Describe explicitly $\mathbb{Z}[\sqrt{2},\sqrt{3}]$

Question

The question is if the following is correct, and if it's redundant or it is easier than what I've done.

I'm using this definition of ring adjunction in Bill Dubuque answer (which is the usual used for example, also here.)

Then $\mathbb{Z}[\sqrt{2},\sqrt{3}] := R$ it's the minimal ring (minimal by inclusion obviously, the smallest possible) (inside the ring extension $\mathbb{R}$ containing both $\mathbb{Z}$, $\sqrt{2}$ and $\sqrt{3}$. Well, using this answer, I will try to extend it to two variables. I can say that one explicit description is all polynomials over $\mathbb{Z}$, on two variables X and Y, with X evaluated in $\sqrt{2}$, $\sqrt{3}$.

Now, that explicit description is the minimal subring we are searching, because 1) It is a subring since it is closed under product (product of polynomials evaluated in the same parameters in the evaluation of the product) and also closed by addition (by the same argument as product) and also contains inverses for addition (negating the coefficients of the polynomial, they are still in $\mathbb{Z}$ and they yield minus the evaluation of the non-negated), in the exact same way as detailed in the answer that inspired this.

2) Any ring that contains this elements, must contain products and sum of elements of $\mathbb{Z}$ $\sqrt{2}$ $\sqrt{3}$, and thus those polynomials, being sums and products of this things, must be inside.

Now, instead of looking at ALL the evaluations, we can use the fact that $(\sqrt{2})^2=2$ wich is in $\mathbb{Z}$, and the same for $(\sqrt{3})^2=3$ wich is also in $\mathbb{Z}$. Then, that means that when evaluating the polynomials, we don't need exponents larger than 1 for X or Y, since they will return integers, so no "new" number will come out of that.

You also need $\sqrt6$ as that can be made by multiplying $\sqrt2 \sqrt3$. — Kaynex, Mar 03 '17 at 15:41
@Kaynex Yes, that's in my solution, it's on the polynomial XY evaluated in ($\sqrt2, \sqrt3$) — Santropedro, Mar 03 '17 at 15:54
Oh. Well, then what is the question? It seems you are perfectly capable of describing this ring. — Kaynex, Mar 03 '17 at 16:05
@Kaynex The question is indeed, if I'm doing it correctly, and what can be said. I'm new to this!. Thanks for reaching out and helping! — Santropedro, Mar 03 '17 at 16:06

egreg · Answer 1 · 2017-03-03T15:43:34.713

2

If you have $\sqrt{2}$ and $\sqrt{3}$ in a subring of $\mathbb{C}$, then also $\sqrt{6}$ belongs to it.

Now, the set $$ \{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}:a,b,c,d\in\mathbb{Z}\} $$ is a subring of $\mathbb{C}$ containing $\sqrt{2}$ and $\sqrt{3}$. Hence…

An alternative, but not so explicit, description is: take all elements of the form $P(\sqrt{2},\sqrt{3})$, for $P\in\mathbb{Z}[X,Y]$.

edited Mar 03 '17 at 15:43

answered Mar 03 '17 at 15:41

egreg

238,574

Suggest: Insert $d$ just before $\sqrt{6}.$ – coffeemath Mar 03 '17 at 15:43
@coffeemath Yes, it slipped off! – egreg Mar 03 '17 at 15:44
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@egreg Why as a subring of $\mathbb{C}$ instead of $\mathbb{R}$? I admit, I didn't specify that in the question, my bad. You can replace with $\mathbb{R}$ right? Obviously, the two are going to be isomorphic with the inclusion. – Santropedro Mar 03 '17 at 15:57
@Santropedro Yes, of course you can. No difference, in this case. – egreg Mar 03 '17 at 16:35
Isn't $\Bbb Z[x,y]/(x^2-2, y^2-3)$ sufficient? – Marc Bogaerts Mar 03 '17 at 23:19
@MarcBogaerts Yes, but is it “explicit”? – egreg Mar 03 '17 at 23:26
Do you mean that $x^2y^2-6$ should also be mentioned as an element of the ideal? – Marc Bogaerts Mar 03 '17 at 23:29
@MarcBogaerts No, I was thinking to the question's title. – egreg Mar 03 '17 at 23:30

Describe explicitly $\mathbb{Z}[\sqrt{2},\sqrt{3}]$

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