2
$\begingroup$

If I consider the field extension $K=Q(\sqrt{5})$ and choose a basis , for example {$1,\sqrt{5}$} . I do not really understand the relation between calculating the discriminant of this basis and considering the discriminant of the minimal polynomial of $\sqrt{5}$ , i.e. $x^2-5$ . Is there any relation ? The discriminant is 20 .

Now the discriminant changes if I take another basis , for example the integral basis , {$1,\frac{1+\sqrt{5}}{2}$} . Then the discriminant is 5 .

or is it always if I consider the standard basis of K that the discriminant of the standard basis is equal to the discriminant of the minimal polynomial ?

I think it follows from Vandermonde determinant .

Is my problem clear ?

$\endgroup$
  • 1
    $\begingroup$ Are you talking about $\mathbb{Q}(\sqrt{2})$ or $\mathbb{Q}(\sqrt{5})$? You seem to use $\sqrt{2}$ and $\sqrt{5}$ interchangeably. Note that $\{1, \sqrt{5}\}$ is not a basis for $\mathbb{Q}(\sqrt{2})$. In fact, $\sqrt{5}$ is not even contained in $\mathbb{Q}(\sqrt{2})$. $\endgroup$ – Tob Ernack Jul 5 '19 at 16:12
  • $\begingroup$ sorry , I was a bit confused , I mean $\mathbb{Q(\sqrt{5})}$ $\endgroup$ – AnabolicHorse Jul 5 '19 at 19:19
  • $\begingroup$ Given $\mathbb Q(\sqrt d)$, the discriminant is $d$ itself if $d \equiv 1 \pmod 4$, otherwise it's $4d$. I think Alaca & Williams explains this best, but I don't have that tome at my fingertips at the moment. $\endgroup$ – Robert Soupe Jul 6 '19 at 0:24
  • 1
    $\begingroup$ Remember that your quadratic field is also $\Bbb Q(\sqrt{20}\,)$. When you take the discriminant of the minimal polynomial of this generator, you get $80$. You need a basis of the ring of algebraic integers within your quadratic field, and you want to calculate the discriminant of that basis. $\endgroup$ – Lubin Jul 6 '19 at 1:57
3
$\begingroup$

In this first part I am just presenting the general theory regarding discriminants. This is fairly well explained in textbooks and lecture notes on Algebraic Number Theory (see Robert Ash's notes or Milne's notes).

Given a number field $K$ with $n$ distinct embeddings in $\mathbb{C}$, you can define the discriminant of any set $\{b_1, b_2, \ldots, b_n\}$ of elements in $K$ (not necessarily a basis) as:

$$\text{disc}(b_1, b_2, \ldots, b_n) = \begin{vmatrix} \sigma_1(b_1) & \sigma_1(b_2) & \cdots & \sigma_1(b_n) \\ \sigma_2(b_1) & \sigma_2(b_2) & \cdots & \sigma_2(b_n) \\ \vdots & \vdots & \ddots & \vdots \\ \sigma_n(b_1) & \sigma_n(b_2) & \cdots & \sigma_n(b_n) \end{vmatrix}^2$$

Note that the discriminant is nonzero if and only if $\{b_1, b_2, \ldots, b_n\}$ is a basis of $K$. From now on, I only consider sets that are bases for $K$ (so they are linearly independent over $\mathbb{Q}$ and they span $K$).

For a different set $\{b_1', b_2', \ldots, b_n'\}$ of elements in $K$ related to $\{b_1, b_2, \ldots, b_n\}$ by the change-of-basis matrix $A$ with entries $a_{ij}$ in $\mathbb{Q}$, (i.e. $b_i = \sum\limits_{j=0}^n a_{ij}b_j'$ for all $i = 1, 2, \ldots, n$), you can relate the discriminants by the change of basis formula:

$$\text{disc}(b_1, b_2, \ldots, b_n) = \left(\det A\right)^2\text{disc}(b_1', b_2', \ldots, b_n')$$

Now we restrict to integral bases. A basis for $K$ given by $\{b_1, b_2, \ldots, b_n\}$ is an integral basis if it spans $\mathcal{O}_K$ as a $\mathbb{Z}$-module (i.e. every element of $\mathcal{O}_K$ is expressible uniquely as a linear combination of the $b_i$ where each coefficient is an integer).

You can show that the discriminant of any integral basis is the same. This follows from the previous formula, because the change-of-basis matrix $A$ must be invertible while having coefficients in $\mathbb{Z}$. Its determinant must then be $1$ or $-1$, and the square is always equal to $1$.

Therefore it makes sense to define the discriminant of the number field $K$, called $d_K$, as the discriminant of any of its integral bases.

If the integral basis is a power-basis (i.e., the integral basis is given by a set $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$) then you can show that the discriminant $d_K$ is the same as the discriminant of the minimal polynomial of $\alpha$ (applying the general formula for the discriminant, you end up with a Vandermonde determinant, just as you said). Note however that not every number field $K$ has an integral power basis, so you cannot always use discriminants of polynomials to calculate the discriminant of the number field.

In general, given a primitive element $\alpha$ of $K$ (i.e. $K = \mathbb{Q}(\alpha))$ and $\alpha \in \mathcal{O}_K$, you can say that the discriminant of the minimal polynomial of $\alpha$ is a square multiple of $d_K$. This follows from the change-of-basis formula applied to the set $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ and any integral basis of $\mathcal{O}_K$. Only if the powers of $\alpha$ form an integral basis can you say that the discriminant of the minimal polynomial equals $d_K$.


To address your specific questions: the set $\{1, \sqrt{5}\}$ is a basis for $K = \mathbb{Q}(\sqrt{5})$. You can calculate the discriminant as:

$$\text{disc}(1, \sqrt{5}) = \begin{vmatrix} 1 & \sqrt{5} \\ 1 & -\sqrt{5} \end{vmatrix}^2 = 20$$

(it also follows from the discriminant of $X^2 - 5$ since this is a power basis).

However $\{1, \sqrt{5}\}$ is not an integral basis of $\mathcal{O}_K$. The ring of integers has an integral basis $\{1, \frac{1 + \sqrt{5}}{2}\}$. The discriminant of this basis is actually equal to $5$, which is a divisor of $20$. You can notice that the change-of-basis matrix from $\{1, \frac{1 + \sqrt{5}}{2}\}$ to $\{1, \sqrt{5}\}$ is given by $A = \begin{pmatrix}1 & 0 \\ -1 & 2\end{pmatrix}$ and the square of its determinant is $4$, so this follows from the change of basis formula.

One last thing I can add, is that a useful criterion for an integral basis is that if the discriminant of a basis (consisting of elements in $\mathcal{O}_K$) is square-free, then it must be an integral basis. This follows again from the change-of-basis formula, and the fact that any basis (consisting of elements in $\mathcal{O}_K$) which is not an integral basis will have a corresponding change-of-basis matrix $A$ with integer determinant larger than $1$ in magnitude. This allows us to prove that $\{1, \frac{1+\sqrt{5}}{2}\}$ is in fact an integral basis of $\mathcal{O}_K$, since its discriminant is $5$ which is square-free. Note that the converse is not necessarily true, though. You can have integral bases whose discriminant is not square-free.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

You can consider this to be an inessential addendum to the excellent answer given by @TobEmack.

There are many equivalent definitions of the discriminant. The one that I like best is that it’s the determinant of the trace pairing on the integers of the algebraic number field $K$ in question. Let’s call our ring of integers $\mathcal O$. Then the field-theoretic trace of an element of $\mathcal O$ will always be in $\Bbb Z$, and because the field extension is separable, the trace pairing is non degenerate.

This means that if $a\in K$ nonzero, then there is $b\in K$ for which $\text{Tr}^K_{\Bbb Q}(ab)\ne0$. In particular, the $\Bbb Q$-bilinear pairing $$ (a,b)\mapsto\text{Tr}(ab)\in\Bbb Q $$ from $K\times K$ to $\Bbb Q$ is non degenerate. The determinant of the pairing (with respect to a $\Bbb Q$-basis $\{\beta_1,\cdots\beta_n\}$ of $K$) will always be nonzero. In particular, if you take a $\Bbb Z$-basis of $\mathcal O$, you get a nonzero number that is an invariant of $\mathcal O$, and thus of $K$.

Let’s use this definition to calculate $\text{disc}^{\mathcal O}_{\Bbb Z}$ for $\mathcal O$ the ring of integers of $\Bbb Q(\sqrt5\,)$. Use the basis with $\beta_1=1$ and $\beta_2=\frac{1+\sqrt5}2$. The matrix we want the determinant of is the one with $(i,j)$-entry equal to $\text{Tr}(\beta_i\beta_j)$. We’ll need the value of $\beta_2^2=\frac{3+\sqrt5}2$. Since $\text{Tr}(\beta_1)=2$, $\text{Tr}(\beta_2)=1$ and $\text{Tr}(\beta_2^2)=3$, we need $$ \det\begin{pmatrix}2&1\\1&3\end{pmatrix}=5\,, $$ and there you have the discriminant.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.