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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 $\alpha$

Bill Dubuque
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    Trying squaring $\sqrt 2 + \sqrt 3$ to get started. Once you have $\sqrt{2\cdot 3}$ as a polynomial function (over $\mathbb Q$, to be pedantic about this), see what happens when you multiply that times $\sqrt 2 + \sqrt 3$. Now it's a bit of linear algebra to get either $\sqrt 2$ or $\sqrt 3$ as a polynomial function of $\sqrt 2 + \sqrt 3$. – hardmath Oct 11 '20 at 16:49
  • @hardmath what about my other questions? –  Oct 11 '20 at 16:52
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    They are reasonable questions, though I suspect you just need a bit of confidence in your ability to reason through the first part to answer them yourself. I'm not trying to answer your Question, just trying to give you some momentum to pursue the ideas. – hardmath Oct 11 '20 at 16:54
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    See https://math.stackexchange.com/a/3832301/589 – lhf Oct 11 '20 at 20:26

3 Answers3

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$(\sqrt 2 + \sqrt 3)^k$ 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 $f(\sqrt 2 + \sqrt 3) = \sqrt 3$.

====

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).

fleablood
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  • Do we have to prove that each of $\mathbb{Q}[\sqrt{2} + \sqrt{3}]$ and $ \mathbb{Q}[\sqrt{2},\sqrt{3}]$ are really fields? –  Oct 11 '20 at 20:41
  • So the given $g$ in the statement of the question above is wrong..... correct? –  Oct 15 '20 at 02:26
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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 $x \in \mathbb{Q}[\sqrt2 + \sqrt3]$.

JLMF
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  • What about question 1? –  Oct 11 '20 at 16:53
  • I think what @hardmath gave is a pretty solid idea. – JLMF Oct 11 '20 at 16:55
  • Ok, thank you Jason! –  Oct 11 '20 at 16:56
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    Note that $\mathbb Q[\sqrt 2 + \sqrt 3]$ need not be closed under taking reciprocals. If it were, then your last argument would make things much easier. The problem is to construct polynomials (rather than rational expressions) of $\sqrt 2 + \sqrt 3$ to get the individual surds. – hardmath Oct 11 '20 at 16:59
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    @hardmath Since $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ is a finite dimensional vector space over $\mathbb{Q}$, and since multiplication by any nonzero element of $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ is a linear map with trivial kernel, it follows that $1$ is in the image of multiplication by any nonzero element of $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. So $\mathbb{Q}[\sqrt{2}+\sqrt{3}]\setminus{0}$ is colsed under inverses. – 2'5 9'2 Oct 11 '20 at 17:04
  • @hardmath so do we need to prove that both are subfields? –  Oct 11 '20 at 17:17
  • @alex.jordan why the kernel is trivial? –  Oct 11 '20 at 17:19
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    @MathIgnorance Because to be otherwise would mean that two nonzero elements multiply together to make $0$. And all this is still taking place within $\mathbb{R}$, after all. – 2'5 9'2 Oct 11 '20 at 17:20
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    @JasonF Your proof of 2. is not correct : an element of $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ is not necessarily of the form described (actually $\sqrt{2}$ is not of the form described while being in this field). Similarily in 3., an element of $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is not necessarily of the form $a+b\sqrt{2}+c\sqrt{3}$ as claimed. – Roland Oct 11 '20 at 17:22
  • @Roland so can you give us a correct proof for 2 in a separate answer please? –  Oct 15 '20 at 02:22
  • @MathIgnorance I wrote an answer for 2. and 3. – Roland Oct 15 '20 at 08:34
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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 :

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

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

Roland
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