Find integral closure of $\mathbb{Q}(\sqrt{2},i)$

Question

I am having trouble finding a basis for the integral closure of $\mathbb{Q}(\sqrt{2},i)$. How would I approach such a task? I also need to show that the closure is not $\mathbb{Z}[\sqrt{2},i]$ and I'm having trouble finding a field element not in the smaller ring, but is satisfied by some monic polynomial in the smaller ring. Can anyone help? Thanks.

Answer 1

Let $A$ be the ring of integers of $\Bbb Q(\sqrt 2, i)$.

We know that every element of $A$ can be written as $a + b\sqrt 2$, where $a, b$ are elements of $\Bbb Q(i)$.

Since the ring of integers of $\Bbb Q(i)$ is $B = \Bbb Z[i]$ and $A$ is the integral closure of $B$, it follows that, for $a, b \in \Bbb Q(i)$, the element $a + b\sqrt 2$ is in $A$ if and only if its minimal polynomial over $\Bbb Q(i)$ has coefficients in $B$, which is again equivalent to saying that its trace and norm are in $B$.

Therefore we get $$a + b\sqrt 2\in A \iff 2a \in B, a^2 - 2b^2 \in B.$$

Now if $a \in B$ and $b \in \frac{1 + i}2B$, then we obviously have $2a \in B$ and $a^2 - 2b^2 \in B$. This tells us that $A$ contains the subset $B + \frac{(1 + i)\sqrt2}2B$.

In particular, the element $\frac{(1 + i)\sqrt 2}{2} = \frac{\sqrt2 + \sqrt2 i}{2}$ is an example of an element in the ring of integers while not in $\Bbb Z[\sqrt 2, i]$, which has $\Bbb Z$-basis $\{1, \sqrt 2, i, \sqrt 2 i\}$.

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We now show that every element in $A$ is in the set $B + \frac{(1 + i)\sqrt2}2B$.

So let $a, b\in \Bbb Q(i)$ satisfy the conditions $2a \in B$ and $a^2 - 2b^2 \in B$. It follows that $8b^2 = (2a)^2 - 4(a^2 - 2b^2) \in B$, hence $(2(1 + i)b)^2 \in B$ and $2(1 + i)b \in B$, since $B$ is integrally closed.

Write $x = 2a$ and $y = 2(1 + i)b$, so that $x, y\in B$. We then have $a^2 - 2b^2 = \frac{x^2}4 - 2\frac{y^2}{8i} = \frac{x^2+ iy^2}4$. Hence $x^2 + iy^2 \in 4B$.

If $x$ is not in $(1 + i)B$, then neither does $y$. But then $x^2 - 1$ is in $2B$, and $y^2 - 1$ also. It follows that $x^2 + iy^2 \in 1 + i + 2B$, contradiction.

Hence both $x$ and $y$ are in $(1 + i)B$. Writing $x_1 = x / (1 + i)$ and $y_1 = y / (1 + i)$, we see that $x_1, y_1\in B$ and $x_1^2 + iy_1^2 \in 2B$. The same argument as above shows that both $x_1$ and $y_1$ are in $(1 + i)B$. Hence both $x$ and $y$ are in fact in $2B$.

So $a = \frac x 2 \in B$ and $(1 + i)b = \frac y 2 \in B$, which implies $b \in \frac{1 + i}2B$. We get $a + b\sqrt 2 \in B + \frac{(1 + i)\sqrt 2}2 B$, as desired.

Answer 2

This is rather a question in ANT, a domain in which we have at our disposal quite an arsenal of tried and proved methods to determine the ring of integers $A_K$ of a number field $K$ (the most well known being the trace, norm and discriminant). Here $K$ is a biquadratic field, say $K=\mathbf Q(\sqrt m,\sqrt n)$, where $m,n$ are distinct square free integers $\neq 1$. Then $A_K$ is entirely known, although the calculations (and the statements!) are a bit lengthy, see e.g. D. Marcus "Number Fields", chap.2, ex.42. The results depend on the congruences mod $4$ of $m,n$ and $k=mn/(m,n)^2$. Recall that the 3 quadratic subfields of $K$ are $\mathbf Q(\sqrt m),\mathbf Q(\sqrt n)$ and $\mathbf Q(\sqrt k)$. Then:

1) If $m\equiv 3, n\equiv k\equiv 2$ mod $4$, then an integral basis of $A_k$ is {$1,\sqrt m, \sqrt n, (\sqrt n + \sqrt k)/2$} ; 2) If $m\equiv 1, n\equiv k\equiv 2$ or $3$ mod $4$, an integral basis is {$ 1 ,(1+\sqrt m)/2,\sqrt n, (\sqrt n + \sqrt k)/2$} ; 3) If $m\equiv n\equiv k\equiv 1$ mod $4$, an integral basis is {$1,(1+\sqrt m)/2),(1+\sqrt n)/2, ((1+\sqrt m)/2)(1+\sqrt k)/2)$}.

This covers all cases up to rearrangements of $m,n,k$.,