# Inclusion in the ring of integers

Let $$K= \mathbb Q(\sqrt3,\sqrt7)$$. I am ask to show that $$\mathcal O_K \ne \mathbb Z[\sqrt3,\sqrt7]$$, where $$\mathcal O_K$$ is the ring of integers.

How can i find $$\mathcal O_K$$ is there a general method on how can i find it? I need help, any hints or links similar to this problem would be appreciated!!

My approach on this problem is to show that $$\mathbb Z[\sqrt3,\sqrt7]\subsetneq \mathcal O_K \subseteq \mathbb Q(\sqrt3,\sqrt7)$$. However i can't show the following inclusion and from here im stuck. Any help would do thank you!!

Hint:

You don't have to find $$\mathcal O_K$$. One can just find an element in $$\mathcal O_K$$ but not in $$\mathbb Z[\sqrt3,\sqrt7]$$.

Consider the element $$\frac{1+\sqrt{21}}2$$. It satisfies the equation $$(x-\frac12)^2=\frac{21}4$$, i.e. $$x^2-x-5=0$$.

Hope this helps.

• Thanks a lot this solved the problem!!! – Ralph John Feb 22 at 8:27
• My other Q about this is there a general method on finding $\mathcal O_K$? – Ralph John Feb 22 at 8:29
• In general, no, but if at least you know the algebraic degree you can narrow down the possibilities. – Mr. Brooks Feb 24 at 22:40
• For that other "Q," you probably have to formally ask it as a separate question. Though it is likely that as you type it up, several suggestions come up. – Bill Thomas Feb 28 at 22:35

Yes, there is a general method of finding the ring of integers in biquadratic number fields $$K=\Bbb Q(\sqrt{m},\sqrt{n})$$ over $$\Bbb Q$$. Arturo's answer is very helpful in explaining this and giving further links - see

Of course, in special cases you don't have to determine $$\mathcal{O}_K$$ explicitly, but it is possible and has been studied well. This site has several posts on it. Here are some examples:

$\mathbb{Q}(\sqrt{m}, \sqrt{n})$ : ring of integers, integral basis and discriminant

On the ring of integers of a compositum of number fields

Ring of integers for $\mathbb{Q}(\sqrt{23},\sqrt{3})$

Algebraic Integers of $\mathbb Q[\sqrt{3},\sqrt{5}]$

Ring of integers of $\mathbb{Q}(\sqrt{-3},\sqrt{5})|\mathbb{Q}$ and group of units

• Thank you very much! – Ralph John Feb 22 at 10:29
• You are welcome! – Dietrich Burde Feb 22 at 10:29

As has already been mentioned, it might suffice to find a "half-integer" in the "composite" intermediate field, in this case $$\textbf Q(\sqrt{21})$$.

But then I thought, can an example of degree $$4$$ be found without too much effort? My first try was $$\frac{1}{4} + \frac{\sqrt 3}{4} + \frac{\sqrt 7}{4},$$ but no luck, the minimal polynomial is $$256x^4 - 256x^3 - \ldots$$ you get the idea.

After various stumblings around that I won't bore you with, I hit upon $$-\frac{\sqrt 3}{2} + \frac{\sqrt 7}{2},$$ which has minimal polynomial $$x^4 - 5x^2 + 1$$. By the way, I believe this might be the fundamental unit of the ring. Regardless of that, this number is clearly not in $$\textbf Z[\sqrt 3 + \sqrt 7]$$.

P.S. You might find this helpful: https://www.lmfdb.org/NumberField/4.4.7056.1

One thing that you can try to find the ring of integers $$\mathcal O_K$$ is to look at the intermediate fields, and multiply different combinations of "typical" integers in those fields.

With biquadratic fields, you know that there are three intermediate quadratic fields (see Is a biquadratic ring uniquely determined by two intermediate quadratic rings?).

Thus, given squarefree integers $$a$$ and $$b$$, we know that $$\mathbb Q(\sqrt a + \sqrt b)$$ has intermediates $$\mathbb Q(\sqrt a)$$, $$\mathbb Q(\sqrt b)$$ and $$\mathbb Q(\sqrt c)$$, where $$c$$ is simply $$ab$$ if $$\gcd(a, b) = 1$$. Then figure $$\theta_a$$, $$\theta_b$$ and $$\theta_c$$, where $$\theta_n = \frac{1}{2} + \frac{\sqrt n}{2}$$ if $$n \equiv 1 \pmod 4$$, otherwise $$\theta_n = \sqrt n$$. Next, compute $$\theta_a \theta_b$$, $$\theta_a \theta_c$$ and $$\theta_b \theta_c$$.

I tried for example $$Q(\sqrt{-3} + \sqrt{-7})$$. From there, I computed $$\left(\frac{1}{2} + \frac{\sqrt{-3}}{2}\right) \left(\frac{1}{2} + \frac{\sqrt{-7}}{2}\right) = \frac{1}{4} + \frac{\sqrt{-3}}{4} + \frac{\sqrt{-7}}{4} + \frac{\sqrt{21}}{4}.$$ I seem to have made a mistake somewhere along the way: this number's minimal polynomial has a coefficient of $$16$$ rather than $$1$$ for $$x^4$$.

Wait, I see my error: I forgot the simple fact that $$i^2 = -1$$. Correcting, I find that $$\left(\frac{1}{2} + \frac{\sqrt{-3}}{2}\right) \left(\frac{1}{2} + \frac{\sqrt{-7}}{2}\right) = \frac{1}{4} + \frac{\sqrt{-3}}{4} + \frac{\sqrt{-7}}{4} - \frac{\sqrt{21}}{4},$$ which has minimal polynomial $$x^4 - x^3 - x^2 - 2x + 4$$.

While this is insufficient to characterize $$\mathcal O_{Q(\sqrt{-3} + \sqrt{-7})}$$, it is sufficient to demonstrate that 's not Z r-3 r-7 same can do for \$\mathcal O_{Q(\sqrt{3} + \sqrt{7