# Calculating the discriminant

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 ?

• 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 $\mathbb{Q}(\sqrt{2})$. – Tob Ernack Jul 5 '19 at 16:12
• sorry , I was a bit confused , I mean $\mathbb{Q(\sqrt{5})}$ – AnabolicHorse Jul 5 '19 at 19:19
• 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. – Robert Soupe Jul 6 '19 at 0:24
• 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. – 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.