Quadratic Integers in $\mathbb Q$

Can someone tell me if $\frac{3}{5}$, $2+3\sqrt{-5}$, $\frac{3+8\sqrt{-5}}{2}$, $\frac{3+8\sqrt{-5}}{5}$, $i\sqrt{-5}$ are all quadratic integers in $\mathbb Q[\sqrt{-5}]$. And if so why are they in $\mathbb Q[\sqrt{-5}]$.

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You can tell if they are integers based on the description in terms of the integer basis, $\{1,\sqrt{-5}\}$. Remember all algebraic integers in this field are given by

$$a+b\sqrt{-5},\quad a,b\in\Bbb Z.$$

Examining the choices by labeling your five numbers $x_1,\ldots, x_5$ Then we have

$$\begin{cases}
a_1 = {3\over 5}, b_1=0 \\
a_2=2, b_2=3 \\
a_3= {3\over 2}, b_2=4 \\
a_4={3\over 5}, b_4={8\over 5} \\
\end{cases}$$

Now for $x_i$ to be an integer, we need $a_i,b_i\in\Bbb Z$, so we see only $x_2$ works. Note that $x_5\not\in\Bbb Q(\sqrt{-5})$, so it certainly isn’t an integer!

Only one of them is. There are a number of different ways to tell, and of these the easiest are probably the minimal polynomial and the algebraic norm.

In a quadratic extension of $\mathbb{Q}$, the minimal polynomial of an algebraic number $z$ looks something like $a_2 x^2 + a_1 x + a_0$, and if $a_2 = 1$, we have an algebraic integer.

If $z = m + n\sqrt{d}$, then the norm is $N(z) = m^2 – dn^2$. In $\mathbb{Q}(\sqrt{-5})$ for $z = m + n\sqrt{-5}$, this works out to $N(z) = m^2 + 5n^2$. If $z$ is an algebraic integer, then $N(z) \in \mathbb{Z}$.

Let’s take the five numbers one by one:

  • $\frac{3}{5}$ is obviously in $\mathbb{Q}$ so it’s also in the extension $\mathbb{Q}(\sqrt{-5})$. But its minimal polynomial is $5x – 3$, so $a_1 = 5$ but $a_2 = 0$. Also, the norm is $\frac{9}{25}$. Clearly $\frac{3}{5}$ is not an algebraic integer.
  • $2 + 3\sqrt{-5}$ has a minimal polynomial of $x^2 – 4x + 49$, so our $a_2$ here is indeed 1. Also, its norm is 49. Here we have an algebraic integer in $\mathbb{Q}(\sqrt{-5})$.
  • $\frac{3}{2} + 4\sqrt{-5}$ has a minimal polynomial of $4x^2 – 12x + 329$, so $a_2 = 4$. And the norm is $\frac{329}{4}$, which is not an integer. So $\frac{3}{2} + 4\sqrt{-5}$ is not an algebraic integer in $\mathbb{Q}(\sqrt{-5})$. (Side note: $\frac{3}{2} + 4\sqrt{5}$ is not an algebraic integer in $\mathbb{Q}(\sqrt{5})$ either, but $\frac{3}{2} + \frac{7\sqrt{5}}{2}$ is).
  • $\frac{3}{5} + \frac{8\sqrt{-5}}{5}$… you can do this one on your own.
  • $i \sqrt{-5}$ works out to $-\sqrt{5}$, which is an algebraic integer, but it comes from a different domain.