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I have no idea how to do this.

To find the minimal polynomial of say $\sqrt2 + \sqrt3$, we need to find the monic polynomial $p \in \mathbb Q$ (correct if I am wrong but monic polynomial is when the coefficient of the highest degree term is $1$) of the smallest possible degree such that $\sqrt2 + \sqrt3$ is a root of $p$.

If we let $u=\sqrt2 + \sqrt3$ then $u ^2 = 5+ 2 \sqrt6 \iff u^2 - 5 = 2 \sqrt6 $, then $(u^2 - 5)^2=24 \iff u^4 -10u^2 +1=0$

All I did was keep squaring until all of the irrational terms go away. But what next? Am I doing this correctly and what do we do next if I am?

Martin Sleziak
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snowman
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3 Answers3

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You know that the minimal polynomial for $\sqrt3$ over $\Bbb Q$ is $X^2-3$, and we’ll believe that this is still the minimal polynomial for $\sqrt3$ over $\Bbb Q(\sqrt2\,)$. This means that the polynomial for $\sqrt3+\sqrt2$ over $\Bbb Q(\sqrt2\,)$ is $(X-\sqrt2\,)^2-3$. Expand this out, and multiply it by its “conjugate” (replacing $\sqrt2$ by $-\sqrt2\,$) and get a $\Bbb Q$-polynomial. That’s it.

Lubin
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    So we have $p=u^4 -10u^2 +1$ which is monic, $p(u)=0$ but we don't know if it is irreducible over the rationals. To check this, can we just simply factorise it completely and find the roots and then you can see that none of the roots belong in $\mathbb Q$, hence it is irreducible so it is a minimum polynomial. – snowman Feb 19 '16 at 21:15
  • No. For a quartic, it’s not enough to show that the roots are not in $\Bbb Q$, as the example $X^4+4=(X^2-2X+2)(X^2+2X+2)$ shows. Rather, if you had a $\Bbb Q$-factorization, then that would be a $\Bbb Q(\sqrt2,)$-factorization, different from the factorization of the form $f\bar f$ that you already saw. Contradicts uniqueness of factorization for polynomials over that quadratic field. – Lubin Feb 19 '16 at 21:29
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    OK so to conclude that it is a minimal polynomial do we find what $Q(\sqrt2 , sqrt3 ): Q]$ is or do we find $[Q(\sqrt2 + \sqrt3 ) : Q] $ or does it simply not matter since $\sqrt2 + \sqrt3 $ is in both of them? – snowman Feb 19 '16 at 21:33
  • It depends only on my (unproved) claim that $X^2-3$ is still irreducible over $\Bbb Q(\sqrt2,)$, which you verify by checking that $\sqrt3\notin\Bbb Q(\sqrt2,)$. – Lubin Feb 19 '16 at 22:48
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    OK I see. Also if we are given the fact that $\mathbb Q(\sqrt2 + \sqrt3) = \mathbb Q (\sqrt2 , \sqrt3 )$, how does this help us in this question? – snowman Feb 19 '16 at 23:20
  • I actually did not claim that the two fields were the same, but that the polynomial in question would be the minimal for $\sqrt2+\sqrt3$. That it is irreducible I (regrettably, I guess) omitted, but is covered in my first comment. – Lubin Feb 19 '16 at 23:27
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    How would we prove that the polynomial I found is irreducible? I know there are other ways to prove that what I found is minimum which everyone seems to have said but how do you directly prove that it is irreducible?? – snowman Feb 20 '16 at 15:05
  • You don’t like the proof I gave in the comment starting “No. For…”? Then here’s another. Put $f=x^4-10x^2+1$, and look at $f_1(x)=f(x+1)$. Now let $f_2=f_1(2x)/8$, clearly irreducible by Eisenstein. If you don’t see that, look at the polynomial whose roots are the reciprocals of the roots of $f_2$, namely $f_3(x)=x^4f_2(1/x)$. – Lubin Feb 20 '16 at 16:01
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    Finding roots of $x^4 - 10x^2 + 1 =0$, Let $y=x^2$, Then $y^2 - 10y + 1 = 0$, $y^2 - 10y = -1$, then plus $25$ to both sides:

    $$y^2 - 10y + 25 = -1 + 25 \iff

    (y - 5)^2 = 24 \iff y = 5 \pm 2\sqrt6$$

    Then $x^2 = 5 \pm 2\sqrt6$ (*)

     – snowman Feb 20 '16 at 22:04
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    Then $$(1) , , , x = \pm \sqrt{5 + 2\sqrt6} $$ $$(2) , , , x = \pm \sqrt{5 - 2\sqrt6}$$ But for these values of $x$, we don't know for certain that these values are irrational but we can manipulate them: Notice for (1): $$5 + 2\sqrt6 = 2 + 2\sqrt6 + 3 = (4 + 4\sqrt6 + 6)/2= (4 + 4\sqrt6 + (\sqrt6)^2 )/2 = ((2 + \sqrt6)^2)/2$$ So

    $$(1) , , , x = \pm (2 + \sqrt6)/\sqrt2$$

    and similar manipulation for (2) gives:

    $$(2) , , , x = \pm (\sqrt6 - 2)/\sqrt2$$ And we can clearly see that none of these roots are rational.

     – snowman Feb 20 '16 at 22:08
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    From (*), we can see that

    $x^2 = 5 + 2\sqrt6 $
    and $x^2 = 5 - 2\sqrt6$. Need to prove $x^4 -10x^2 + 1$ is irreducible. If it is, then $$x^4 -10x^2 + 1 = (a + bx + x^2)(c + dx + x^2)$$

    where $a,b,c,d$ are rationals.

    But we can rewrite our expression as $$(-2\sqrt6 - 5 + x^2)(2\sqrt6 - 5 + x^2)$$

    in which we can see that $b=d=0$ but

    $a = -2\sqrt6 - 5$ is not rational

    and $b = 2\sqrt6 - 5$ is not rational which is a contradiction. Hence it is irreducible.

     – snowman Feb 20 '16 at 22:11
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    Is this OK for a proof that it is irreducible? – snowman Feb 20 '16 at 22:12
  • You went long way round Robin Hood’s barn, since $5+2\sqrt6=(\sqrt2+\sqrt3,)^2$. – Lubin Feb 21 '16 at 04:58
  • I'm going to make a new thread about this irreducibility specifically because there is a lot more questions on this I want to ask. – snowman Feb 21 '16 at 11:41
  • In general, for distinct primes $p$, $q$, the minimal polynomial of $\sqrt{p} + \sqrt{q}$ over $\mathbb{Q}$ is $x^4+2x^2(p+q)+(p-q)^2$. – RKR Jan 17 '17 at 18:04
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First, show that $\sqrt{3}$ is not in the quadratic extension generated by $\sqrt{2}.$ That means that the degree of the extension is at least $4.$ But you have found a polynomial of degree $4,$ so it must be minimal.

Igor Rivin
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Suppose $\;\sqrt2\in\Bbb Q(\sqrt3)\;$ , then there exist $\;a,b\in\Bbb Q\;$ such that

$$\sqrt2=a+b\sqrt3\implies 2=a^2+3b^2+2ab\sqrt3\implies\sqrt3\in\Bbb Q\;,\;\;\text{contradiction}$$

Thus, $\;x^2-2\;$ must be irreducible in $\;\Bbb Q(\sqrt3)[x]\;$ , so that $\;\sqrt2+\sqrt3\;$ must belong to an extension of $\;\Bbb Q\;$ of at least degree $\;4\;$ . Since you already found a rational polynomial of degree four which vanishes at $\;\sqrt2+\sqrt3\;$ you finished, as then this must be an irreducible polynomial (otherwise this sum of square roots would belong to an extension of degree less than four).

DonAntonio
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  • So why did you suppose $\sqrt2 \in \mathbb Q (\sqrt3)$ and not $\sqrt3 \in \mathbb Q (\sqrt2)$? – snowman Feb 18 '16 at 22:10
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    also why do we even show such thing? What is the use? – snowman Feb 18 '16 at 22:12
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    Also how did you get $x^2 -2$??? – snowman Feb 18 '16 at 22:14
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    @snowman (1) You can suppose either, (2) The use is to show that $;\sqrt2+\sqrt3;$ doesn't belong neither to $;\Bbb Q(\sqrt3);$ nor to $;\Bbb Q(\sqrt2);$ , (3) That polynomial is the minimal one of $;\sqrt3;$ over the rationals, and it remains irreducible over $;\Bbb Q(\sqrt2;$ as explained. – DonAntonio Feb 18 '16 at 22:22
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    @snowman: You can suppose $\sqrt 3\in \mathbb Q(\sqrt 2)$ as well, you will get (as well) a contradiction. Notice that $a\notin\mathbb Q(b)\iff b\notin \mathbb Q(a)$. The important thing with $\sqrt 2\notin \mathbb Q(\sqrt 3)$, it's that $\mathbb Q(\sqrt 2,\sqrt 3)\neq \mathbb Q(\sqrt 2)$ so the minimal polynomial of $\sqrt 2$ over $\mathbb Q(\sqrt 3)$ is at least $2$. Since $x^2-2\in \mathbb Q(\sqrt 3)$ has $\sqrt 2$ as root and it's irreducible over $\mathbb Q(\sqrt 3)$ (since $\mathbb Q(\sqrt 2,\sqrt 3)\neq \mathbb Q(\sqrt 2)$ or in other word, since $\sqrt 2\notin \mathbb Q(\sqrt 3)$) – Surb Feb 18 '16 at 22:23
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    @Joanpemo So where does the $4$ come from? How did you deduce that the min polynomial has degree $4$ by what you did above? – snowman Feb 18 '16 at 22:36
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    The degree of $;\Bbb Q(\sqrt3);$ over $;\Bbb Q;$ is two, and since the minimal polynomial of $;\sqrt2;$ over the former field is two, then the degree of$;\Bbb Q(\sqrt3)(\sqrt2)=\Bbb Q(\sqrt3,\sqrt2);$ over the rationals is four. The sum $;\sqrt2+\sqrt3;$ is already contained in the last field. – DonAntonio Feb 18 '16 at 22:40
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    @Joanpemo So $x^2-2$ is irreducible in $\mathbb Q (\sqrt3) [x]$ and $x^2-3$ is irreducible in $\mathbb Q (\sqrt2) [x]$. Both of these have degree $2$. Why is it that we multiply these minimal orders together with apparently gives us the minimal order of polynomials of $\sqrt2 + \sqrt3$. Also I know that $Q (\sqrt2, \sqrt3) = Q (\sqrt2 + \sqrt3)$ but can we assume that we don't know this please. – snowman Feb 18 '16 at 22:49
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    @snowman Not that I said, but I use the multiplicativity of fields towers: $$[\Bbb Q(\sqrt3,\sqrt2):\Bbb Q]=[\Bbb Q(\sqrt3,\sqrt2):\Bbb Q(\sqrt2)]\cdot [\Bbb Q(\sqrt2):\Bbb Q] =2\cdot2=4$$ – DonAntonio Feb 18 '16 at 22:52
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    @Joanpemo OK so this finds the degree. I understand how $[\mathbb Q (\sqrt2) : \mathbb Q ]=2$ but how does $[\mathbb Q (\sqrt2 , \sqrt3) : \mathbb Q (\sqrt2 ) ]=2$ aswell??? This is like the final thing that I am confused on!!! – snowman Feb 18 '16 at 23:30
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    There is contradiction where you pointed it out, but I think that you need just a few more words to justify it, since it might be that $ab=0$. – Lubin Feb 19 '16 at 22:57
  • @Lubin Please, whom are you addressing and for what part? – DonAntonio Feb 19 '16 at 23:05
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    @Joanpemo, my apologies: I assumed that responses without explicit addressee were directed to the answerer of the question that the comment appeared in. In other words, to you. And you used the word “contradiction” only once. – Lubin Feb 19 '16 at 23:22
  • @DonAntonio don't we need to show that the minimal polynomial can't have degree less than 4, something which is a divisor of 4. Since $\sqrt{2}+\sqrt{3}$ is not in $\mathbb{Q}$, this degree can't be 1, but it can be 2. Don't we need to prove that it can't even be 2? – Prince Kumar May 24 '20 at 18:15
  • @PrinceKumar If we already know it has a subfield of degree two and since its degree is not two, then it must be four . And it can't be two since the degree of $;\Bbb Q(\sqrt3);$ over the rationals is two and $;\sqrt2;$ does not belong to it.... – DonAntonio May 24 '20 at 20:47