Discriminant of a number field is arguably its most important numerical invariant. Quite closely connected to it is the different ideal. Here we discuss the most basic properties of these two concepts. The discussion mostly follows this expository paper by K. Conrad, but also takes from a series of exercises in chapter 3 of Number Fields.

## Discriminant

Let $$K$$ be a number field of degree $$n$$ over $$\mathbb Q$$. By standard results of field theory there are precisely $$n$$ embeddings of $$K$$ into $$\mathbb C$$, call them $$\sigma_1,\dots,\sigma_n$$. We recall a standard definition:

Definition: For any $$n$$ elements $$\alpha_1,\dots,\alpha_n\in K$$ we define the discriminant of these elements to be the square of the determinant of $$M=(\sigma_j(\alpha_i))_{i,j}$$. We denote it by $$\mathrm{disc}(\alpha_1,\dots,\alpha_n)$$.

It’s easy to see $$\mathrm{disc}(\alpha_1,\dots,\alpha_n)$$ doesn’t depend on the order of $$\alpha_i$$ nor the order of $$\sigma_j$$. Also, $$\det(M)^2=\det(MM^T)=\det((T(\alpha_i\alpha_j)_{i,j})$$, where $$T$$ denotes the trace. From there it’s straightforward that the discriminant lies in $$\mathbb Q$$ , and we can also deduce that $$\mathrm{disc}(\alpha_1,\dots,\alpha_n)\neq 0$$ iff $$\alpha_1,\dots,\alpha_n$$ are linearly independent over $$K$$. Lastly, if $$\beta_1,\dots,\beta_n$$ are elements which are $$K$$-linear combinations of $$\alpha_i$$ represented by a matrix $$V$$, then we easily see $$\mathrm{disc}(\beta_1,\dots,\beta_n)=(\det V)^2\mathrm{disc}(\alpha_1,\dots,\alpha_n)$$. In particular, if $$\alpha_1,\dots,\alpha_n$$ and $$\beta_1,\dots,\beta_n$$ are two bases of the same additive group, then they have the same discriminant. Therefore, it makes sense to speak of the discriminant of an additive subgroup $$\mathrm{disc}(A)$$ to be the discriminant of any of its bases.

The most important additive subgroup of a number field is its ring of integers $$\mathcal O_K$$. The discriminant of this ring will also be sometimes called the discriminant of the field $$K$$ and denoted by $$\mathrm{disc}(K)$$.

## Dual basis and codifferent

Consider an additive subgroup $$A$$ generated by a basis $$\alpha_1,\dots,\alpha_n$$ of $$K$$. Then the matrix $$(T(\alpha_i\alpha_j))_{i,j}$$ is invertible (since its determinant is nonzero discriminant). Considering the columns of its inverse as coefficients of a linear combination of $$\alpha_j$$, so constructed elements, call them $$\alpha_1^*,\dots,\alpha_n^*$$, satisfy $$T(\alpha_i\alpha_j^*)=\begin{cases} 1 & \text{if }i=j\\ 0 & \text{otherwise} \end{cases}$$. Moreover, by uniqueness of matrix inverse, these elements are defined uniquely. We verify that the are linearly independent, hence form a basis: if $$a_1\alpha_1^*+\dots+a_n\alpha_n^*=0$$, then $$0=T(0)=T(\alpha_j*(a_1\alpha_1^*+\dots+a_n\alpha_n^*))=a_1T(\alpha_j^*\alpha_1)+\dots+a_nT(\alpha_j^*\alpha_n)=a_j$$, so the linear combination is trivial.

Definition: Given a basis $$\alpha_1,\dots,\alpha_n$$, we call the basis $$\alpha_1^*,\dots,\alpha_n^*$$ its dual basis. We call the additive group generated by them the dual group and is denoted by $$A^*$$.

Note it’s not immediately clear that this definition is independent of the basis of $$A$$ we choose. The first result which we properly state and prove will imply this.

Proposition 1: $$A^*$$ is precisely the set of $$\alpha\in K$$ such that $$T(\alpha A)\subseteq\mathbb Z$$.

Proof: Let $$\alpha\in K$$. Since dual basis is a basis, we can write $$\alpha=a_1\alpha_1^*+\dots+a_n\alpha_n^*$$, and $$\alpha\in A^*$$ iff $$a_1,\dots,a_n\in\mathbb Z$$. At the same time, $$a_i=T(\alpha\alpha_i)$$. It clearly follows that if $$T(\alpha A)\subseteq\mathbb Z$$, then $$a_i\in\mathbb Z$$. Conversely, if $$a_1,\dots,a_n\in\mathbb Z$$, then for any $$\beta=b_1\alpha_1+\dots+b_n\alpha_n\in A,b_1,\dots,b_n\in\mathbb Z$$ we have $$T(\alpha\beta)=a_1b_1+\dots+a_nb_n\in\mathbb Z$$, so $$T(\alpha A)\subseteq\mathbb Z$$. $$\square$$

It is possible to explicitly give the dual basis if the basis is of the form $$1,\alpha,\dots,\alpha^{n-1}$$ with $$\alpha\in\mathcal O_K$$, i.e. its minimal polynomial over $$\mathbb Q$$ has integer coefficients.

Proposition 2: Let $$f(x)=(x-\alpha)(c_{n-1}x^{n-1}+\dots+c_1x+c_0)$$ be the minimal polynomial of $$\alpha$$. Then $$\frac{c_0}{f'(\alpha)},\dots,\frac{c_{n-1}}{f'(\alpha)}$$ is the dual basis of $$1,\alpha,\dots,\alpha^{n-1}$$. Moreover, $$(\mathbb Z[\alpha])^*=\frac{1}{f'(\alpha)}\mathbb Z[\alpha]$$.

Proof: Let $$\alpha_1=\alpha,\alpha_2,\dots,\alpha_n$$ be the conjugates of $$\alpha$$ in $$\mathbb C$$. It’s easy to see $$c_i=c_i(\alpha)$$ is a monic polynomial in $$\alpha$$ of degree $$i$$, and if we divided $$f(x)$$ by $$x-\alpha_j$$, the coefficients would be $$c_i(\alpha_j)$$. For $$k=0,\dots,n-1$$ consider the polynomial

$$\displaystyle\sum_{i=1}^n\frac{\alpha_i^k}{f'(\alpha_i)}\frac{f(x)}{x-\alpha_i}$$.

It’s easy to see that each term is $$1$$ for $$x=\alpha_i^k$$ and $$0$$ for $$x=\alpha_j,j\neq i$$. Hence this polynomial of degree smaller than $$n$$ agrees with polynomial $$x^k$$ at $$n$$ places, so the polynomials must be equal. Comparing coefficient of $$x^j$$ we get

$$\displaystyle\sum_{i=1}^n\frac{\alpha_i^k}{f'(\alpha_i)}c_j(\alpha_i)=\begin{cases} 1 & \text{if }j=k\\ 0 & \text{otherwise} \end{cases}$$,

but the left hand side is precisely $$T\alpha_k\frac{c_j(\alpha)}{f'(\alpha)})$$, showing the first claim. To see the second claim, recall that $$c_j$$ are monic polynomials of degree $$j$$, so we can show by induction that $$\frac{\alpha^j}{f'(\alpha)}\in(\mathbb Z[\alpha])^*$$. We omit the details. $$\square$$

The construction of dual additive group also preserves the property of being a fractional ideal. More precisely:

Proposition 3: Let $$\frak a$$ be a fractional ideal. Then $$\frak a^*$$ (considered as the dual of the additive group) is also a fractional ideal. Moreover, $$\frak a^*=\frak a^{-1}\mathcal O_K^*$$. [recall that $$\frak a^{-1}$$ is defined as the set of these $$\alpha\in K$$ for which $$\alpha\frak a\subseteq\mathcal O_K$$. In this post we establish that $$\frak a\frak a^{-1}=\mathcal O_K$$]

Proof: Fix any any $$\beta\mathcal O_K$$. For $$\alpha\in\frak a^*$$ we have $$T(\alpha\frak a)\subseteq\mathbb Z$$. But, since $$\frak a$$ is a fractional ideal, $$\beta\frak a\subseteq\frak a$$, so $$T(\beta\alpha\frak a)=T(\alpha(\beta\frak a))\subseteq T(\alpha\frak a)\subseteq\mathbb Z$$, so $$\beta\alpha\in\frak a^*$$. This shows $$\frak a^*$$ is a fractiona ideal.

For the second part, suppose first $$\alpha\in\frak a^*$$. For any $$\beta\in\frak a$$ we have $$\beta\mathcal O_K\subseteq\frak a$$, so $$T(\alpha\beta\mathcal O_K)\subseteq T(\alpha\frak a)\subseteq\mathbb Z$$, so $$\alpha\beta\in\mathcal O_K^*$$. Hence, $$\alpha\frak a\subseteq\mathcal O_K^*$$. Hence $$\alpha\in\alpha\frak a\frak a^{-1}\subseteq\frak a^{-1}\mathcal O_K^*$$. For the converse, pretty much this argument in reverse works. $$\square$$

## Different ideal

Previous proposition shows that duals work a bit like inverses. By taking duals inverse, we get another important ideal.

Definition: Let $$\frak a$$ be a fractional ideal. We define the different of $$\frak a$$ to be $$\mathrm{diff}\frak a=(\frak a^*)^{-1}$$. In particular, we call the different of $$\mathcal O_K$$ the different of $$K$$ $$\mathrm{diff} K$$.

Note that $$\mathcal O_K\subseteq\mathcal O_K^*$$, so $$\mathrm{diff} K$$ is an ideal in $$\mathcal O_K$$. From proposition 3 we immediately have $$\mathrm{diff} \frak a=\frak a\mathrm{diff} K$$, hence for the most part we only have to focus our attention of $$\mathrm{diff} K$$. It takes particularly simple form when $$\mathcal O_K=\mathbb Z[\alpha]$$ – by proposition 2 we then have $$\mathrm{diff} K=(f'(\alpha))$$, $$f$$ being the minimal polynomial of $$\alpha$$.

Recall the definition of the norm of an ideal: $$N(\frak a)=[\mathcal O_K:\frak a]=|\mathcal O_K/\frak a|$$.

Theorem 1: $$N(\mathrm{diff} K)=|\mathrm{disc} K|$$

Proof: First we note that for fractional ideals $$\frak a\supseteq\frak b$$ and $$\frak c$$ we have an isomorphism of rings $$\frak{ac}/\frac{bc} \cong \frak a/\frak b$$ (this is quite straightforward to establish). In particular, taking $$\frak a=\mathcal O_K^*=(\mathrm{diff} K)^{-1},\frak b=\mathcal O_K,\frak c=\mathrm{diff} K$$ this gives $$\mathcal O_K/\mathrm{diff} K\cong\mathcal O_K^*/\mathcal O_K$$. In particular, $$N(\mathrm{diff} K)=[\mathcal O_K:\mathrm{diff}K]=[\mathcal O_K^*:\mathcal O_K]$$. It is well-known that for two free abelian groups $$A\supseteq B$$ of the same rank, $$[A:B]$$ is the absolute value of the determinant of a transformation taking basis of $$A$$ to the basis of $$B$$. In our case, take $$\alpha_1,\dots,\alpha_n$$ an integral basis of $$\mathcal O_K$$ and $$\alpha_1^*,\dots,\alpha_n^*$$ its dual basis. We write $$\alpha_i=a_{i1}\alpha_1^*+\dots+a_{in}\alpha_n^*$$. Then $$a_{ij}=T((a_{i1}\alpha_1^*+\dots+a_{in}\alpha_n^*)\alpha_j)=T(\alpha_i\alpha_j)$$. In other words, the transformation matrix is precisely the matrix $$(T(\alpha_i\alpha_j)_{ij}$$, whose determinant is $$\mathrm{disc} K$$. This establishes the theorem. $$\square$$

The different is important when working with ramification of primes in a number field. As will be proven in the future post, different ideal is divisible precisely by prime ideals which which are ramified in $$K$$. In the next blog post we shall establish, among other things, this result in normal extensions, together with precise formula for the exponent of this prime.

As a closing remark, it is worth poining out that the whole theory of discriminants and differents can be built in extensions $$L/K$$ for $$K$$ different from the field of rational numbers, although things get a lot more technical, since, for example, discriminant has to be considered as an ideal and not a single element. I hope to one day cover the theory of general discriminants and different ideals in another blog post or two.