Show $\phi(a+b\sqrt2) = a-b\sqrt2$ is an isomorphism of $\mathbb{Q(\sqrt2)}$ with itself.
proof: I already showed $\phi[(a_1 + b_1\sqrt2)(a_2 + b_2\sqrt2) ] = \phi(a_1 + b_1\sqrt2)\phi(a_2 + b_2\sqrt2)$ and that
$\phi[(a_1 + b_1\sqrt2)+ (a_2 + b_2\sqrt2) ] = \phi(a_1 + b_1\sqrt2)+ \phi(a_2 + b_2\sqrt2)$
so $\phi$ is a ring homomorphism.
So now I need to show $\phi$ is surjective and injective.
$\phi$ is onto since for every $a+ b\sqrt2 \in \mathbb{Q(\sqrt2)}$ , there is a $a-b\sqrt2 \in\mathbb{Q(\sqrt2)} $ such that $a+b\sqrt2 = \phi(a-b\sqrt2)$.
And $\phi$ is one to one, since by definition $ker\phi = $ {$a+b\sqrt2 \in \mathbb{Q\sqrt2} : \phi(a + b\sqrt2) = 0 $}.
Then this happens when $a, b = 0$. So $ker\phi = $ {$0+0\sqrt2 \in \mathbb{Q\sqrt2} : \phi(0 + 0\sqrt2) = 0 $}.
So since the $ker\phi$ is zero , then $\phi$ is injective.
Thus, $\phi$ is an isomorphism .
Can someone please verify , my approach in showing $\phi$ is surjective and injective function? Some feedback would really help. Thank you!
Best Answer
This looks pretty good! Here are a couple changes I might suggest:
I would rephrase this as:
The reason is that you shouldn't really say "there is a $a-b\sqrt{2}$"; when you say "there is a...", the next thing you say should be a new, undefined variable, not something that already has a defined value. The way you write it is like saying "not all odd numbers are prime because there is a $9$ which is not prime".
In this part, you could give a bit more explanation. Why does this only happen when $a,b=0$?