Field extensions, proving an element is in a field

[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

 

[Tedjn]

Solve for the square root of 2.

 

[Firepanda]

Solve what exactly?

 

[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.

 

[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))

 

[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.

 

[Firepanda]

hmm what's the difference?

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

 

[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.

 

[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}}

 

[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?

 

[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]

 

[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].

 

[Firepanda]

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

Is this point crucial to the solution?

 

[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.