial is irreducible.
Now take $a=sqrt[2019]{2}$ and $b=sqrt[2019]{3}$. How to prove that the degree of $a+b$ is:
greater than 2019? (if it can be done with less advanced methods then the second part)
equal $2019^2$? (if this value is correct)
algebraic-number-theory minimal-polynomials
algebraic-number-theory minimal-polynomials
asked Jan 26 at 23:19
larry01larry01
960416
asked Jan 26 at 23:19
larry01larry01
960416
asked Jan 26 at 23:19
larry01larry01
960416
asked Jan 26 at 23:19
larry01larry01
960416
asked Jan 26 at 23:19
larry01larry01
960416
960416
3
$begingroup$
You have a simple algorithm for finding the minimal polynomial of such a number.
$endgroup$
– ÍgjøgnumMeg
Jan 26 at 23:25
2
$begingroup$
@ÍgjøgnumMeg: really? We have a simple algorithm for finding a polynomial (with degree $2019^2$) in $mathbb{Z}[x]$ which vanishes at $a+b$, but what is ensuring the irreducibility of such polynomial?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:34
add a comment |
3
$begingroup$
You have a simple algorithm for finding the minimal polynomial of such a number.
$endgroup$
– ÍgjøgnumMeg
Jan 26 at 23:25
2
$begingroup$
@ÍgjøgnumMeg: really? We have a simple algorithm for finding a polynomial (with degree $2019^2$) in $mathbb{Z}[x]$ which vanishes at $a+b$, but what is ensuring the irreducibility of such polynomial?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:34
3
3
$begingroup$
You have a simple algorithm for finding the minimal polynomial of such a number.
$endgroup$
– ÍgjøgnumMeg
Jan 26 at 23:25
$begingroup$
You have a simple algorithm for finding the minimal polynomial of such a number.
$endgroup$
– ÍgjøgnumMeg
Jan 26 at 23:25
2
2
$begingroup$
@ÍgjøgnumMeg: really? We have a simple algorithm for finding a polynomial (with degree $2019^2$) in $mathbb{Z}[x]$ which vanishes at $a+b$, but what is ensuring the irreducibility of such polynomial?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:34
$begingroup$
@ÍgjøgnumMeg: really? We have a simple algorithm for finding a polynomial (with degree $2019^2$) in $mathbb{Z}[x]$ which vanishes at $a+b$, but what is ensuring the irreducibility of such polynomial?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I claim the order is $2019^2$, but I'm not going to prove some of my sub-claims (and I'm not 100% certain they're all true). So it's not quite a full answer, but maybe it's a new way for you to look at the problem.
Throughout, let $zeta$ denote a primitive $2019^{tinymbox{th}}$ root of unity, let $alpha = sqrt[2019]{2}$, and let $beta = sqrt[2019]{3}$.
The first sub-claim is that $mathbb{Q}(alpha, beta)$ has degree $2019^2$ over $mathbb{Q}$. Let $f(x) = x^{2019}-2$ and $g(x)=x^{2019}-3$. If $g(x)$ splits over $mathbb{Q}(alpha)$, then, since all roots of $g$ in $mathbb{C}$ have the form $zeta^k beta$, then $beta in mathbb{Q}(alpha)$ (which I sub-claim to be false without proof), as the constant term of any factor of $g$ will be of the form $zeta^ell beta$, which is not real unless $ell = 0 pmod{2019}$, in which case $beta in mathbb{Q}(alpha)$. So $g(x)$ is the minimal polynomial of $beta$ over $mathbb{Q}(alpha)$, and thus $mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
The second sub-claim is that $alpha+beta$ generates $mathbb{Q}(alpha,beta)$ over $mathbb{Q}$, i.e., is a primitive element. This can be done by showing that
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} ne 1,$$
for any choice of $m,n$ (which I sub-claim to be true without proof). To see why, see the constructive proof of the primitive element theorem. This implies that $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
Note, as @ÍgjøgnumMeg indicates in the comments, there is an algorithm for finding the minimal polynomial of $alpha + beta$. Since it lies in $mathbb{Q}(alpha,beta)$, which has basis ${alpha^ibeta^j : 0le i,jle 2018}$ (this hinges upon my first sub-claim), and so you can raise $alpha+beta$ to higher and higher powers, writing these powers in terms of the basis, and look for linear dependency among the thus-far-computed powers until there is one.
answered Jan 27 at 1:34
mathprincessmathprincess
313
$endgroup$
$begingroup$
(+1) Nice work so far. Do you agree that checking the invertibility of a $2019^2times 2019^2$ matrix is not really a simple approach?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:47
1
$begingroup$
Absolutely. Do you know an easy way to show $betanotin mathbb{Q}(alpha)$? I feel like the second sub-claim has a chance to be doable, but I'm not so confident about the doability of the first sub-claim.
$endgroup$
– mathprincess
Jan 27 at 2:16
$begingroup$
In order to show that $betainmathbb{Q}(alpha)$, I would invoke that for infinite primes of the form $p=2019k+1$ we have that $2^kequiv 1$ and $3^knotequiv 1pmod{p}$ (and vice-versa), sort of extension of quadratic reciprocity and Dirichlet's theorem on APs. In other terms, $mathbb{Q}(alpha,beta)$ has to be an actual extension of $mathbb{Q}(alpha)$ since in infinite cases $mathbb{F}_p(alpha)=mathbb{F}_p$ but $mathbb{F}_p(beta)neqmathbb{F}_p$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:22
$begingroup$
Another way is to compute the discriminants of $x^{2019}-2$ and $x^{2019}-3$. Except for the common factor $-2019^{2019}$ they have distinct factorizations, so the quadratic subfields of $mathbb{Q}(alpha)$ and the quadratic subfields of $mathbb{Q}(beta)$ are different things and we are sure that $betanotinmathbb{Q}(alpha)$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:28
1
$begingroup$
Interesting! How come you need there to be infinitely many such primes?
$endgroup$
– mathprincess
Jan 27 at 2:33
|
show 1 more comment
$begingroup$
Let $alpha=2^{1/2019}$, $beta=3^{1/2019}$, $zeta=exp(2pi i/2019)$, $K=mathbb{Q}(alpha)$, and $L=mathbb{Q}(zeta,beta)$.
Here, we prove the following statement: $alphanotin L$ and the degree of $alpha$ over $L$ is $2019$.
Then it implies the validity of the second claim of mathprincess since
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} = 1,$$
cannot hold if $alphanotin L$. This also leads to the validity of the first claim and a full solution to this question.
It is known that $[L:mathbb{Q}]=2019phi(2019)$ (see Jacobson, Velez cited in this answer: Computing the Galois group of polynomials $x^n-a in mathbb{Q}[x]$). Thus, the basis of $mathbb{Q}(zeta)$ over $mathbb{Q}$ are also a basis of $L$ over $mathbb{Q}(beta)$. Then
$$
textrm{disc}_{mathbb{Q}(beta)}L=textrm{disc}_{mathbb{Q}}{mathbb{Q}(zeta)} biggvert (2019)^{phi(2019)}.
$$
It is well-known that a rational prime $p$ does not divide the discriminant $textrm{disc}_{mathbb{Q}}L$ if and only if $p$ is unramified in $L$. Then we have by (Exrcise 23b of Marcus 'Number Fields'),
$$begin{align}
textrm{disc}_{mathbb{Q}}L &biggvert (textrm{disc}_{mathbb{Q}}{mathbb{Q}(beta)})^{phi(2019)} N_{mathbb{Q}}^{mathbb{Q}(beta)}textrm{disc}_{mathbb{Q(beta)}}L\
&biggvert (3^{2018}cdot 2019^{2019})^{phi(2019)}((2019)^{phi(2019)})^{2019}.end{align}
$$
This gives $2nmid textrm{disc}_{mathbb{Q}}L$.
Thus, the rational prime $2$ is unramified in $L$. Now, take any prime $mathcal{P}subseteq L$ lying above $2$. Then $mathcal{P}$ divides any coefficient of $x^{2019}-2$ except for the leading coefficient, and $mathcal{P}^2$ does not divide the constant coefficient $2$ since $2$ is unramified in $L$. Hence, by Eisenstein, $x^{2019}-2$ over $L$ is irreducible. This shows that the degree of $alpha$ over $L$ must be $2019$. Moreover, the same argument also yields that the degree of $alpha$ over $mathbb{Q}(beta)$ is $2019$.
As in mathprincess's answer, we may proceed with $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$. Then
$$begin{align}
[mathbb{Q}(alpha+beta):mathbb{Q}]&=[mathbb{Q}(alpha+beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]\
&=[mathbb{Q}(alpha,beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]
=2019^2.end{align}
$$
answered Jan 27 at 17:05
i707107i707107
12.5k21647
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I claim the order is $2019^2$, but I'm not going to prove some of my sub-claims (and I'm not 100% certain they're all true). So it's not quite a full answer, but maybe it's a new way for you to look at the problem.
Throughout, let $zeta$ denote a primitive $2019^{tinymbox{th}}$ root of unity, let $alpha = sqrt[2019]{2}$, and let $beta = sqrt[2019]{3}$.
The first sub-claim is that $mathbb{Q}(alpha, beta)$ has degree $2019^2$ over $mathbb{Q}$. Let $f(x) = x^{2019}-2$ and $g(x)=x^{2019}-3$. If $g(x)$ splits over $mathbb{Q}(alpha)$, then, since all roots of $g$ in $mathbb{C}$ have the form $zeta^k beta$, then $beta in mathbb{Q}(alpha)$ (which I sub-claim to be false without proof), as the constant term of any factor of $g$ will be of the form $zeta^ell beta$, which is not real unless $ell = 0 pmod{2019}$, in which case $beta in mathbb{Q}(alpha)$. So $g(x)$ is the minimal polynomial of $beta$ over $mathbb{Q}(alpha)$, and thus $mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
The second sub-claim is that $alpha+beta$ generates $mathbb{Q}(alpha,beta)$ over $mathbb{Q}$, i.e., is a primitive element. This can be done by showing that
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} ne 1,$$
for any choice of $m,n$ (which I sub-claim to be true without proof). To see why, see the constructive proof of the primitive element theorem. This implies that $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
Note, as @ÍgjøgnumMeg indicates in the comments, there is an algorithm for finding the minimal polynomial of $alpha + beta$. Since it lies in $mathbb{Q}(alpha,beta)$, which has basis ${alpha^ibeta^j : 0le i,jle 2018}$ (this hinges upon my first sub-claim), and so you can raise $alpha+beta$ to higher and higher powers, writing these powers in terms of the basis, and look for linear dependency among the thus-far-computed powers until there is one.
answered Jan 27 at 1:34
mathprincessmathprincess
313
$endgroup$
$begingroup$
(+1) Nice work so far. Do you agree that checking the invertibility of a $2019^2times 2019^2$ matrix is not really a simple approach?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:47
1
$begingroup$
Absolutely. Do you know an easy way to show $betanotin mathbb{Q}(alpha)$? I feel like the second sub-claim has a chance to be doable, but I'm not so confident about the doability of the first sub-claim.
$endgroup$
– mathprincess
Jan 27 at 2:16
$begingroup$
In order to show that $betainmathbb{Q}(alpha)$, I would invoke that for infinite primes of the form $p=2019k+1$ we have that $2^kequiv 1$ and $3^knotequiv 1pmod{p}$ (and vice-versa), sort of extension of quadratic reciprocity and Dirichlet's theorem on APs. In other terms, $mathbb{Q}(alpha,beta)$ has to be an actual extension of $mathbb{Q}(alpha)$ since in infinite cases $mathbb{F}_p(alpha)=mathbb{F}_p$ but $mathbb{F}_p(beta)neqmathbb{F}_p$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:22
$begingroup$
Another way is to compute the discriminants of $x^{2019}-2$ and $x^{2019}-3$. Except for the common factor $-2019^{2019}$ they have distinct factorizations, so the quadratic subfields of $mathbb{Q}(alpha)$ and the quadratic subfields of $mathbb{Q}(beta)$ are different things and we are sure that $betanotinmathbb{Q}(alpha)$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:28
1
$begingroup$
Interesting! How come you need there to be infinitely many such primes?
$endgroup$
– mathprincess
Jan 27 at 2:33
|
show 1 more comment
$begingroup$
I claim the order is $2019^2$, but I'm not going to prove some of my sub-claims (and I'm not 100% certain they're all true). So it's not quite a full answer, but maybe it's a new way for you to look at the problem.
Throughout, let $zeta$ denote a primitive $2019^{tinymbox{th}}$ root of unity, let $alpha = sqrt[2019]{2}$, and let $beta = sqrt[2019]{3}$.
The first sub-claim is that $mathbb{Q}(alpha, beta)$ has degree $2019^2$ over $mathbb{Q}$. Let $f(x) = x^{2019}-2$ and $g(x)=x^{2019}-3$. If $g(x)$ splits over $mathbb{Q}(alpha)$, then, since all roots of $g$ in $mathbb{C}$ have the form $zeta^k beta$, then $beta in mathbb{Q}(alpha)$ (which I sub-claim to be false without proof), as the constant term of any factor of $g$ will be of the form $zeta^ell beta$, which is not real unless $ell = 0 pmod{2019}$, in which case $beta in mathbb{Q}(alpha)$. So $g(x)$ is the minimal polynomial of $beta$ over $mathbb{Q}(alpha)$, and thus $mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
The second sub-claim is that $alpha+beta$ generates $mathbb{Q}(alpha,beta)$ over $mathbb{Q}$, i.e., is a primitive element. This can be done by showing that
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} ne 1,$$
for any choice of $m,n$ (which I sub-claim to be true without proof). To see why, see the constructive proof of the primitive element theorem. This implies that $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
Note, as @ÍgjøgnumMeg indicates in the comments, there is an algorithm for finding the minimal polynomial of $alpha + beta$. Since it lies in $mathbb{Q}(alpha,beta)$, which has basis ${alpha^ibeta^j : 0le i,jle 2018}$ (this hinges upon my first sub-claim), and so you can raise $alpha+beta$ to higher and higher powers, writing these powers in terms of the basis, and look for linear dependency among the thus-far-computed powers until there is one.
answered Jan 27 at 1:34
mathprincessmathprincess
313
$endgroup$
$begingroup$
(+1) Nice work so far. Do you agree that checking the invertibility of a $2019^2times 2019^2$ matrix is not really a simple approach?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:47
1
$begingroup$
Absolutely. Do you know an easy way to show $betanotin mathbb{Q}(alpha)$? I feel like the second sub-claim has a chance to be doable, but I'm not so confident about the doability of the first sub-claim.
$endgroup$
– mathprincess
Jan 27 at 2:16
$begingroup$
In order to show that $betainmathbb{Q}(alpha)$, I would invoke that for infinite primes of the form $p=2019k+1$ we have that $2^kequiv 1$ and $3^knotequiv 1pmod{p}$ (and vice-versa), sort of extension of quadratic reciprocity and Dirichlet's theorem on APs. In other terms, $mathbb{Q}(alpha,beta)$ has to be an actual extension of $mathbb{Q}(alpha)$ since in infinite cases $mathbb{F}_p(alpha)=mathbb{F}_p$ but $mathbb{F}_p(beta)neqmathbb{F}_p$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:22
$begingroup$
Another way is to compute the discriminants of $x^{2019}-2$ and $x^{2019}-3$. Except for the common factor $-2019^{2019}$ they have distinct factorizations, so the quadratic subfields of $mathbb{Q}(alpha)$ and the quadratic subfields of $mathbb{Q}(beta)$ are different things and we are sure that $betanotinmathbb{Q}(alpha)$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:28
1
$begingroup$
Interesting! How come you need there to be infinitely many such primes?
$endgroup$
– mathprincess
Jan 27 at 2:33
|
show 1 more comment
$begingroup$
I claim the order is $2019^2$, but I'm not going to prove some of my sub-claims (and I'm not 100% certain they're all true). So it's not quite a full answer, but maybe it's a new way for you to look at the problem.
Throughout, let $zeta$ denote a primitive $2019^{tinymbox{th}}$ root of unity, let $alpha = sqrt[2019]{2}$, and let $beta = sqrt[2019]{3}$.
The first sub-claim is that $mathbb{Q}(alpha, beta)$ has degree $2019^2$ over $mathbb{Q}$. Let $f(x) = x^{2019}-2$ and $g(x)=x^{2019}-3$. If $g(x)$ splits over $mathbb{Q}(alpha)$, then, since all roots of $g$ in $mathbb{C}$ have the form $zeta^k beta$, then $beta in mathbb{Q}(alpha)$ (which I sub-claim to be false without proof), as the constant term of any factor of $g$ will be of the form $zeta^ell beta$, which is not real unless $ell = 0 pmod{2019}$, in which case $beta in mathbb{Q}(alpha)$. So $g(x)$ is the minimal polynomial of $beta$ over $mathbb{Q}(alpha)$, and thus $mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
The second sub-claim is that $alpha+beta$ generates $mathbb{Q}(alpha,beta)$ over $mathbb{Q}$, i.e., is a primitive element. This can be done by showing that
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} ne 1,$$
for any choice of $m,n$ (which I sub-claim to be true without proof). To see why, see the constructive proof of the primitive element theorem. This implies that $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
Note, as @ÍgjøgnumMeg indicates in the comments, there is an algorithm for finding the minimal polynomial of $alpha + beta$. Since it lies in $mathbb{Q}(alpha,beta)$, which has basis ${alpha^ibeta^j : 0le i,jle 2018}$ (this hinges upon my first sub-claim), and so you can raise $alpha+beta$ to higher and higher powers, writing these powers in terms of the basis, and look for linear dependency among the thus-far-computed powers until there is one.
answered Jan 27 at 1:34
mathprincessmathprincess
313
$endgroup$
I claim the order is $2019^2$, but I'm not going to prove some of my sub-claims (and I'm not 100% certain they're all true). So it's not quite a full answer, but maybe it's a new way for you to look at the problem.
Throughout, let $zeta$ denote a primitive $2019^{tinymbox{th}}$ root of unity, let $alpha = sqrt[2019]{2}$, and let $beta = sqrt[2019]{3}$.
The first sub-claim is that $mathbb{Q}(alpha, beta)$ has degree $2019^2$ over $mathbb{Q}$. Let $f(x) = x^{2019}-2$ and $g(x)=x^{2019}-3$. If $g(x)$ splits over $mathbb{Q}(alpha)$, then, since all roots of $g$ in $mathbb{C}$ have the form $zeta^k beta$, then $beta in mathbb{Q}(alpha)$ (which I sub-claim to be false without proof), as the constant term of any factor of $g$ will be of the form $zeta^ell beta$, which is not real unless $ell = 0 pmod{2019}$, in which case $beta in mathbb{Q}(alpha)$. So $g(x)$ is the minimal polynomial of $beta$ over $mathbb{Q}(alpha)$, and thus $mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
The second sub-claim is that $alpha+beta$ generates $mathbb{Q}(alpha,beta)$ over $mathbb{Q}$, i.e., is a primitive element. This can be done by showing that
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} ne 1,$$
for any choice of $m,n$ (which I sub-claim to be true without proof). To see why, see the constructive proof of the primitive element theorem. This implies that $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$ has degree $2019^2$ over $mathbb{Q}$.
Note, as @ÍgjøgnumMeg indicates in the comments, there is an algorithm for finding the minimal polynomial of $alpha + beta$. Since it lies in $mathbb{Q}(alpha,beta)$, which has basis ${alpha^ibeta^j : 0le i,jle 2018}$ (this hinges upon my first sub-claim), and so you can raise $alpha+beta$ to higher and higher powers, writing these powers in terms of the basis, and look for linear dependency among the thus-far-computed powers until there is one.
answered Jan 27 at 1:34
mathprincessmathprincess
313
edited Jan 27 at 1:41
answered Jan 27 at 1:34
mathprincessmathprincess
313
answered Jan 27 at 1:34
mathprincessmathprincess
313
answered Jan 27 at 1:34
mathprincessmathprincess
313
313
$begingroup$
(+1) Nice work so far. Do you agree that checking the invertibility of a $2019^2times 2019^2$ matrix is not really a simple approach?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:47
1
$begingroup$
Absolutely. Do you know an easy way to show $betanotin mathbb{Q}(alpha)$? I feel like the second sub-claim has a chance to be doable, but I'm not so confident about the doability of the first sub-claim.
$endgroup$
– mathprincess
Jan 27 at 2:16
$begingroup$
In order to show that $betainmathbb{Q}(alpha)$, I would invoke that for infinite primes of the form $p=2019k+1$ we have that $2^kequiv 1$ and $3^knotequiv 1pmod{p}$ (and vice-versa), sort of extension of quadratic reciprocity and Dirichlet's theorem on APs. In other terms, $mathbb{Q}(alpha,beta)$ has to be an actual extension of $mathbb{Q}(alpha)$ since in infinite cases $mathbb{F}_p(alpha)=mathbb{F}_p$ but $mathbb{F}_p(beta)neqmathbb{F}_p$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:22
$begingroup$
Another way is to compute the discriminants of $x^{2019}-2$ and $x^{2019}-3$. Except for the common factor $-2019^{2019}$ they have distinct factorizations, so the quadratic subfields of $mathbb{Q}(alpha)$ and the quadratic subfields of $mathbb{Q}(beta)$ are different things and we are sure that $betanotinmathbb{Q}(alpha)$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:28
1
$begingroup$
Interesting! How come you need there to be infinitely many such primes?
$endgroup$
– mathprincess
Jan 27 at 2:33
|
show 1 more comment
$begingroup$
(+1) Nice work so far. Do you agree that checking the invertibility of a $2019^2times 2019^2$ matrix is not really a simple approach?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:47
1
$begingroup$
Absolutely. Do you know an easy way to show $betanotin mathbb{Q}(alpha)$? I feel like the second sub-claim has a chance to be doable, but I'm not so confident about the doability of the first sub-claim.
$endgroup$
– mathprincess
Jan 27 at 2:16
$begingroup$
In order to show that $betainmathbb{Q}(alpha)$, I would invoke that for infinite primes of the form $p=2019k+1$ we have that $2^kequiv 1$ and $3^knotequiv 1pmod{p}$ (and vice-versa), sort of extension of quadratic reciprocity and Dirichlet's theorem on APs. In other terms, $mathbb{Q}(alpha,beta)$ has to be an actual extension of $mathbb{Q}(alpha)$ since in infinite cases $mathbb{F}_p(alpha)=mathbb{F}_p$ but $mathbb{F}_p(beta)neqmathbb{F}_p$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:22
$begingroup$
Another way is to compute the discriminants of $x^{2019}-2$ and $x^{2019}-3$. Except for the common factor $-2019^{2019}$ they have distinct factorizations, so the quadratic subfields of $mathbb{Q}(alpha)$ and the quadratic subfields of $mathbb{Q}(beta)$ are different things and we are sure that $betanotinmathbb{Q}(alpha)$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:28
1
$begingroup$
Interesting! How come you need there to be infinitely many such primes?
$endgroup$
– mathprincess
Jan 27 at 2:33
$begingroup$
(+1) Nice work so far. Do you agree that checking the invertibility of a $2019^2times 2019^2$ matrix is not really a simple approach?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:47
$begingroup$
(+1) Nice work so far. Do you agree that checking the invertibility of a $2019^2times 2019^2$ matrix is not really a simple approach?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:47
1
1
$begingroup$
Absolutely. Do you know an easy way to show $betanotin mathbb{Q}(alpha)$? I feel like the second sub-claim has a chance to be doable, but I'm not so confident about the doability of the first sub-claim.
$endgroup$
– mathprincess
Jan 27 at 2:16
$begingroup$
Absolutely. Do you know an easy way to show $betanotin mathbb{Q}(alpha)$? I feel like the second sub-claim has a chance to be doable, but I'm not so confident about the doability of the first sub-claim.
$endgroup$
– mathprincess
Jan 27 at 2:16
$begingroup$
In order to show that $betainmathbb{Q}(alpha)$, I would invoke that for infinite primes of the form $p=2019k+1$ we have that $2^kequiv 1$ and $3^knotequiv 1pmod{p}$ (and vice-versa), sort of extension of quadratic reciprocity and Dirichlet's theorem on APs. In other terms, $mathbb{Q}(alpha,beta)$ has to be an actual extension of $mathbb{Q}(alpha)$ since in infinite cases $mathbb{F}_p(alpha)=mathbb{F}_p$ but $mathbb{F}_p(beta)neqmathbb{F}_p$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:22
$begingroup$
In order to show that $betainmathbb{Q}(alpha)$, I would invoke that for infinite primes of the form $p=2019k+1$ we have that $2^kequiv 1$ and $3^knotequiv 1pmod{p}$ (and vice-versa), sort of extension of quadratic reciprocity and Dirichlet's theorem on APs. In other terms, $mathbb{Q}(alpha,beta)$ has to be an actual extension of $mathbb{Q}(alpha)$ since in infinite cases $mathbb{F}_p(alpha)=mathbb{F}_p$ but $mathbb{F}_p(beta)neqmathbb{F}_p$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:22
$begingroup$
Another way is to compute the discriminants of $x^{2019}-2$ and $x^{2019}-3$. Except for the common factor $-2019^{2019}$ they have distinct factorizations, so the quadratic subfields of $mathbb{Q}(alpha)$ and the quadratic subfields of $mathbb{Q}(beta)$ are different things and we are sure that $betanotinmathbb{Q}(alpha)$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:28
$begingroup$
Another way is to compute the discriminants of $x^{2019}-2$ and $x^{2019}-3$. Except for the common factor $-2019^{2019}$ they have distinct factorizations, so the quadratic subfields of $mathbb{Q}(alpha)$ and the quadratic subfields of $mathbb{Q}(beta)$ are different things and we are sure that $betanotinmathbb{Q}(alpha)$.
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:28
1
1
$begingroup$
Interesting! How come you need there to be infinitely many such primes?
$endgroup$
– mathprincess
Jan 27 at 2:33
$begingroup$
Interesting! How come you need there to be infinitely many such primes?
$endgroup$
– mathprincess
Jan 27 at 2:33
|
show 1 more comment
$begingroup$
Let $alpha=2^{1/2019}$, $beta=3^{1/2019}$, $zeta=exp(2pi i/2019)$, $K=mathbb{Q}(alpha)$, and $L=mathbb{Q}(zeta,beta)$.
Here, we prove the following statement: $alphanotin L$ and the degree of $alpha$ over $L$ is $2019$.
Then it implies the validity of the second claim of mathprincess since
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} = 1,$$
cannot hold if $alphanotin L$. This also leads to the validity of the first claim and a full solution to this question.
It is known that $[L:mathbb{Q}]=2019phi(2019)$ (see Jacobson, Velez cited in this answer: Computing the Galois group of polynomials $x^n-a in mathbb{Q}[x]$). Thus, the basis of $mathbb{Q}(zeta)$ over $mathbb{Q}$ are also a basis of $L$ over $mathbb{Q}(beta)$. Then
$$
textrm{disc}_{mathbb{Q}(beta)}L=textrm{disc}_{mathbb{Q}}{mathbb{Q}(zeta)} biggvert (2019)^{phi(2019)}.
$$
It is well-known that a rational prime $p$ does not divide the discriminant $textrm{disc}_{mathbb{Q}}L$ if and only if $p$ is unramified in $L$. Then we have by (Exrcise 23b of Marcus 'Number Fields'),
$$begin{align}
textrm{disc}_{mathbb{Q}}L &biggvert (textrm{disc}_{mathbb{Q}}{mathbb{Q}(beta)})^{phi(2019)} N_{mathbb{Q}}^{mathbb{Q}(beta)}textrm{disc}_{mathbb{Q(beta)}}L\
&biggvert (3^{2018}cdot 2019^{2019})^{phi(2019)}((2019)^{phi(2019)})^{2019}.end{align}
$$
This gives $2nmid textrm{disc}_{mathbb{Q}}L$.
Thus, the rational prime $2$ is unramified in $L$. Now, take any prime $mathcal{P}subseteq L$ lying above $2$. Then $mathcal{P}$ divides any coefficient of $x^{2019}-2$ except for the leading coefficient, and $mathcal{P}^2$ does not divide the constant coefficient $2$ since $2$ is unramified in $L$. Hence, by Eisenstein, $x^{2019}-2$ over $L$ is irreducible. This shows that the degree of $alpha$ over $L$ must be $2019$. Moreover, the same argument also yields that the degree of $alpha$ over $mathbb{Q}(beta)$ is $2019$.
As in mathprincess's answer, we may proceed with $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$. Then
$$begin{align}
[mathbb{Q}(alpha+beta):mathbb{Q}]&=[mathbb{Q}(alpha+beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]\
&=[mathbb{Q}(alpha,beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]
=2019^2.end{align}
$$
answered Jan 27 at 17:05
i707107i707107
12.5k21647
$endgroup$
add a comment |
$begingroup$
Let $alpha=2^{1/2019}$, $beta=3^{1/2019}$, $zeta=exp(2pi i/2019)$, $K=mathbb{Q}(alpha)$, and $L=mathbb{Q}(zeta,beta)$.
Here, we prove the following statement: $alphanotin L$ and the degree of $alpha$ over $L$ is $2019$.
Then it implies the validity of the second claim of mathprincess since
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} = 1,$$
cannot hold if $alphanotin L$. This also leads to the validity of the first claim and a full solution to this question.
It is known that $[L:mathbb{Q}]=2019phi(2019)$ (see Jacobson, Velez cited in this answer: Computing the Galois group of polynomials $x^n-a in mathbb{Q}[x]$). Thus, the basis of $mathbb{Q}(zeta)$ over $mathbb{Q}$ are also a basis of $L$ over $mathbb{Q}(beta)$. Then
$$
textrm{disc}_{mathbb{Q}(beta)}L=textrm{disc}_{mathbb{Q}}{mathbb{Q}(zeta)} biggvert (2019)^{phi(2019)}.
$$
It is well-known that a rational prime $p$ does not divide the discriminant $textrm{disc}_{mathbb{Q}}L$ if and only if $p$ is unramified in $L$. Then we have by (Exrcise 23b of Marcus 'Number Fields'),
$$begin{align}
textrm{disc}_{mathbb{Q}}L &biggvert (textrm{disc}_{mathbb{Q}}{mathbb{Q}(beta)})^{phi(2019)} N_{mathbb{Q}}^{mathbb{Q}(beta)}textrm{disc}_{mathbb{Q(beta)}}L\
&biggvert (3^{2018}cdot 2019^{2019})^{phi(2019)}((2019)^{phi(2019)})^{2019}.end{align}
$$
This gives $2nmid textrm{disc}_{mathbb{Q}}L$.
Thus, the rational prime $2$ is unramified in $L$. Now, take any prime $mathcal{P}subseteq L$ lying above $2$. Then $mathcal{P}$ divides any coefficient of $x^{2019}-2$ except for the leading coefficient, and $mathcal{P}^2$ does not divide the constant coefficient $2$ since $2$ is unramified in $L$. Hence, by Eisenstein, $x^{2019}-2$ over $L$ is irreducible. This shows that the degree of $alpha$ over $L$ must be $2019$. Moreover, the same argument also yields that the degree of $alpha$ over $mathbb{Q}(beta)$ is $2019$.
As in mathprincess's answer, we may proceed with $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$. Then
$$begin{align}
[mathbb{Q}(alpha+beta):mathbb{Q}]&=[mathbb{Q}(alpha+beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]\
&=[mathbb{Q}(alpha,beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]
=2019^2.end{align}
$$
answered Jan 27 at 17:05
i707107i707107
12.5k21647
$endgroup$
add a comment |
$begingroup$
Let $alpha=2^{1/2019}$, $beta=3^{1/2019}$, $zeta=exp(2pi i/2019)$, $K=mathbb{Q}(alpha)$, and $L=mathbb{Q}(zeta,beta)$.
Here, we prove the following statement: $alphanotin L$ and the degree of $alpha$ over $L$ is $2019$.
Then it implies the validity of the second claim of mathprincess since
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} = 1,$$
cannot hold if $alphanotin L$. This also leads to the validity of the first claim and a full solution to this question.
It is known that $[L:mathbb{Q}]=2019phi(2019)$ (see Jacobson, Velez cited in this answer: Computing the Galois group of polynomials $x^n-a in mathbb{Q}[x]$). Thus, the basis of $mathbb{Q}(zeta)$ over $mathbb{Q}$ are also a basis of $L$ over $mathbb{Q}(beta)$. Then
$$
textrm{disc}_{mathbb{Q}(beta)}L=textrm{disc}_{mathbb{Q}}{mathbb{Q}(zeta)} biggvert (2019)^{phi(2019)}.
$$
It is well-known that a rational prime $p$ does not divide the discriminant $textrm{disc}_{mathbb{Q}}L$ if and only if $p$ is unramified in $L$. Then we have by (Exrcise 23b of Marcus 'Number Fields'),
$$begin{align}
textrm{disc}_{mathbb{Q}}L &biggvert (textrm{disc}_{mathbb{Q}}{mathbb{Q}(beta)})^{phi(2019)} N_{mathbb{Q}}^{mathbb{Q}(beta)}textrm{disc}_{mathbb{Q(beta)}}L\
&biggvert (3^{2018}cdot 2019^{2019})^{phi(2019)}((2019)^{phi(2019)})^{2019}.end{align}
$$
This gives $2nmid textrm{disc}_{mathbb{Q}}L$.
Thus, the rational prime $2$ is unramified in $L$. Now, take any prime $mathcal{P}subseteq L$ lying above $2$. Then $mathcal{P}$ divides any coefficient of $x^{2019}-2$ except for the leading coefficient, and $mathcal{P}^2$ does not divide the constant coefficient $2$ since $2$ is unramified in $L$. Hence, by Eisenstein, $x^{2019}-2$ over $L$ is irreducible. This shows that the degree of $alpha$ over $L$ must be $2019$. Moreover, the same argument also yields that the degree of $alpha$ over $mathbb{Q}(beta)$ is $2019$.
As in mathprincess's answer, we may proceed with $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$. Then
$$begin{align}
[mathbb{Q}(alpha+beta):mathbb{Q}]&=[mathbb{Q}(alpha+beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]\
&=[mathbb{Q}(alpha,beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]
=2019^2.end{align}
$$
answered Jan 27 at 17:05
i707107i707107
12.5k21647
$endgroup$
Let $alpha=2^{1/2019}$, $beta=3^{1/2019}$, $zeta=exp(2pi i/2019)$, $K=mathbb{Q}(alpha)$, and $L=mathbb{Q}(zeta,beta)$.
Here, we prove the following statement: $alphanotin L$ and the degree of $alpha$ over $L$ is $2019$.
Then it implies the validity of the second claim of mathprincess since
$$frac{(zeta^m - 1)alpha}{(1-zeta^n)beta} = 1,$$
cannot hold if $alphanotin L$. This also leads to the validity of the first claim and a full solution to this question.
It is known that $[L:mathbb{Q}]=2019phi(2019)$ (see Jacobson, Velez cited in this answer: Computing the Galois group of polynomials $x^n-a in mathbb{Q}[x]$). Thus, the basis of $mathbb{Q}(zeta)$ over $mathbb{Q}$ are also a basis of $L$ over $mathbb{Q}(beta)$. Then
$$
textrm{disc}_{mathbb{Q}(beta)}L=textrm{disc}_{mathbb{Q}}{mathbb{Q}(zeta)} biggvert (2019)^{phi(2019)}.
$$
It is well-known that a rational prime $p$ does not divide the discriminant $textrm{disc}_{mathbb{Q}}L$ if and only if $p$ is unramified in $L$. Then we have by (Exrcise 23b of Marcus 'Number Fields'),
$$begin{align}
textrm{disc}_{mathbb{Q}}L &biggvert (textrm{disc}_{mathbb{Q}}{mathbb{Q}(beta)})^{phi(2019)} N_{mathbb{Q}}^{mathbb{Q}(beta)}textrm{disc}_{mathbb{Q(beta)}}L\
&biggvert (3^{2018}cdot 2019^{2019})^{phi(2019)}((2019)^{phi(2019)})^{2019}.end{align}
$$
This gives $2nmid textrm{disc}_{mathbb{Q}}L$.
Thus, the rational prime $2$ is unramified in $L$. Now, take any prime $mathcal{P}subseteq L$ lying above $2$. Then $mathcal{P}$ divides any coefficient of $x^{2019}-2$ except for the leading coefficient, and $mathcal{P}^2$ does not divide the constant coefficient $2$ since $2$ is unramified in $L$. Hence, by Eisenstein, $x^{2019}-2$ over $L$ is irreducible. This shows that the degree of $alpha$ over $L$ must be $2019$. Moreover, the same argument also yields that the degree of $alpha$ over $mathbb{Q}(beta)$ is $2019$.
As in mathprincess's answer, we may proceed with $mathbb{Q}(alpha+beta)=mathbb{Q}(alpha,beta)$. Then
$$begin{align}
[mathbb{Q}(alpha+beta):mathbb{Q}]&=[mathbb{Q}(alpha+beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]\
&=[mathbb{Q}(alpha,beta):mathbb{Q}(beta)][mathbb{Q}(beta):mathbb{Q}]
=2019^2.end{align}
$$
answered Jan 27 at 17:05
i707107i707107
12.5k21647
edited Jan 27 at 22:20
answered Jan 27 at 17:05
i707107i707107
12.5k21647
answered Jan 27 at 17:05
i707107i707107
12.5k21647
answered Jan 27 at 17:05
i707107i707107
12.5k21647
12.5k21647
add a comment |
add a comment |
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$begingroup$
You have a simple algorithm for finding the minimal polynomial of such a number.
$endgroup$
– ÍgjøgnumMeg
Jan 26 at 23:25
2
$begingroup$
@ÍgjøgnumMeg: really? We have a simple algorithm for finding a polynomial (with degree $2019^2$) in $mathbb{Z}[x]$ which vanishes at $a+b$, but what is ensuring the irreducibility of such polynomial?
$endgroup$
– Jack D'Aurizio
Jan 27 at 1:34