# Proving $\sqrt{6}$ is not part of a field

I'm currently in a beginning analysis course, and I am asked to prove that $F =\{a+b\sqrt2 +c\sqrt3 :a,b,c∈Q\}$ is not a field.

I know that this violates the first multiplication axiom, that if $x,y \in F$ then $xy \in F$. However, I don't know how to prove that $\sqrt6$ cannot be written in the form $a+b\sqrt2 +c\sqrt3,$ where $a,b,c∈Q$. Is there a way to show this using elementary algebra, and not go into field extensions?

• If it was a field element, it would be invertible. Perhaps you can write $\sqrt{6} (a + b \sqrt{2} + c \sqrt{3}) = 1$, and try to find some constaints on the values of $a,b,c$. If these constraints are too restrictive, then there would be no inverse. – Alfred Yerger Sep 25 '16 at 1:45
• I'm still having trouble. Can you give me a hint? – lithium123 Sep 25 '16 at 1:52
• Are you allowed any field theory? E.g. field automorphisms? – peter a g Sep 25 '16 at 3:06
• Basic field theory shows that the smallest field containing $\sqrt3$, $\sqrt2$ and the rationals has dimension 4 as a vector space over the rationals. Do you know any field theory: the tower theorem, for example? If you do it suffices to show that $\sqrt3$ is not in $\Bbb{Q} [\sqrt2]$. – Vik78 Sep 25 '16 at 3:09
• Using only algebra is a valid constraint. – jnyan Sep 25 '16 at 3:34

Suppose $\sqrt{6}=a+b\sqrt{2}+c\sqrt{3}$ with $a,b,c$ rational. Then also $$\sqrt{6}-a=b\sqrt{2}+c\sqrt{3}$$ and when you square both sides of this, the only surd around will be $\sqrt{6}$, which makes your life a lot easier. You should be able to manipulate the result to show that $$(b^2-3)(c^2-2)=0$$ which contradicts the original rationality assumption.
• Is it possible to get rid of $\sqrt{6}$ without squaring $\sqrt{6} = \frac{6 + a^{2} - 3b^{2} - 3c}{a + bc}$? – User 1234 Dec 7 '16 at 5:12
Assume $a+b\sqrt 2 + c\sqrt 3=\sqrt 6, a,b,c\in \Bbb Q$. Then $\sqrt 6-b \sqrt 2=a+c\sqrt 3.$ Square both sides and you are down to two radicals. Isolate them and square again and you have one.