Proving $\Bbb Q[\sqrt 2,\sqrt 3] = \{ a + b\sqrt{2} + c\sqrt 3 + d\sqrt 6\ \!: a,b,c,d\in \Bbb Q\}$

Question

Let ${\mathbb Q}[\sqrt 2, \sqrt 3]$ denote the smallest subring of ${\mathbb R}$ which contains ${\mathbb Q}, \sqrt 2$ and $\sqrt 3$. Show that this consists exactly of the real numbers of the form $a+b\sqrt2+c\sqrt 3+d\sqrt6$ where $a,b,c,d \in {\mathbb Q}$.

I'm given that it suffices to show that $(a)$ the set of elements of the form $a+b\sqrt2+c\sqrt 3+d\sqrt6$ does in fact form a ring and that $(b)$ every ring containing ${\mathbb Q},\sqrt 2$ and $\sqrt3$ has to contain all elements of the form $a+b\sqrt2+c\sqrt 3+d\sqrt6$.

$(a)$ is fine as we merely show that both $0$ and $1$ are of this form and that the addition, subtraction and multiplication of any two elements of this form can be rearranged back into this form. $(b)$ confused me, however. I looked at the solution which said:

"That $(b)$ is true follows from the fact that a ring is closed under addition and multiplication and that $\sqrt 2 \sqrt 3=\sqrt 6$"

I can't really see this, could anyone expand on it a little?

Thank you.

Answer 1

Well, suppose you take any element of the form $a+b\sqrt{2} + c\sqrt{3} + d\sqrt{6} = a + b\sqrt{2} + c\sqrt{3} + (d\sqrt{2})\sqrt{3}$. Isn't this clearly in $\mathbb{Q}(\sqrt{2},\sqrt{3})$?

Answer 2

Hint $ $ The smallest subring of $\,\Bbb R\,$ containing $\,\Bbb Q,\ r,\ s$ is $\,\Bbb Q[r,s],\,$ i.e. the set of all polynomials in $\,r,s\,$ with rational coefficients: clearly they form a ring, and must be contained in any subring containing $\,\Bbb Q,\ r,\ s.\,$ But since $\,r,s = \sqrt{2},\,\sqrt{3}\,$ satisfy $\,r^2 =2,\, s^2=3,\,$ all $\,r^k,s^k$ with $\,k\ge 2\,$ can be eliminated, hence every such polynomial is equal to one of the form $\ a + b r + c s + d rs.$

Remark $\ $ If you are familiar with evaluation homomorphisms then it is more instructive to view the above as follows: the universal way of adjoining $\,r,s\,$ to $\,\Bbb Q\,$ arises by specializing the adjunction of indeterminates $\,\Bbb Q[x,y]\,$ via the evaluation homomorphism $\,x,y\mapsto r,s.\,$ See here for further discussion on the universal view of ring adjunctions.