Finding a polynomial that has solution (root) as the sum of roots

[swampwiz]

AIUI, an algebraic is defined as a number that can be the solution (root) of some integer polynomial, and is any number that can be constructed via any binary arithmetic operation or unary root operation with arguments that are themselves algebraic numbers. I have been able to prove this for almost all cases - with the unproven case being where the number is defined as the addition of a pair of algebraic numbers that happen to have a root operation in each.

Here is an example:

x = a^1/2 + b^1/3

( x - a^1/2 ) = b^1/3

( x - a^1/2 )³ = b

I can't figure out how to continue to de-root a.

 

[fresh_42]

The proof of the statement is easier than showing the polynomial. If $a,b$ are algebraic over $K$ then $b$ is algebraic over $K(a)$ and $a+b\in K(a)(b)=K(a,b)$ is algebraic over $K.$

In order to get the polynomial, write $(x-a)p(x)=0$ and $(x-b)q(x)=0.$ Then $(x-(a+b)/2)p(x)q(x)=0$ and with $(a+b)/2$ algebraic, we have $a+b$ algebraic, too. But you will need the other roots to set up $p(x)$ and $q(x)$. That's the only way I see.

 

[martinbn]

AIUI, an algebraic is defined as a number that can be the solution (root) of some integer polynomial, and is any number that can be constructed via any binary arithmetic operation or unary root operation with arguments that are themselves algebraic numbers. I have been able to prove this for almost all cases - with the unproven case being where the number is defined as the addition of a pair of algebraic numbers that happen to have a root operation in each.

Here is an example:

x = a^1/2 + b^1/3

( x - a^1/2 ) = b^1/3

( x - a^1/2 )³ = b

I can't figure out how to continue to de-root a.

Expand, solve for root a and square.

$x^3-3x^2\sqrt{a}+3ax-a\sqrt{a} = b$

$\sqrt{a}(3x^2+a) = x^3+3ax-b$

$a(3x^2+a)^2 = (x^3+3ax-b)^2$

 

[swampwiz]

I think I was looking at how to do this for the general case of any number of root terms. Only having 2 terms makes it easy to separate, but that can't be done for 3 terms.

x = a^1/2 + b^1/3 + c^1/5

 

[fresh_42]

I think I was looking at how to do this for the general case of any number of root terms.

Why?

 

[swampwiz]

Why?

I'd to prove that any value made of arithmetic & root operations can generate a polynomial that has a solution equal to that value - i.e., this would prove that this type of value is an algebraic number.

 

[fresh_42]

I'd to prove that any value made of arithmetic & root operations can generate a polynomial that has a solution equal to that value - i.e., this would prove that this type of value is an algebraic number.

This makes no sense. You do not need to know a polynomial for the proof. And without a specific example, you cannot calculate one. You ask for one algorithm that works for any field, for any number of algebraic elements. No way.

 

[swampwiz]

This makes no sense. You do not need to know a polynomial for the proof. And without a specific example, you cannot calculate one. You ask for one algorithm that works for any field, for any number of algebraic elements. No way.

If it can be done for any number, then surely there must be some systematic way to do it.

 

[fresh_42]

If it can be done for any number, then surely there must be some systematic way to do it.

What is any number? Heck, what is even a number? Which fields? Separable extensions? Known Galois group, in case it is Galois? Determine your splitting fields, and calculate all roots. Then apply what I have written in post #2 or proceed as @martinbn suggested in post #3. All you have in the end is Viète.

Your question comes down to: Given any algebraic field extension $L/K$ and an element $a\in L.$ Find its minimal polynomial over $K$. This means: calculate all powers of $a$ up to $\dim_K L$ and reassemble them with coefficients in $K$ to get $0.$

That was done in post #3.

 

[martinbn]

If it can be done for any number, then surely there must be some systematic way to do it.

There are nonconstructive proves, there doesn't have to be a systematic way to do it.

On the other hand the first result after a search is this one
https://math.stackexchange.com/ques...product-of-two-algebraic-numbers-is-algebraic

 

[swampwiz]

The proof of the statement is easier than showing the polynomial. If $a,b$ are algebraic over $K$ then $b$ is algebraic over $K(a)$ and $a+b\in K(a)(b)=K(a,b)$ is algebraic over $K.$

In order to get the polynomial, write $(x-a)p(x)=0$ and $(x-b)q(x)=0.$ Then $(x-(a+b)/2)p(x)q(x)=0$ and with $(a+b)/2$ algebraic, we have $a+b$ algebraic, too. But you will need the other roots to set up $p(x)$ and $q(x)$. That's the only way I see.

So what you are saying is:

Because a & b are each algebraic

0 = ( x - a ) = ( x - b )
then for any polynomials p( x ) & q( x )

The proof of the statement is easier than showing the polynomial. If $a,b$ are algebraic over $K$ then $b$ is algebraic over $K(a)$ and $a+b\in K(a)(b)=K(a,b)$ is algebraic over $K.$

In order to get the polynomial, write $(x-a)p(x)=0$ and $(x-b)q(x)=0.$ Then $(x-(a+b)/2)p(x)q(x)=0$ and with $(a+b)/2$ algebraic, we have $a+b$ algebraic, too. But you will need the other roots to set up $p(x)$ and $q(x)$. That's the only way I see.

OK, that's a nice little proof that gets me over the hump!

 

[fresh_42]

So what you are saying is:

Because a & b are each algebraic

0 = ( x - a ) = ( x - b )

?

I mean: because $\alpha,\beta$ are algebraic, $K\subseteq K(\alpha ,\beta )$ is algebraic. And since $\alpha +\beta \in K(\alpha ,\beta )$ is also algebraic. I would do it in steps: $K\subseteq K(\alpha )\subseteq K(\alpha ,\beta )$.

then for any polynomials p( x ) & q( x )

OK, that's a nice little proof that gets me over the hump!

 

[swampwiz]

?

I mean: because $\alpha,\beta$ are algebraic, $K\subseteq K(\alpha ,\beta )$ is algebraic. And since $\alpha +\beta \in K(\alpha ,\beta )$ is also algebraic. I would do it in steps: $K\subseteq K(\alpha )\subseteq K(\alpha ,\beta )$.

I had accidentally posted before finishing the post.

0 = ( x - a ) = ( x - b ) = ( x - a ) p( x ) = ( x - b ) q( x ) = ( x - a ) p( x ) q( x ) = ( x - b ) p( x ) q( x )

Summing half of each

0 = ( x - ( a + b ) / 2 ) p( x ) q( x )

thus ( ( a + b ) / 2 ) and as well ( a + b ) must be algebraic.

 

[swampwiz]

OK, I think I know what I mean by "an algorithm". If I could prove that any addition, multiplication or root operation having algebraic numbers as arguments results in an algebraic number, then I'm done.

I will now prove why a power, root or product of algebraic numbers is an algebraic number.

GIVEN: a, b -> ALGEBRAIC

THUS: ( a + b ) -> ALGEBRAIC

THUS: ∃ [ f( x ) = 0 = ( x - a ) p( x ) ] &

[ g₂( x ) = 0 = ( x + a ) ( x - a ) p( x ) = ( x² - a² ) p( x )

[ g₃( x ) = 0 = ( x² + a x + a² ) ( x - a ) p( x ) = ( x³ - a³ ) p( x )

NOTE: The root operation should be presumed to be a consistent deMoivre root index.

[ g_1/2( x ) = 0 = ( x^1/2 - a^1/2 ) ( x^1/2 + a^1/2 ) p( x ) = ( x^1/2 - a^1/2 ) p_1/2( x )

[ g_1/3( x ) = 0 = ( x^1/3 - a^1/3 ) ( x^2/3 + a^1/3 x^1/3 + a^2/3 ) p( x ) = ( x^1/3 - a^1/3 ) p_1/3( x )

LET: u = xⁿ

THUS: 0 = ( u - aⁿ ) p( x( u ) )

THUS: aⁿ -> ALGEBRAIC

LET: u = x^{( 1 / n )}

THUS: 0 = ( u - a^{( 1 / n )} ) p_1/n( x( u ) )

THUS: a^{( 1 / n )} -> ALGEBRAIC

( a b ) = [ ( a + b )² - ( a² + b² ) ] => SUM OF POWERS OF ALGEBRAICS => SUM OF ALGEBRAICS

THUS: ( a b ) -> ALGEBRAIC

QED