# Rings and Fields Question

1. Nov 14, 2009

### mikethemike

1. The problem statement, all variables and given/known data:

Hey guys, I'm a new user so my semantics might be difficult to read...

Let A={a+b(square root(2)) ; a,b in Q}

(i)Describe A as a factor ring of Q[x] ( The polynomial Ring)

(ii) Show A is a field

3. The attempt at a solution

(i) Let x be in C (the complex plane) then A= Q[x]/Q[c in C/{root (2)},

I don't think this is correct however, as x is always indeterminate.

(ii) I'm pretty stumped at this. I can't even convince myself that every a of A has multiplicitive inverse...

Any help would be much apprecciated! Thanks

Mike

2. Nov 14, 2009

### rasmhop

a) You're right that your approach as given doesn't work because you simply adjoin $\sqrt{2}$ to Q which is a legal operation, but not a polynomial as you observe.

The key idea here as you observed is that we want to identify x with $\sqrt{2}$. Consider the map $f : Q[x] \to A$:
$$f\left(\sum_{i=0}^n a_ix^i\right) = \sum_{i=0}^n a_i \sqrt{2}^i$$
which you can verify is a homomorphism.
Or a bit more concisely if you have developed the characteristic property of polynomials: Let $f : Q[x] \to A$ be the unique homomorphism sending a rational number r to r and x to $\sqrt{2}$. You can confirm that this is surjective by considering polynomials of degree 0 or 1. Now what does the first isomorphism theorem of rings tell you? If you want a more explicit description you can identify x with $\sqrt{2}$ by considering the ideal $I=(x-\sqrt{2})$ and then confirm that Q[x]/I works (this is also what using the first isomorphism theorem gives you).

b) For a moment let's pretend it's a field to get ideas. That is given $a+b\sqrt{2} \in A$ it has a multiplicative inverse $c+d\sqrt{2}\in A$ such that:
$$(a+b\sqrt{2})(c+d\sqrt{2}) = 1$$
Multiplying these we get:
$$ac+2bd+(bc+ad)\sqrt{2} = 1$$
so we want:
$$ac+2bd=1$$
$$bc+ad = 0$$
Now try to find c,d here in terms of a and b. If you succeed you can use this to get an explicit formula for $(a+b\sqrt{2})^{-1}$ since you know exactly what it must be, and then it's easy to confirm that it works (this is pretty similar to how you showed that the complex numbers have inverses).

EDIT: For b I assumed you had showed that A is a commutative ring and only need the multiplicative inverses. That it's a commutative ring should be pretty easy to confirm, but if you have trouble with any particular aspect of it just post.

3. Nov 15, 2009

### mikethemike

Hey Rasmhop,

That's perfect and very helpful. Thank you!