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?

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Sep 25, 2016 at 1:45
  • $\begingroup$ I'm still having trouble. Can you give me a hint? $\endgroup$
    – lithium123
    Commented Sep 25, 2016 at 1:52
  • $\begingroup$ Are you allowed any field theory? E.g. field automorphisms? $\endgroup$
    – peter a g
    Commented Sep 25, 2016 at 3:06
  • $\begingroup$ 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 $\Bbb{Q} [\sqrt2]$. $\endgroup$
    – Vik78
    Commented Sep 25, 2016 at 3:09
  • $\begingroup$ Using only algebra is a valid constraint. $\endgroup$
    – jnyan
    Commented Sep 25, 2016 at 3:34

2 Answers 2


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.

  • $\begingroup$ A nice solution, which I suspect was the intended one. +1 $\endgroup$ Commented Sep 25, 2016 at 3:13
  • $\begingroup$ Is it possible to get rid of $\sqrt{6}$ without squaring $\sqrt{6} = \frac{6 + a^{2} - 3b^{2} - 3c}{a + bc}$? $\endgroup$
    – User 1234
    Commented Dec 7, 2016 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.


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