Show that Z[root 7] is not isomorphic to Z[root 13].

How can i show this by contradiction?

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- Mar 7th 2010, 07:16 AMBecksIsomorphic
Show that Z[root 7] is not isomorphic to Z[root 13].

How can i show this by contradiction? - Mar 7th 2010, 10:16 AMclic-clac
Hi

Assume there is a ring isomorphism $\displaystyle f:\mathbb{Z}[\sqrt{7}]\rightarrow\mathbb{Z}[\sqrt{13}].$

Using the morphism properties, try to find a contradiction with $\displaystyle f(\sqrt{7}).$ (Remember that, by definition, $\displaystyle f(1)=1$ hence $\displaystyle f(7)=?$ ) - Mar 7th 2010, 10:34 AMBecks
What are the morphism properties?

- Mar 7th 2010, 10:52 AMclic-clac
What appears in its definition, for instance, for any $\displaystyle x,y,\ f(xy)=f(x)f(y)$ and $\displaystyle f(x+y)=f(x)+f(y).$

- Mar 8th 2010, 08:10 AMBecks
Could you start me off please?

- Mar 8th 2010, 11:13 AMclic-clac
We started a proof by contradiction, assuming there was an isomorphism $\displaystyle f:\mathbb{Z}[\sqrt{7}]\rightarrow\mathbb{Z}[\sqrt{13}]$

Well, consider $\displaystyle f(\sqrt{7}),$ you can see it is an element of $\displaystyle \mathbb{Z}[\sqrt{13}]$ whose square is $\displaystyle 7$ (**why?**)

What we have to do now is to prove there is no squared root of $\displaystyle 7$ in $\displaystyle \mathbb{Z}[\sqrt{13}],$ assume we already know that, then it's over because we obtained a contradiction.

To show the missing part, we can also use a proof by contradiction: suppose there is a $\displaystyle x\in\mathbb{Z}[\sqrt{13}]$ such that $\displaystyle x^2=7.$

$\displaystyle x=a+b\sqrt{13}$ for some $\displaystyle a,b\in\mathbb{Z}$ and we get $\displaystyle 7=a^2+13b^2+2ab\sqrt{13}\ ;$ since $\displaystyle 7-a^2+13b^2$ belongs to $\displaystyle \mathbb{Z}$ while $\displaystyle 2ab\sqrt{13}$ is irrational or $\displaystyle 0,$ that means it is $\displaystyle 0,$ i.e. $\displaystyle ab=0.$

$\displaystyle x$ being irrational, we can say $\displaystyle a=0$ and we finally have $\displaystyle x=b\sqrt{13}\ ;$ impossible: consider their squares and conclude.