1

I'm trying to find a general unit/inverse formula for $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \lbrace a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \ \vert \ a,b,c,d\in\mathbb{Q}\rbrace$ and then plug in my specific unit $1+\sqrt{2}+\sqrt{3}+\sqrt{6}$ to prove is it a unit/inverse:

$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} = (a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}$$

let $x=a+b\sqrt{2}$ and let $y=c+d\sqrt{2}$,

$$(x+y\sqrt{3})(x-y\sqrt{3}) = x^2-3y^2$$ $$\Rightarrow (x+y\sqrt{3})\left(\frac{x-y\sqrt{3}}{x^2-3y^2}\right) = 1$$

this is in the form $a\cdot a^{-1} = 1$ which means, $$a^{-1} = \frac{x-y\sqrt{3}}{x^2-3y^2} = \frac{(a+b\sqrt{2})-(c+d\sqrt{2})\sqrt{3}}{(a+b\sqrt{2})^2-3(c+d\sqrt{2})^2} = \frac{a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}}{a^2+2\sqrt{2}ab+2b^2-3c^2-6\sqrt{2}cd-6d^2}$$

expanding unit/inverse expression into the form $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$,

$$\frac{a}{a^2+2\sqrt{2}ab+2b^2-3c^2-6\sqrt{2}cd-6d^2} + \left(\frac{b}{a^2+2\sqrt{2}ab+2b^2-3c^2-6\sqrt{2}cd-6d^2}\right)\sqrt{2} + \left(\frac{-c}{a^2+2\sqrt{2}ab+2b^2-3c^2-6\sqrt{2}cd-6d^2}\right)\sqrt{3} + \left(\frac{-d}{a^2+2\sqrt{2}ab+2b^2-3c^2-6\sqrt{2}cd-6d^2}\right)\sqrt{6}$$

plugging in our claimed unit values which is $a=b=c=d=1$

$$\frac{1}{-6-4\sqrt{2}} + \left(\frac{1}{-6-4\sqrt{2}}\right)\sqrt{2} + \left(\frac{-1}{-6-4\sqrt{2}}\right)\sqrt{3} + \left(\frac{-1}{-6-4\sqrt{2}}\right)\sqrt{6}$$

and since $a^{-1} \in \mathbb{Q}(\sqrt{2},\sqrt{3}) \text{ this means }a,b,c,d \in \mathbb{Q} \Rightarrow a,b,c,d = \frac{p}{q} \text{ such that } p,q \in \mathbb{Z} \text{ and } q \neq 0$. However, the denominators of $a,b,c,d = -6-4\sqrt{2}$ which is irrational. But I know $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field which implies the existence of units/inverses for all non-zero elements in the ring.

What is my mistake?

lambdaserb
  • 29
  • 3
  • 2
    When people say "unit" in this context, they very likely mean a unit relative to the ring of algebraic integers sitting inside the given number field. – user43208 Feb 23 '24 at 18:38
  • $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field. Every nonzero element is a unit. The element $1+\sqrt{2}+\sqrt{3}+\sqrt{6}$ is not zero, and lies in the field, so it is a unit. – Arturo Magidin Feb 23 '24 at 18:56
  • 1
    I agree with @user43208. Because $\Bbb{Q}(\sqrt2,\sqrt3)$ is a field, the question of invertibility in the field is non-interesting (though still a useful exercise to figure out what the inverse is). Anyway, whatever the goal actually is, I would try and use the trick $$1+\sqrt2+\sqrt3+\sqrt6=(1+\sqrt2)(1+\sqrt3).$$ Here the products $$(\sqrt2-1)(\sqrt2+1)=2-1=1$$ and $$(\sqrt3-1)(\sqrt3+1)=3-1=2$$ come to the fore. – Jyrki Lahtonen Feb 23 '24 at 18:58
  • 1
    Hi :) we have $(1+\sqrt 2+\sqrt 3+\sqrt 6) ^{-1}=(1+\sqrt 2) ^{-1}(1+\sqrt 3) ^{-1}=(\sqrt 2-1) \cdot\frac 12(\sqrt 3-1) $ – Jochen Feb 23 '24 at 18:59
  • It has an obvious factorization (as in the first dupe), and each quadratic number factor can be inverted by rationalizing the denominator as in in the second dupe (follow the links there for algorithms to invert higher degree algebraics). – Bill Dubuque Feb 23 '24 at 19:11
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Feb 23 '24 at 19:32
  • @BillDubuque This may well a duplicate, and I'm all for marking it as such, but using those two is ... a stretch, if you ask me. – Jyrki Lahtonen Feb 23 '24 at 19:36
  • @Jyrki Please read my closing comment above. The answer follows immediately from the linked dupes, as I explained in the closing comment. – Bill Dubuque Feb 23 '24 at 19:38
  • @BillDubuque I did read it. I'm still uncomfortable using them. The factorization is obvious to us, but not to all the readers at the level of this question. An earlier incarnation of Pablo's answer would be better IMHO. – Jyrki Lahtonen Feb 23 '24 at 19:43
  • @Jyrki Don't you have the capability to add further duplicates? If not let me know and I will add them for you. But I don't understand why you think that is necessary. If you want a (general) link for the factorization, it can be viewed as the special case $,a=b=c=1,$ of completing a product, yielding $,xy+x+y+1 = (x+1)(y+1).,$ OP is case $,x,y = \sqrt2,,\sqrt 3\ \ $ – Bill Dubuque Feb 23 '24 at 19:54
  • Calculation of the inverse is straightforward but quite tedious. You can prove that $X=1+\sqrt{2}+\sqrt{3}+\sqrt{6}$ is invertible proving that its norm is distinct of $0$. For this,with the simple galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ you do have $$N(X)=(1+\sqrt{2}+\sqrt{3}+\sqrt{6})(1+\sqrt{2}-\sqrt{3}-\sqrt{6})(1-\sqrt{2}+\sqrt{3}-\sqrt{6})(1-\sqrt{2}-\sqrt{3}+\sqrt{6})$$ We get $N(X)=4\ne0$ so $X$ is invertible.

    Note that $X$ is a unit in the sens of to be invertible but $X$ is not a unit of the ring of integers of $\mathbb{Q}(\sqrt{2},\sqrt{3})$.

     – Piquito Feb 23 '24 at 20:15

1 Answers1

1

I got $$ ( 1 + \sqrt 6)^2 - ( \sqrt 2 + \sqrt 3)^2 = 2 $$ so that

Will Jagy
  • 139,541
  • Either this or the factorization Jochen and I pointed out in the comments. It all comes to the same... – Jyrki Lahtonen Feb 23 '24 at 19:01
  • @JyrkiLahtonen It All Ends the Same song by musical group made up of Hollywood actors, producers, https://www.youtube.com/watch?v=wKPaDy57iMg some slang from the 1930's apparently... http://vaudandthevillains.com/ – Will Jagy Feb 23 '24 at 19:05
  • 1
    So the square roots of all of $5\pm2\sqrt6$ and $7\pm2\sqrt6$ simplify ($\approx$ denest). Surely this can be used to cook up a trick question or two :-) – Jyrki Lahtonen Feb 23 '24 at 19:05
  • @JyrkiLahtonen I've seen lots of questions here on factoring a quartic, intended as difference of squares $q^2 - b l^2$ where $q$ is quadratic, $l$ is linear, b constant rational, a few times $b$ has also been negative – Will Jagy Feb 23 '24 at 19:11
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Feb 23 '24 at 19:14