Isomorphic

**Becks** · March 7th 2010, 07:16 AM

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

How can i show this by contradiction?

**clic-clac** · March 7th 2010, 10:16 AM

Hi

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

Using the morphism properties, try to find a contradiction with $f(\sqrt{7}).$ (Remember that, by definition, $f(1)=1$ hence $f(7)=?$ )

**Becks** · March 7th 2010, 10:34 AM

What are the morphism properties?

**clic-clac** · March 7th 2010, 10:52 AM

What appears in its definition, for instance, for any $x,y,\ f(xy)=f(x)f(y)$ and $f(x+y)=f(x)+f(y).$

**Becks** · March 8th 2010, 08:10 AM

Could you start me off please?

**clic-clac** · March 8th 2010, 11:13 AM

We started a proof by contradiction, assuming there was an isomorphism $f:\mathbb{Z}[\sqrt{7}]\rightarrow\mathbb{Z}[\sqrt{13}]$
Well, consider $f(\sqrt{7}),$ you can see it is an element of $\mathbb{Z}[\sqrt{13}]$ whose square is $7$ (why?)
What we have to do now is to prove there is no squared root of $7$ in $\mathbb{Z}[\sqrt{13}],$ 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 $x\in\mathbb{Z}[\sqrt{13}]$ such that $x^2=7.$
$x=a+b\sqrt{13}$ for some $a,b\in\mathbb{Z}$ and we get $7=a^2+13b^2+2ab\sqrt{13}\ ;$ since $7-a^2+13b^2$ belongs to $\mathbb{Z}$ while $2ab\sqrt{13}$ is irrational or $0,$ that means it is $0,$ i.e. $ab=0.$
$x$ being irrational, we can say $a=0$ and we finally have $x=b\sqrt{13}\ ;$ impossible: consider their squares and conclude.

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