# On Algebraic Integers

1. Mar 29, 2009

### e(ho0n3

The problem statement, all variables and given/known data
If a is an algebraic integer with a^3 + a + 1 = 0 and b is an algebraic integer with b^2 + b - 3 = 0, prove that both a + b and ab are algebraic integers.

2. Relevant equations
An algebraic number is said to be an algebraic integer if it satisfies an equation of the form x^m + c_{m-1}x^{m-1} + ... + c_0 = 0, where the c's are integers.

3. The attempt at a solution
Since the algebraic numbers form a field, ab and a + b satisfy some polynomial of the form c_mx^m + ... + c_0, where the c's are integers and m <= 6. The problem here is that c_m may not equal 1. I don't know how to get around this. Any tips?

2. Mar 30, 2009

### Hurkyl

Staff Emeritus
You could compute it.

3. Mar 30, 2009

### e(ho0n3

What do you mean compute it? Are you implicitly suggest that c_m will always be 1?

4. Mar 30, 2009

### matt grime

No, he's suggesting that you explicitly compute a polynomial that is monic that has a+b as a root (and one for ab). It is obviously not true that c_m will always be 1, but it is true that c_m can be 1 if you do it properly.

5. Mar 30, 2009

### e(ho0n3

OK. I guess the next question is how do I compute such a thing. Expanding (a+b)^6 + c_5(a+b)^5 + ... + c_0 and then guessing at the possible c's doesn't seem like the smart way to do this.

6. Mar 30, 2009

### Hurkyl

Staff Emeritus
Then use a non-guessing method to solve the equation.

7. Mar 30, 2009

### e(ho0n3

Ah, of course! ... But I don't know of any non-guessing method.

8. Mar 30, 2009

### matt grime

Find a monic poly with sqrt(2)+sqrt(3) as a root.

9. Mar 30, 2009

### e(ho0n3

Why sqrt(2) + sqrt(3), may I ask?

10. Mar 30, 2009

### matt grime

Why not? If you don't understand a more general case try an example. This is the simplest example I could think of.

11. Mar 30, 2009

### e(ho0n3

Good idea. You should have mentioned that in the first place: Let u = (sqrt{2} + sqrt{3})^2 and then solve u^2 + bu + c = 0 to get that x^4 - 10x^2 + 1.

So with my problem, I let u = (alpha + beta)^3 and solve u^2 + bu + c = 0 ... I don't know any integers b and c that will make this work.

12. Mar 31, 2009

### matt grime

Probably because there aren't any. But why would u have to satisfy a quadratic equation? Try thinking about (a+b)^6 and smaller powers of a+b (i.e. not u).

Incidentally, why should *I* have to tell you that if you're struggling with a problem then you should try an easier example? Bye (as in that's the end of my contribution).