Finding the minimal polynomial over Q

[PsychonautQQ]

Homework Statement

Find the minimal polynomial of a = i*(2)^1/2 + (3)^1/2

Homework Equations

The Attempt at a Solution

Well, I know the minimal polynomial will have degree four, and that's about it. Will it help if I look at the linear factors of the minimal polynomial in some splitting field and multiply them all together?

Could i take a general monic quartic polynomial, plug in a, and solve for the coefficients?

any insight would help, I'm pretty lost.

 

[Ray Vickson]

Homework Statement

Find the minimal polynomial of a = i*(2)^1/2 + (3)^1/2

Homework Equations

The Attempt at a Solution

Well, I know the minimal polynomial will have degree four, and that's about it. Will it help if I look at the linear factors of the minimal polynomial in some splitting field and multiply them all together?

Could i take a general monic quartic polynomial, plug in a, and solve for the coefficients?

any insight would help, I'm pretty lost.

If $r = \sqrt{3} + \sqrt{2}\, i$, then $r^2 = 1+2 \sqrt{6}\, i$, $r^3 = -3 \sqrt{3} + 7 \sqrt{2}\, i$, etc.

There is no real rational polynomial of the form $p_1 = a_0 + a_1 r$ that is zero, because (i) the real and imaginary parts do not cancel out; and (ii) $\sqrt{2}$ and $\sqrt{3}$ are irrational numbers, and neither is a rational multiple of the other.

Next, try $p_2 = a_0 + a_1 r + a_2 r^2$ with rational $a_0,a_1,a_2$. We have
$$p_2 = a_0 + \sqrt{3} a_1 + a_2 + i \left(\sqrt{2} a_1 + 2 \sqrt{6} a_2 \right) $$
There are no rational $a_0,a_1,a_2$ that can make $p_2$ vanish.

Now go on to $p_3 = a_0 + a_1 r + a_2 r^2 + a_3 r^3$ and see if it can possibly equal zero. If so, you have a minimal polynomial for $r$. If not, go to the next possibility $p_4 = p_3 + a_4 r^3$, etc.

 

[fresh_42]

Homework Statement

Find the minimal polynomial of a = i*(2)^1/2 + (3)^1/2

Homework Equations

The Attempt at a Solution

Well, I know the minimal polynomial will have degree four, and that's about it. Will it help if I look at the linear factors of the minimal polynomial in some splitting field and multiply them all together?

If you manage to guess other roots, yes. Hint: consider conjugates.

Could i take a general monic quartic polynomial, plug in a, and solve for the coefficients?

any insight would help, I'm pretty lost.

Yes. But I would try to avoid this. Too many chances for mistakes.

How do you know, that the degree has to be four? Or to put it another way: how do you guarantee, that the polynomial you will have found, is irreducible?

 

[PsychonautQQ]

If you manage to guess other roots, yes. Hint: consider conjugates.

Yes. But I would try to avoid this. Too many chances for mistakes.

How do you know, that the degree has to be four? Or to put it another way: how do you guarantee, that the polynomial you will have found, is irreducible?

I suppose I will know it's irreducible by Eisenstein Criterion? I know it will have degree four because [Q((i*2^1/2),(3^1/2) : Q] = 4 and the square root of 3 and the square root of 2 multiplied by i each bring new elements to the extension so it won't be 2.

 

[PsychonautQQ]

If $r = \sqrt{3} + \sqrt{2}\, i$, then $r^2 = 1+2 \sqrt{6}\, i$, $r^3 = -3 \sqrt{3} + 7 \sqrt{2}\, i$, etc.

There is no real rational polynomial of the form $p_1 = a_0 + a_1 r$ that is zero, because (i) the real and imaginary parts do not cancel out; and (ii) $\sqrt{2}$ and $\sqrt{3}$ are irrational numbers, and neither is a rational multiple of the other.

Next, try $p_2 = a_0 + a_1 r + a_2 r^2$ with rational $a_0,a_1,a_2$. We have
$$p_2 = a_0 + \sqrt{3} a_1 + a_2 + i \left(\sqrt{2} a_1 + 2 \sqrt{6} a_2 \right) $$
There are no rational $a_0,a_1,a_2$ that can make $p_2$ vanish.

Now go on to $p_3 = a_0 + a_1 r + a_2 r^2 + a_3 r^3$ and see if it can possibly equal zero. If so, you have a minimal polynomial for $r$. If not, go to the next possibility $p_4 = p_3 + a_4 r^3$, etc.

won't it be near impossible to find the rational coefficients on the fourth polynomial that make it vanish? Should I use the general quartic formula in reverse somehow because I will know the roots? I believe on exists, does it not?

 

[Ray Vickson]

won't it be near impossible to find the rational coefficients on the fourth polynomial that make it vanish? Should I use the general quartic formula in reverse somehow because I will know the roots? I believe on exists, does it not?

It is not "near impossible"; it is actually very easy but algebraically messy. First, the real and imaginary parts must both vanish. Next, the coefficients of $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{6}$ must vanish. That gives you a set of easy linear equations.

I have actually done it, and it works. However, in my old age I have gotten lazy and so let Maple do all the heavy algebraic lifting; but if you are young and energetic you can readily do it all by hand.

 

[fresh_42]

You have $\sqrt{3}+i\sqrt{2}$. Remember that the minimal polynomial of $i$ is $(x^2+1)=(x-i)(x+i)$, i.e. the conjugate of $i$, which is $-i$, is also a root. So if this works for $i$, why shouldn't it work for $i\sqrt{2}$ and $\sqrt{3}\;$?
At the same time, you can learn here, how the automorphism group of a field extension works, i.e. which automorphisms there are.

 

[PsychonautQQ]

It is not "near impossible"; it is actually very easy but algebraically messy. First, the real and imaginary parts must both vanish. Next, the coefficients of $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{6}$ must vanish. That gives you a set of easy linear equations.

I have actually done it, and it works. However, in my old age I have gotten lazy and so let Maple do all the heavy algebraic lifting; but if you are young and energetic you can readily do it all by hand.

Okay, thanks for the encouragement! I'll get my youthful vigor started on this right away!

 

[PsychonautQQ]

You have $\sqrt{3}+i\sqrt{2}$. Remember that the minimal polynomial of $i$ is $(x^2+1)=(x-i)(x+i)$, i.e. the conjugate of $i$, which is $-i$, is also a root. So if this works for $i$, why shouldn't it work for $i\sqrt{2}$ and $\sqrt{3}\;$?
At the same time, you can learn here, how the automorphism group of a field extension works, i.e. which automorphisms there are.

Hm, I see what you are saying. So i know complex conjugation will be one of the mappings in the automorphism group, and I think the other automorphism will flip the sign of (3)^1/2,

So, the automorphism group is isomorphic to Z2 x Z2. If I apply each of these automorphism's to (a) then I will get four different complex numbers, varying only by flipping the signs in different places. Would multiplying these all together give me some important information?

 

[fresh_42]

Hm, I see what you are saying. So i know complex conjugation will be one of the mappings in the automorphism group, and I think the other automorphism will flip the sign of (3)^1/2,

So, the automorphism group is isomorphic to Z2 x Z2. If I apply each of these automorphism's to (a) then I will get four different complex numbers, varying only by flipping the signs in different places. Would multiplying these all together give me some important information?

Why don't you multiply $(x-a_i)$ instead, say $[(x-\sqrt{3})(x+\sqrt{3})][(x-i\sqrt{2})(x+i\sqrt{2})]$?

 

[PsychonautQQ]

Why don't you multiply $(x-a_i)$ instead, say $[(x-\sqrt{3})(x+\sqrt{3})][(x-i\sqrt{2})(x+i\sqrt{2})]$?

But I need to find the minimal polynomial over Q? Well i bet when I do the multiplication the imaginary parts will all cancel each other out.

When you say multiply the (x-a_i) together, but then you have each a_i term being a single part of the expression, ie (+/-)i*2^(1/2) or (+/-)3^(1/2). Woudln't each (x-a_i) be of the form (x(+/-)i*2^(1/2)(+/-)3^(1/2)). Like, the automorphism wouldn't make the real or imaginary part go away, would it? Sorry this is so messy, you're a boss.

 

[fresh_42]

But I need to find the minimal polynomial over Q? Well i bet when I do the multiplication the imaginary parts will all cancel each other out.

When you say multiply the (x-a_i) together, but then you have each a_i term being a single part of the expression, ie (+/-)i*2^(1/2) or (+/-)3^(1/2). Woudln't each (x-a_i) be of the form (x(+/-)i*2^(1/2)(+/-)3^(1/2)). Like, the automorphism wouldn't make the real or imaginary part go away, would it? Sorry this is so messy, you're a boss.

Try to think less complicated. You are looking for a polynomial with rational coefficients, i.e. a $p(x) \in \mathbb{Q}[x]$.
$a_1=\sqrt{3}$ has to be one root of $p(x)$ which means $p(a_1)=0$, resp. $(x-a_1)$ divides $p(x)$ in $\mathbb{Q}(a_1)[x]=\mathbb{Q}(\sqrt{3})[x]$.

Therefore $p(x) = (x-a_1) \cdot q(x) = (x -\sqrt{3}) \cdot q(x)$ has to hold for another polynomial $q(x)$, that has all the rest of the $a_i$ as zeros. This means $q(x)=(x-a_2)(x-a_3)(x-a_4)$ because you already know that the degree is four by your observation that $\sqrt{3}$ and $i\sqrt{2}$ have to both contribute to the minimal polynomial. In which field $q(x)$ has its coefficients isn't important here. (But you can find out, if you consider where $p(x)$ and $(x-a_1)=(x-\sqrt{3})$ are from.)
So in total we get $p(x)=(x -\sqrt{3}) \cdot (x-a_2)\cdot(x-a_3)\cdot(x-a_4)$.

Now the binomial formula $(A^2-B^2)=(A-B)(A+B)$ is a good method to deal with square roots, e.g. in $(x^2+1)=(x-i)(x+i)$ where the two roots $i$ and $-i$ become $+1$ in the polynomial of degree two. The analog here is $+\sqrt{3} \, , \, -\sqrt{3}\, , \, +i\sqrt{2} \, , \, -i\sqrt{2}$ as candidates for the $a_i$. I simply took them and suggested to compute $(x-a_1)\cdot(x-a_2)\cdot(x-a_3)\cdot(x-a_4)$ and you have all required roots in there. The binomial formula guarantees that the square roots vanish, the square root $i = \sqrt{-1}$ included.

 

[PsychonautQQ]

Try to think less complicated. You are looking for a polynomial with rational coefficients, i.e. a $p(x) \in \mathbb{Q}[x]$.
$a_1=\sqrt{3}$ has to be one root of $p(x)$ which means $p(a_1)=0$, resp. $(x-a_1)$ divides $p(x)$ in $\mathbb{Q}(a_1)[x]=\mathbb{Q}(\sqrt{3})[x]$.

Therefore $p(x) = (x-a_1) \cdot q(x) = (x -\sqrt{3}) \cdot q(x)$ has to hold for another polynomial $q(x)$, that has all the rest of the $a_i$ as zeros. This means $q(x)=(x-a_2)(x-a_3)(x-a_4)$ because you already know that the degree is four by your observation that $\sqrt{3}$ and $i\sqrt{2}$ have to both contribute to the minimal polynomial. In which field $q(x)$ has its coefficients isn't important here. (But you can find out, if you consider where $p(x)$ and $(x-a_1)=(x-\sqrt{3})$ are from.)
So in total we get $p(x)=(x -\sqrt{3}) \cdot (x-a_2)\cdot(x-a_3)\cdot(x-a_4)$.

Now the binomial formula $(A^2-B^2)=(A-B)(A+B)$ is a good method to deal with square roots, e.g. in $(x^2+1)=(x-i)(x+i)$ where the two roots $i$ and $-i$ become $+1$ in the polynomial of degree two. The analog here is $+\sqrt{3} \, , \, -\sqrt{3}\, , \, +i\sqrt{2} \, , \, -i\sqrt{2}$ as candidates for the $a_i$. I simply took them and suggested to compute $(x-a_1)\cdot(x-a_2)\cdot(x-a_3)\cdot(x-a_4)$ and you have all required roots in there. The binomial formula guarantees that the square roots vanish, the square root $i = \sqrt{-1}$ included.

Reading this made me smarter, God bless

 

[PsychonautQQ]

Try to think less complicated. You are looking for a polynomial with rational coefficients, i.e. a $p(x) \in \mathbb{Q}[x]$.
$a_1=\sqrt{3}$ has to be one root of $p(x)$ which means $p(a_1)=0$, resp. $(x-a_1)$ divides $p(x)$ in $\mathbb{Q}(a_1)[x]=\mathbb{Q}(\sqrt{3})[x]$.

How do I know that 3^(1/2) has to be a root? I mean, obviously it kind of makes sense, but can you explain it more clearly for me?

 

[fresh_42]

How do I know that 3^(1/2) has to be a root? I mean, obviously it kind of makes sense, but can you explain it more clearly for me?

I don't know it either.
I took your statement

Well, I know the minimal polynomial will have degree four,

from the OP as an assumption. However, it's not so difficult.

We have to adjoint $a=\sqrt{3}+i\sqrt{2}$. So the task is to show, that $i\sqrt{2}\, , \, \sqrt{3} \in \mathbb{Q}(a)$.
If this can be shown, then $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{3}+i\sqrt{2}) \subseteq \mathbb{Q}(\sqrt{3}\, , \, i\sqrt{2}) \subseteq \mathbb{Q}(\sqrt{3}+i\sqrt{2})= \mathbb{Q}(a)$ and equalities hold everywhere plus our field extension is of degree four. (*)

I've done it for $\sqrt{3}$ by playing with the equations. There might be a shorter way than the one I've found, but it will do.
We may assume, that we have $a$ available because of (*). (This is because with the suggested minimal poynomial in post #10 or #12, we would get $\sqrt{3}$ and $i \sqrt{2}$ and (*) shows that this is equivalent to adjoining $a$.)

To show (*) I calculated $a^2=1+2i\sqrt{6}$ and $a \cdot \sqrt{3} = 3 +i\sqrt{6}$ which allows me to eliminate $\sqrt{6}$ and get an algebraic expression of $\sqrt{3}$ with coefficients in $\mathbb{Q}(a)$. (And with that, $i\sqrt{2} = a - \sqrt{3} \in \mathbb{Q}(a)$ as well.)

 

[Ray Vickson]

How do I know that 3^(1/2) has to be a root? I mean, obviously it kind of makes sense, but can you explain it more clearly for me?

I get that the minimal polynomial has roots $\sqrt{3} \pm \sqrt{2}\,i$ and $-\sqrt{3} \pm \sqrt{2} \, i$. Here is what I did: first, look at $r = \sqrt{3} + \sqrt{2} \, i$, $r^2 = 1 + 2 \sqrt{6} \, i$, $r^3 = -3 \sqrt{3} + 7 \sqrt{2} \, i$ and $r^4 = -23 + 4 \sqrt{6} \, i$.

The minimal (monic) polynomial of $r$ is not of degree 1, so let's try degree 2: $p_2(r) = r^2 + a_1 r + a_0$. We have
$$p_2(r) = 1+a_0 + \sqrt{3} a_1 + \left( 2 \sqrt{6} + \sqrt{2} a_1 \right) \, i$$.
The condition $p_2(r) = 0$ requires $1+a_0 + \sqrt{3} a_1 = 0$ and $2 \sqrt{6} + \sqrt{2} a_1 = 0$. There is no rational solution for $(a_0,a_1)$, so the minimal polynomial cannot have degree 2.

Let's try $p_3(r) = r^3 + a_2 r^2 + a_1 r + a_0$. We have
$$p_3(r) = a_0+a_2 + (a_1-3) \sqrt{3} + \left( (a_1+7) \sqrt{2} + 2 \sqrt{6} a_2 \right).$$
Equating the real part to 0 implies that $a_0 + a_2 +(a_1-3) \sqrt{2} = 0$, so rational $a_j$ must satisfy $a_1 = 3$ and $a_0 + a_2 = 0$. Equating the imaginary part to 0 implies that $(a_1+7) \sqrt{2} + 2 \sqrt{6} a_2 = 0$, or $a_1+7 + 2 \sqrt{3} a_2 = 0$. This has no rational solutions, so the minimal polynomial cannot have degree 3.

Now, if we try $p_4(r) = r^4 + a_3 r^3 a_2 r^2 + a_1 r + a_0$, and equate the real and imaginary parts to zero, and further use the facts that $\sqrt{2}$ and $\sqrt{3}$ are irrational, we get four linear equations in the four unknown $a_k$. Those equations have a unique (integer-valued) solution. Thus, the minimal polynomial of $r$ has degree 4, and we can actually write it down. Then we can find its other roots. (Of course, the conjugate $\sqrt{3} - \sqrt{2} \, i$ must be a root, so we need only worry about finding the others.)

Admittedly, this proof is pretty simple-minded, using no sophisticated concepts and theorems, but at least it works and is readily understandable. I make no claims that it can be generalized.

 

[PsychonautQQ]

I get that the minimal polynomial has roots $\sqrt{3} \pm \sqrt{2}\,i$ and $-\sqrt{3} \pm \sqrt{2} \, i$. Here is what I did: first, look at $r = \sqrt{3} + \sqrt{2} \, i$, $r^2 = 1 + 2 \sqrt{6} \, i$, $r^3 = -3 \sqrt{3} + 7 \sqrt{2} \, i$ and $r^4 = -23 + 4 \sqrt{6} \, i$.

The minimal (monic) polynomial of $r$ is not of degree 1, so let's try degree 2: $p_2(r) = r^2 + a_1 r + a_0$. We have
$$p_2(r) = 1+a_0 + \sqrt{3} a_1 + \left( 2 \sqrt{6} + \sqrt{2} a_1 \right) \, i$$.
The condition $p_2(r) = 0$ requires $1+a_0 + \sqrt{3} a_1 = 0$ and $2 \sqrt{6} + \sqrt{2} a_1 = 0$. There is no rational solution for $(a_0,a_1)$, so the minimal polynomial cannot have degree 2.

Let's try $p_3(r) = r^3 + a_2 r^2 + a_1 r + a_0$. We have
$$p_3(r) = a_0+a_2 + (a_1-3) \sqrt{3} + \left( (a_1+7) \sqrt{2} + 2 \sqrt{6} a_2 \right).$$
Equating the real part to 0 implies that $a_0 + a_2 +(a_1-3) \sqrt{2} = 0$, so rational $a_j$ must satisfy $a_1 = 3$ and $a_0 + a_2 = 0$. Equating the imaginary part to 0 implies that $(a_1+7) \sqrt{2} + 2 \sqrt{6} a_2 = 0$, or $a_1+7 + 2 \sqrt{3} a_2 = 0$. This has no rational solutions, so the minimal polynomial cannot have degree 3.

Now, if we try $p_4(r) = r^4 + a_3 r^3 a_2 r^2 + a_1 r + a_0$, and equate the real and imaginary parts to zero, and further use the facts that $\sqrt{2}$ and $\sqrt{3}$ are irrational, we get four linear equations in the four unknown $a_k$. Those equations have a unique (integer-valued) solution. Thus, the minimal polynomial of $r$ has degree 4, and we can actually write it down. Then we can find its other roots. (Of course, the conjugate $\sqrt{3} - \sqrt{2} \, i$ must be a root, so we need only worry about finding the others.)

Admittedly, this proof is pretty simple-minded, using no sophisticated concepts and theorems, but at least it works and is readily understandable. I make no claims that it can be generalized.

And I can find the minimal polynomial by plugging my complex value into r, r^2, r^3 and r^4 and then solving for the coefficients? And then to find the other roots I first divide by (3)^1/2 + i(2)^1/2 and it's conjugate and go from there?

 

[Ray Vickson]

And I can find the minimal polynomial by plugging my complex value into r, r^2, r^3 and r^4 and then solving for the coefficients? And then to find the other roots I first divide by (3)^1/2 + i(2)^1/2 and it's conjugate and go from there?

Yes to both, but remember: you need to use the fact that if you have an equation of the form $A + B \sqrt{2} = 0$ with rational $A,B$, then you must have $A=0$ and $B=0$. Ditto for equations like $A + B \sqrt{3} = 0$.

 

[PsychonautQQ]

Yes to both, but remember: you need to use the fact that if you have an equation of the form $A + B \sqrt{2} = 0$ with rational $A,B$, then you must have $A=0$ and $B=0$. Ditto for equations like $A + B \sqrt{3} = 0$.

Okay. These extra conditions are results of linear independence between basis vectors of the field extension, correct? And you are telling me these relations because they have valuable information about coefficients that I will need to know when solving for my coefficients of the 4rth degree polynomial (although it will be 2 degree's once I divide by the complex number and it's conjugate to get two of the roots), correct?
Edit: i got the solution that the minimal polynomial must be x^4-2x^2+25

 

[PsychonautQQ]

Yes to both, but remember: you need to use the fact that if you have an equation of the form $A + B \sqrt{2} = 0$ with rational $A,B$, then you must have $A=0$ and $B=0$. Ditto for equations like $A + B \sqrt{3} = 0$.

I'm running into a slight difficulty. I plugged in the degree's of r into the fourth degree polynomial and began to separate terms to find the linear equations that I would solve for. I grouped all the terms without and imaginary number or irrational number, then I made a group of all the terms with a (3)^1/2, then I ran into a problem. The final group of coefficients all are being multiplied by an imaginary number, which is okay, but then some of them are also being multiplied by (6)^1/2 whilst others are being multiplied by (2)^1/2. I guess this shouldn't surprise me because the polynomial was of degree four so I should have enough information to solve for 4 coefficients, but I actually have more equations than necessary because I assumed the polynomial to be monic. Is this thinking correct?

 

[Ray Vickson]

Okay. These extra conditions are results of linear independence between basis vectors of the field extension, correct? And you are telling me these relations because they have valuable information about coefficients that I will need to know when solving for my coefficients of the 4rth degree polynomial (although it will be 2 degree's once I divide by the complex number and it's conjugate to get two of the roots), correct?
Edit: i got the solution that the minimal polynomial must be x^4-2x^2+25

I get that polynomial as well.

The linear independence of square roots is not needed in this problem, but may very well be needed in other examples. All we need in THIS problem is the fact that $\sqrt{2}$ and $\sqrt{3}$ are irrational numbers. One of the equations has the form $A \sqrt{2} + B \sqrt{6} = 0$, but we can factor out $\sqrt{2}$ to get $A + B \sqrt{3} = 0$.

In more general cases we might need to use the somewhat nontrivial theorem which states: if $0 < n_1 < n_2 < \cdots < n_k$ are squarefree integers, then $\sqrt{n_1}, \sqrt{n_2}, \ldots, \sqrt{n_k}$ are linealy independent over $\mathbb{Q}$. See "A Question on linear independence of square roots", M. Klazar (2009), available in http://kam.mff.cuni.cz/~klazar/tomek_pr.pdf .

So, for example: 2, 3 and 6 are squarefree, so their square roots are linearly independent over $\mathbb{Q}$. We did not need the full power of the theorem in this problem.

 

[SrayD]

Hey, PsychonautQQ; side question are you fan of the game Psychonauts? Been longtime since I played, your name reminded me.

Anyway, saw your post. It brought back memories from abstract algebra class & all the confusion/over-working with finding minimal polys over a ring and/or field.
The method I used is really simple, and similar to Ray Vickson's post#2. Let r = 3^(1/2) + i*2^(1/2); as he showed r^2 = 1 + 2i*6^(1/2); but consider this instead of any higher powers: (r^2 - 1) = 2i*6^(1/2) then look at squaring both sides, the 'i' & root are gone, then solve for r to get poly in 'r'. It will be the monic irreducible over Q that you need. This method works in general, have to raise to powers & also re-arrange to separate terms that are in the field from those in extension field.

Hope this helps with any other problems of this type.

 

[fresh_42]

Why so complicated? Simply show $\mathbb{Q}(\sqrt{3}+i\sqrt{2}) = \mathbb{Q}(\sqrt{3}\, ,\,i\sqrt{2})$ and the minimal polynomial drops out for free by using either directly $A^2-B^2 = (A-B)(A+B)$ or the argument with the conjugates.
And the equation of the fields is simply shown by $(\sqrt{3}+i\sqrt{2})^2 = 1 - 2i\sqrt{6}$ and $(\sqrt{3}+i\sqrt{2})\,\cdot\,\sqrt{3}=3+i\sqrt{6}$. Done.

 

[PsychonautQQ]

Hey, PsychonautQQ; side question are you fan of the game Psychonauts? Been longtime since I played, your name reminded me.

Anyway, saw your post. It brought back memories from abstract algebra class & all the confusion/over-working with finding minimal polys over a ring and/or field.
The method I used is really simple, and similar to Ray Vickson's post#2. Let r = 3^(1/2) + i*2^(1/2); as he showed r^2 = 1 + 2i*6^(1/2); but consider this instead of any higher powers: (r^2 - 1) = 2i*6^(1/2) then look at squaring both sides, the 'i' & root are gone, then solve for r to get poly in 'r'. It will be the monic irreducible over Q that you need. This method works in general, have to raise to powers & also re-arrange to separate terms that are in the field from those in extension field.

Hope this helps with any other problems of this type.

Hey man, thanks for the response. I actually have not played psychonauts, but I get asked that question all the time (PsychonautQQ is my gamertag for every online game I play :P)... Anyway, thanks for the insight on better ways to find the minimal polynomial of an element.

That being said, you ahve to re-arrange to separate terms that are in the field from those in the extension field, when you had r^2 - 1 = 2i(6)^1/2 and then squared it are you saying that this wouldn't work if you hadn't separated the terms first? Interesting.

I'm actually working on a similar problem right now, I have to find the minimal polynomial of r = c(2)^1/4, where c is a primitive third root of unity. So I know the minimal polynomial over degree will have degree 12 already, I can even take successive powers of r to get all 12 roots. Is multiplying (x-r^1)(x-r^2)*....*(x-r^12) the fastest way to get the minimal polynomial here in this case do you think?

 

[SrayD]

PysychonautQQ is a pretty cool gamer tag. Almost did not see this before I left work; no internet connection available where I live; not yet anyway.
But, towards your question. Yeah it does not always work so well, but it's good to check if method will work since it goes pretty quick.
Gonna try using formatting given here, not how to check if I did it right.
If I'm reading it right your trying to get min poly for r = c*2⁴ where c = (-1 + i*3^1/2)/2 or (-1 - i*3^1/2)/2?
Then look at r⁴ = (c*2^1/4)⁴ = c⁴*2, expand c⁴ and separate terms in your field from those that are not; should work like last one from this point. Will illustrate example from the last problem: from (r^2 - 1) = 2i*6^(1/2) square both sides gives (r^2 -1)^2 = -24, then expand & simplify.
Have to sign off till tomorrow, hope this helps.

 

[PsychonautQQ]

PysychonautQQ is a pretty cool gamer tag. Almost did not see this before I left work; no internet connection available where I live; not yet anyway.
But, towards your question. Yeah it does not always work so well, but it's good to check if method will work since it goes pretty quick.
Gonna try using formatting given here, not how to check if I did it right.
If I'm reading it right your trying to get min poly for r = c*2⁴ where c = (-1 + i*3^1/2)/2 or (-1 - i*3^1/2)/2?
Then look at r⁴ = (c*2^1/4)⁴ = c⁴*2, expand c⁴ and separate terms in your field from those that are not; should work like last one from this point. Will illustrate example from the last problem: from (r^2 - 1) = 2i*6^(1/2) square both sides gives (r^2 -1)^2 = -24, then expand & simplify.
Have to sign off till tomorrow, hope this helps.

Thanks for all the help. I'm a pretty slow learner and I just want to clarify: After I expand c^4 and separate the terms, I then have to cube both sides so I have a polynomial with r^12, correct?

 

[PsychonautQQ]

PysychonautQQ is a pretty cool gamer tag. Almost did not see this before I left work; no internet connection available where I live; not yet anyway.
But, towards your question. Yeah it does not always work so well, but it's good to check if method will work since it goes pretty quick.
Gonna try using formatting given here, not how to check if I did it right.
If I'm reading it right your trying to get min poly for r = c*2⁴ where c = (-1 + i*3^1/2)/2 or (-1 - i*3^1/2)/2?
Then look at r⁴ = (c*2^1/4)⁴ = c⁴*2, expand c⁴ and separate terms in your field from those that are not; should work like last one from this point. Will illustrate example from the last problem: from (r^2 - 1) = 2i*6^(1/2) square both sides gives (r^2 -1)^2 = -24, then expand & simplify.
Have to sign off till tomorrow, hope this helps.

Also, how do I know which numbers are constructible in q(r) where r is the third primitive root of unit multiplied by the fourth real root of 2?

 

[SrayD]

PsychonautQQ

You may have already figured it out, but;
Don't have to cube expression after expanding c⁴; & it happens that product of degrees of elements just give the max possible degree for poly, it can happen to be less than that when roots span common subspace. Once you have only elements from base field in expression your done, don't necessarily need to try for max degree. So, look at, c⁴ = c³*c remember what c³ is, and see you just need to get rid of 'i' & square root in just one term.

As for your other post about constructibility, I really have to dig deep, been a long-time. Think generally if the extension generated by an element is of degree 2 or a power of 2, that it's constructible. Not sure if any special exceptions when roots of unity are involved, that are not of degree a power of 2; should get another opinion to be sure.