0

I have a trouble. I hope someone can do it.

How to prove that $\left\{a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \mid (a,b,c,d)\in \mathbb{Q}^4\right\}$ is a field?

I can't prove the existence of multiplicative inverse. Thanks.

gt245
  • 31

2 Answers2

2

Hint (for proving the existence of multiplicative inverses):

If $z$ is a non-zero element of this form, show that so are $z^2,z^3$ and $z^4$. The set $\{1,z,z^2,z^3,z^4\}$ is linearly dependent over $\Bbb{Q}$. Why? So you get a non-trivial linear relation between them. How can you take advantage?

Try to mimic the following proof for the fact that multiplicative inverses of numbers of the form $z=a+b\sqrt3\neq0$ are of the same form.

We easily see that $z^2=(a^2+3b^2)+2ac\sqrt3$ is of the same form. So is $1=z^0$. Let $a_0,a_1,a_2$ be unknowns. Consider the equation $$ a_0+a_1z+a_2z^2=0. \qquad(*) $$ By looking at the coefficients of $1$ and $\sqrt3$ we see that if $a_0,a_1,a_2)$ is a solution of the linear system of equations $$ \left\{\begin{array}{cccccc} a_0&+&aa_1&+&(a^2+3b^2)a_2&=&0,\\ &&ba_1&+&(2ac)a_2&=&0, \end{array}\right. $$ then equation $(*)$ also holds. Two linear homogeneous equations, three unknowns, so linear algebra tells us that there is a non-trivial solution $(a_0,a_1,a_2)\in\Bbb{Q}^3$. As $1,z,z^2$ are all non-zero, at least two of the $a_i$:s must be non-zero. If $a_0\neq0$, then $$ a_0+a_1z+a_2z^2=0\implies \frac1z=-\frac{a_1}{a_0}-\frac{a_2}{a_0}z $$ simply by dividing the equation by $a_0z\neq0.$ But here the the right hand side is clearly of the form $a+b\sqrt3$, so in our set (soon to be shown a field). If $a_0=0$, then $a_1\neq0\neq a_2$, and we divide the equation $(a_0+)a_1z+a_2z^2=0$ by $a_1z^2$ instead.

This time you will get four equations and five unknowns. The exact form of the powers $z^2,z^3,z^4$ will not matter as long as you convince yourself of the fact that you get something like $$ z^j=a_j+b_j\sqrt2+c_j\sqrt3+d_j\sqrt6 $$ for all $j=1,2,3,4$ and some rational numbers $a_j,b_j,c_j,d_j$.

Jyrki Lahtonen
  • 133,153
  • I can't prove the existence of multiplicative inverse. – gt245 Sep 19 '13 at 12:12
  • I elaborated a bit by showing how you can do the same in the field ${a+b\sqrt3\mid a,b\in\Bbb{Q}}$. You will get a larger system with more unknowns. Can you manage? There are also other methods for showing that this is a field, but this is one of the more down-to-earth ones I think? – Jyrki Lahtonen Sep 19 '13 at 12:47
0

Hint: $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field by definition. Prove $$ [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4 $$

and show $$ \{1,\sqrt{2},\sqrt{3},\sqrt{6}\} $$

form a basis for the extension.

Belgi
  • 23,150