# Finding the minimal polynomial over Q

1. Oct 16, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
Find the minimal polynomial of a = i*(2)^1/2 + (3)^1/2

2. Relevant equations

3. 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.

2. Oct 16, 2016

### Ray Vickson

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.

3. Oct 16, 2016

### Staff: Mentor

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?

4. Oct 16, 2016

### PsychonautQQ

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.

5. Oct 16, 2016

### PsychonautQQ

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?

6. Oct 16, 2016

### Ray Vickson

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.

7. Oct 16, 2016

### Staff: Mentor

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.

8. Oct 17, 2016

### PsychonautQQ

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

9. Oct 17, 2016

### PsychonautQQ

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?

10. Oct 17, 2016

### Staff: Mentor

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

11. Oct 18, 2016

### PsychonautQQ

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.

12. Oct 18, 2016

### Staff: Mentor

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.

13. Oct 18, 2016

### PsychonautQQ

14. Oct 18, 2016

### PsychonautQQ

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?

15. Oct 18, 2016

### Staff: Mentor

I don't know it either.
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.)

Last edited: Oct 18, 2016
16. Oct 18, 2016

### Ray Vickson

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.

17. Oct 19, 2016

### PsychonautQQ

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?

18. Oct 19, 2016

### Ray Vickson

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 $A + B \sqrt{3} = 0$.

19. Oct 19, 2016

### PsychonautQQ

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

Last edited: Oct 19, 2016
20. Oct 19, 2016

### PsychonautQQ

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?