Let us define $R:=\mathbb{Z}$, $\alpha:=\sqrt{-7}$ and $K:=\mathbb{Q}[\alpha]$. Find the "algebric closure of $R$ above $K$". i.e. express explicitly the set $\{p\in K:\exists 0\ne f\in R[x],f\text{ is monic},0=f(p)\}$.

According to the comments I have 2 attempts but none of them is develop for a full solution;

Let $A$ be the required algebric integer closure and let us define $\zeta=\sqrt{-7}$.

Attempt #1:

Let $z\in\mathbb Q(\zeta)$. Let $m\in\mathbb Q[x]$ be the minimal polynomial of $z$ in $\mathbb Q$. We know that $\deg(m)=2$. Hence $\exists p_1,p_2,q_1,q_2\in\mathbb Z$ such that $(p_1,q_1)=(p_2,q_2)=1$ and $m(x)=x^2+{p_1\over q_1}x+{p_2\over q_2}$. We know that $m(z)=0$. Thus, $$0=z^2+{p_1\over q_1}z+{p_2\over q_2}\Rightarrow \\ 0=q_1 q_2z^2+p_1q_2 z+p_2 q_1 $$ Let us define $\mathbb Z[x]\ni l(x)=q_1 q_2x^2+p_1q_2 x+p_2 q_1$. Here I stuck.

Attempt #2:

If $z=a+b\zeta, (a,b\in\mathbb Q)$ is algebric integer over $\mathbb Z$ then $\exists$ a monic polynomial which zeroing in $z$ and in $\bar z$. Hence $$ x^2-2ax+a^2-7b^2=(x-z)(x-\bar z)| f $$

Both attempts, I am not sure how to proceed. Thanks in advence.

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    $\begingroup$ The integers are not a field. You cannot algebraically close a ring, only a field. $\endgroup$ – Parcly Taxel Sep 12 '19 at 15:47
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    $\begingroup$ $O_K$ is the integral closure of $\Bbb{Z}$ in $K$ : it is the set of elements of $K$ algebraic over $\Bbb{Z}$ with monic minimal polynomial, equivalently those that are roots of some monic integer polynomial, equivalently those such that $\Bbb{Z}[\beta]$ is a finitely generated $\Bbb{Z}$-module (from which we obtain $O_K$ is a ring since $\Bbb{Z}[\beta_1,\beta_2]$ is finitely generated) $\endgroup$ – reuns Sep 12 '19 at 15:48
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    $\begingroup$ @ParclyTaxel I've corrected my question. $\endgroup$ – J. Doe Sep 12 '19 at 15:51
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    $\begingroup$ Hint: Try writing down an arbitrary element in $\mathbb{Q}[\sqrt{-7}]$ and see what its minimal polynomial is. Try to proceed from there. $\endgroup$ – rogerl Sep 12 '19 at 15:54
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    $\begingroup$ Any element $z$ of $K$ can be written uniquely in the form $z=a+b\sqrt{-7}$ with $a,b\in\Bbb{Q}$. Your task is to figure out the minimal polynomial of $z$ over $\Bbb{Z}$, and then determine for which choices of $a,b$ the resulting polynomial has integer coefficients. $\endgroup$ – Jyrki Lahtonen Sep 12 '19 at 15:55

Your comment has one grave mistake in the second equation, which would have made life easier.

You should have the pair of equations : $$ \begin{cases} 0 = a^2d^2 - 7b^2c^2 + abd^2s + tb^2d^2 \\ 0 = 2abcd + b^2cds = 2a + bs \tag{*} \end{cases} $$

(you made a mistake in the comment, and let us assume $c \neq 0$, because the first part of your comment covers the case of it being zero i.e which real numbers are in the algebraic closure) which sounds like plenty to work with. However, after the mistake in the comment you get $s = \frac{-2a}{b}$, which when you put in the first equation gives you after simplification that $t = \frac{a^2}{b^2} - \frac{7c^2}{d^2}$.

But, $s$ and $t$ are integers, because $f$ was chosen to have integer coefficients! Thus, the above two quantities must be integers.

Now we will go in hidden steps, because hide and seek is my (second) favourite game.

So $s = \frac{-2a}{b}$ is an integer. Then, $-2a = bs$. Recall $a$ and $b$ are chosen coprime.

  • So $b$ divides $-2a$. What can you conclude from coprimality?

$b$ must divide $2$.

  • Write down a set consisting of every possible value that $b$ can take.

$b$ can be any of $2,1,-1,-2$ i.e. a divisor of $2$.

Look at the answers before going to the next section.

Now note that $t = \frac{a^2}{b^2} - \frac{7c^2}{d^2}$ is an integer. Recall $c,d$ are co prime.

  • Suppose $|b| = 1$. Then $d^2$ divides $7c^2$(why?) so what values can $d$ take?

Well, $d^2$ must divide $7$, but that forces $d = \pm 1$.

  • Suppose $|b| = 2$. Conclude that $d^2$ divides $28$. What values can $d$ take now?

Well, we have $a^2-4t = \frac{28c^2}{d^2}$ after rearrangement, and the RHS must be an integer, now the claim follows from coprimality and $d$ may be any of $2,1,-1,-2$.

  • Show in fact that $|d| = 2$ if $|b| = 2$, by noticing something extra in the argument in the above yellow box.

Well, we have $28$ divides $d^2(a^2 - 4t)$. However, $b$ is coprime to $a$, so $a$ is odd, so $a^2 - 4t$ is odd, and therefore $d^2$ is even, hence $d$ is even so $|d| = 2$.

Thus , we conclude that an algebraic integer is either of the form $a + c\alpha$ or $\frac{a+c \alpha}{2}$ , where $a,c$ are integers in the former case, and odd integers in the latter case. Let us show that these are sufficient conditions.

For this, we must show that every element of any of the above two forms , call it $z$, satisfies some polynomial.

First, I will do a preliminary step.

  • Let $a,c$ be odd integers. Show that $a^2 + 7c^2$ is a multiple of $4$.

Write $a^2 + 7c^2 = (a+c)(a-c) + 8c^2$, and recognize the RHS as the sum of two multiples of $4$.

  • Show that $z + \bar z$ and $z \bar{z}$ are both integers, where $\bar z$ is the conjugate of $z$. (Note that the conjugate of $p+q\alpha$ is $p-q\alpha$ for $p,q $ real).

We have $a -c\alpha + a + c \alpha = 2a$, so clearly $z+\bar z$ is integral in both cases. Their product is $a^2 + 7c^2$ in the first case, of course an integer, and $\frac{a^2+7c^2}{4}$ in the second case, which is an integer by the first step.

Conclude that if $z + \bar z = -s$ and $z\bar z = t$ then $z^2 + sz+t = 0$. Consequently, $z$ satisfies a polynomial with integer coefficients.

Easy verification, let $f(x) = x^2 + sx+t$, then $f(z) = 0$.

This concludes the description of the algebraic closure. Is there a better description.

Exercise : Show that the algebraic closure can be expressed as $\{a + bl : a,b \in \mathbb Z\}$ where $l = \frac{1+\alpha}{2}$. Thus, the algebraic closure is $\mathbb Z[\frac{1+\alpha}{2}]$


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