Find $f$ and $g$ by trial and error and a rigorous proof for showing that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}]$

Question

Here is the question I am trying to solve:

Find polynomials $f(x), g(x) \in \mathbb{Q}[x]$ such that $\sqrt{2} = f(\sqrt{2} + \sqrt{3})$ and $\sqrt{3} = g(\sqrt{2} + \sqrt{3}).$ Deduce the equality of fields: $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$

And I found the following solution online:

a formula for $f$ and $g$

$\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}]$

My questions are:

1- How did we know that $f$ and $g$ look like that? what are the trials that lead to this?

2- Is there a more rigorous way of proving that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2},\sqrt{3}]$? or what is written is enough?

3- the stated reason for showing that $ \mathbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2} + \sqrt{3}]$ is not that much clear for me ......could anyone explain it in details for me please?

Note: $F[\alpha]$ is a symbol for subring while $F(\alpha)$ is a symbol for a subfield for any $\alpha$

Answer 1

2 - Any arbitrary element in $\mathbb{Q}[\sqrt2 + \sqrt3]$ looks like $x=a+b(\sqrt2+\sqrt3)$, where $a,b\in\mathbb{Q}$. That should give you why $x\in\mathbb{Q}[\sqrt2,\sqrt3]$.

3 - Consider the reciprocal of $\sqrt2 + \sqrt3$, $\frac{1}{\sqrt2 + \sqrt 3} = \sqrt3-\sqrt2$. This element is also in $\mathbb{Q}[\sqrt2 + \sqrt3]$ since it needs to be closed. Now you should be able to show that both $\sqrt2$ and $\sqrt3$ are in $\mathbb{Q}[\sqrt2 + \sqrt3]$. Note that any element in $\mathbb{Q}[\sqrt2,\sqrt3]$ looks like $x=a+b\sqrt2 + c\sqrt3$, where $a,b,c \in \mathbb{Q}$, so $x \in \mathbb{Q}[\sqrt2 + \sqrt3]$.

Answer 2

$(\sqrt 2 + \sqrt 3)^k$ will be a linear combination of $\sqrt 2$, $\sqrt 3$, and $\sqrt 6$ so any polynomial $f(\sqrt 2 + \sqrt 3)$ will yield result of $a\sqrt 6 + b \sqrt 2 +c \sqrt 3 + d$ so we need a polynomial where the yielded values are $a=c=d=0$ and $b = 1$ (and for $g$, $a=b=d=0; c=1$). And, for simplicity, we want the least power.

Now if we have a power of $k$ and $k+1$ coefficients, $e_k$ we will end up with the $4$ variables, $a,b,c,d$ in $a\sqrt 6 + b \sqrt 2 + c\sqrt 3 + d$ being linear combinations of $e_j$. That is we will have $4$ equations of combinations of $e_j$ equaling each of $a,b,c,d$. If we have fewer than $4$ coefficients, $e_j$, we must have linear dependence which are likely to result in inconsistancies. If we have more than $4$ coefficients we will have superflous solutions.

Best guess is to try a polyinomial of $k+1 = 4$ or of power $k = 3$.

A power of $3$ with $mx^3 + nx^2 + px + q$ will yield $m(2\sqrt 2 + 6\sqrt 3 + 9\sqrt 2+3\sqrt 3) + n(5 + 2\sqrt 6) + p(\sqrt 2 + \sqrt 3) + q$ so we'd need:

$2m +9m+p =11m + p = 1; 6m + 3m+p =9m + p= 0; 2n = 0; 5n+q = 0$ which is a far more promising set of equations.

$n=q=0$ and $11m + p = 1$ and $9m + p =0$. So $m= \frac 12$ and $p=-\frac 92$.

So $f(x) = \frac 12x^3 -\frac 92x$ will yeild $f(\sqrt 2+ \sqrt 3) = \sqrt 2$.

TO find $g$ is much the same but we must solve $11m + p = 0$ and $9m+p = 1$ so $m=-\frac 12$ and $p=\frac {11}2$.

ANd $g(x) = -\frac 12x^3 + \frac {11}2 x$ will yeild $f(\sqrt 2 + \sqrt 3) = \sqrt 3$.

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Note we could have tried powers of $k < 3$ but the $4$ equations and fewer than $4$ unknown all lead to inconsistencies.

$k = 0$ and $f(x) = e_0$ requires $e_0 = \sqrt 2 \in \mathbb Q$. Impossible. $k = 1$ and $f(x) = e_1x + e_0$ requires $e_1 = 1; e_1=0;e_0=0$. Impossible.

$k = 2$ and $f(x)=e_2x^2 + e_1x + e_0$ when evaluated for $x=\sqrt 2 + \sqrt 3$ would yield

$e_2(5+2 \sqrt 6) + e_1(\sqrt 2 +\sqrt 3) +e_0$ which to equal $\sqrt 2$ would require $2e_2 = 0; e_1=1; e_1=0; 5e_1+e_0 = 0$ which is of course impossible ($4$ equations and $3$ unknowns requires linear dependence, and in this case inconsistently so).

Answer 3

As requested, I give another answer for 2. and 3. which is more accurate than the currently accepted.

First I start with two general claims which are worth knowing :

• $\mathbb{Q}[x_1,x_2,...,x_n]$ is the smallest $\mathbb{Q}$-algebra containing $x_1,...,x_n$. It follows that if $A$ is a $\mathbb{Q}$-algebra such that $x_1,...,x_n\in A$, then $\mathbb{Q}[x_1,x_2,...,x_n]\subset A$.
• If $A$ is a $\mathbb{Q}$-algebra and $f\in\mathbb{Q}[X]$ a polynomial, then for any $a\in A, f(a)\in A$ (this is simply because a $\mathbb{Q}$-algebra is stable by taking power, sums, and multiplication by a rational number).

Now the proof of the questions :

2. $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is a $\mathbb{Q}$-algebra containing $\sqrt{2}+\sqrt{3}$. From the first claim : $\mathbb{Q}[\sqrt{2}+\sqrt{3}]\subset\mathbb{Q}[\sqrt{2},\sqrt{3}]$.

3. From question 1., we have polynomials $f,g\in\mathbb{Q}[X]$ such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ and $g(\sqrt{2}+\sqrt{3})=\sqrt{3}$. Hence, from the second claim, we have $\sqrt{2},\sqrt{3}\in\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. Therefore, using the first claim $\mathbb{Q}[\sqrt{2},\sqrt{3}]\subset \mathbb{Q}[\sqrt{2}+\sqrt{3}]$.