# Field extensions, proving an element is in a field

1. May 18, 2010

### Firepanda

I can do all the question apart from the 'Conclude rt(2) in an element of Q(rt(2) + rt(3))'

That's all I need help on.

I tried

rt(2) = a + b(rt(2) + rt(3))

squaring it and equating you get

2 = a2 + 5b2 + (some other terms of rt(2) rt(3) rt(5))

but this cannot be true for a, b in Q, how else are you supposed to do this?

Thanks

2. May 18, 2010

### Tedjn

Solve for the square root of 2.

3. May 18, 2010

### Firepanda

Solve what exactly?

4. May 18, 2010

### Tedjn

You know that $(\sqrt{2}+\sqrt{3})^2 - 2\sqrt{2}(\sqrt{2}+\sqrt{3}) - 1 = 0$. Here, we're fortunate because there is a $\sqrt{2}$ available that we can isolate.

5. May 18, 2010

### Firepanda

I tried that before but it left me with something pretty messy when all i wanted to show was it could be written as a + b(rt(2) + rt(3))

i ended up with rt(2) = -(2 + rt(5)) / (rt(2) + rt(3))

6. May 18, 2010

### Tedjn

Remember that this is $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, not $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. Also, I don't see from where the $\sqrt{5}$ comes.

7. May 18, 2010

### Firepanda

hmm whats the difference?

guess I was slacking when jotting down notes and didn't realise when they were curvy brackets and not :(

8. May 18, 2010

### Tedjn

The former is the smallest field containing (root 2 + root 3), which means there are inverses. I've slacked before in math classes, so I know the feeling.

9. May 18, 2010

### Firepanda

Then that sort of ruins my answer for part i) of

isn't k = {0, 1, T, 1+T} since k = {a + bT | a, b in {0,1}}?

and so g is irreducible since no element of k is a root

but you're saying k isn't {a + bT | a, b in {0,1}}

10. May 18, 2010

### Tedjn

It turns out that F2(T) contains all quotients of polynomials in F2[T]. Interpret it as the field of fractions of F2[T]. What have you learned that you can apply relating irreducible polynomials over the two?

11. May 18, 2010

### Firepanda

Probably missed your point but ill take a stab

if it's irreducible in Fp[T], p prime, then it's irreducible in Z[X]

12. May 18, 2010

### Tedjn

It can be tough to wrap your head around, but the indeterminate variable is X. So from what you proved we know that our polynomial g(X) is irreducible in (F2[T])[X]. It is not even in Z[X]. However, we would like to say something about its irreducibility in (F2(T))[X]. This might remind you of the connection between irreducibility in Z[X] and Q[X].

13. May 18, 2010

### Firepanda

If it's irreducible in Q[X] then it's irreducible in Z[X] is gauss's lemma, and them im stumped

Is this point crucial to the solution?

14. May 18, 2010

### Tedjn

I think it's the opposite, that irreducibility in Z[X] implies irreducibility in Q[X]. I've also just finished studying these kinds of things, so I can't be sure that this is crucial, but it seems so to me. We can use Gauss's lemma to conclude what you need passing from the ring to the field of fractions.