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## Homework Statement

The question : http://gyazo.com/5372336302b5ef289b305172bcd16a2a

## Homework Equations

First Isomorphism theorem.

## The Attempt at a Solution

Define [itex]\phi : \mathbb{Q}[x]/<x^2-2> → Q[ \sqrt{2} ] \space | \space \phi (f(x)) = f( \sqrt{2})[/itex]

So showing phi is a homomorphism is quite easy so I'll skip those details.

My question lies in my argument for phi being a bijection. I COULD show it's 1-1 and onto which would mean it's an isomorphism, but I need practice with the first isomorphism theorem so I'm going to try it and hope I know what's going on.

So, I believe I need to show phi is onto and then argue about the kernel.

To show phi is onto, suppose (a+b√2) is in Q[√2] and f(x) is in Q[x]/<x

^{2}-2> so that f(x) = ax+b.

Now, [itex]\phi (f(x)) = f( \sqrt{2}) = a + b \sqrt{2}[/itex]. Hence for every a+b√2 in Q[√2] there exists f(x) in Q[x]/<x

^{2}-2> such that phi(f(x)) = a+b√2.

Hence phi is onto.

Now consider that ker(phi) = { f(x) in Q[x]/<x

^{2}-2> | f(√2) = 0 }. Since x

^{2}-2 is in ker(phi) and x

^{2}-2 is of the smallest degree, we conclude that ker(phi) = x

^{2}-2 and hence by the first isomorphism theorem, Q[x]/<x

^{2}-2> is isomorphic to Q[√2].

This is my first time trying to apply the theorem rather than doing it the long way of showing 1-1 and onto correspondence.

If anyone could tell me where I may have went wrong or if it looks good, it would be much appreciated.