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!!



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.

  • $\begingroup$ Thanks a lot this solved the problem!!! $\endgroup$ – Ralph John Feb 22 at 8:27
  • $\begingroup$ My other Q about this is there a general method on finding $\mathcal O_K$? $\endgroup$ – Ralph John Feb 22 at 8:29
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    $\begingroup$ In general, no, but if at least you know the algebraic degree you can narrow down the possibilities. $\endgroup$ – Mr. Brooks Feb 24 at 22:40
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    $\begingroup$ 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. $\endgroup$ – 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

Integers in biquadratic extensions

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

  • $\begingroup$ Thank you very much! $\endgroup$ – Ralph John Feb 22 at 10:29
  • $\begingroup$ You are welcome! $\endgroup$ – 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


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