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 ?

  • 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 $\mathbb{Q(\sqrt{5})}$ $\endgroup$ 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. $\endgroup$ 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

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.


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.


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