Norm, Trace and Discriminant in a Number Field

This post offers a summary note on the norm, trace, characteristic polynomial and discriminant of an element $\alpha \in K$, a number field of degree $n$ over $\mathbb{Q}$.

$\blacktriangleright \; \mathbb{Q}$-linear transformation: For any $\alpha \in K$, we can define a $\mathbb{Q}$-linear transformation $m_{\alpha}: K \rightarrow K$, such that \begin{equation}\label{Q_lin_trans} m_{\alpha}(x) = \alpha{x},\end{equation} for some $x \in K$. Such an $m_{\alpha}$ can be represented as a matrix (with respect to a basis for $K$ over $\mathbb{Q}$), denoted by $\left[m_{\alpha}\right]$. We define the $\emph{trace}$ and $\emph{norm}$ of $\alpha$ as follows: \begin{align}\label{trace_norm} & T_{K/\mathbb{Q}}(\alpha) = trace\left[m_{\alpha}\right]\\ & N_{K/\mathbb{Q}}(\alpha) = det\left[m_{\alpha}\right]\end{align} The characteristic polynomial $char(\alpha)$ is defined as, \begin{equation}\label{char_poly} char_{K/\mathbb{Q}}(\alpha) = det\left(XI - \left[m_{\alpha}\right]\right),\end{equation} where $X$ is the indeterminate and $I$ is an $n{\times}n$ identity matrix.

Example-1: When we exactly know the representation of the algebraic number $\theta$ with respect to a basis in the number field $\mathbb{Q}[\sqrt[3]{3}]$, \begin{equation}\label{trace_norm_ex1} \theta = q_0 + q_1\sqrt[3]{3} + q_2\sqrt[3]{9},\end{equation} where $q_{i} \in \mathbb{Q}$, $\forall i \in \{0, 1, 2\}$. $\theta$ acts on the basis elements $\{1, \sqrt[3]{3} , \sqrt[3]{9}\}$ as, \begin{align*}& \theta . 1 = q_0 + q_1\sqrt[3]{3} + q_2\sqrt[3]{9}\\ & \theta . \sqrt[3]{3} = 3q_2 + q_0\sqrt[3]{3} + q_1\sqrt[3]{9}\\ & \theta . \sqrt[3]{9} = 3q_1 + 3q_2\sqrt[3]{3} + q_0\sqrt[3]{9},\end{align*} which gives, \begin{align*}& \left[m_{\theta}\right] = \begin{bmatrix} q_0 & 3q_2 & 3q_1\\ q_1 & q_0 & 3q_2\\ q_2 & q_1 & q_0\\ \end{bmatrix} \implies T(\theta) = 3q_0, \; N(\theta) = det \left[m_{\theta}\right] \; \text{(figure this out!)}\end{align*}

Example-2: When we do not know the algebraic number $\theta$ in terms of its basis elements, but know its minimal polynomial over $\mathbb{Q}$, denoted as $min(\theta,\mathbb{Q})$. Let $\theta$ be the root of the irreducible polynomial $X^{3}-3X+1$. What are the trace and norm of $\theta$? Since $X^{3}-3X+1$ is irreducible, let’s choose the basis $\{1, \theta, \theta^{2}\}$. $\theta$ acts on these basis elements as, \begin{align*}& \theta . 1 = 0 + \theta + 0\\ & \theta . \theta = 0 + 0 + \theta^{2}\\ & \theta . \theta^{2} = -1 + 3\theta + 0,\end{align*} which gives, \begin{align*}& \left[m_{\theta}\right] = \begin{bmatrix} 0 & 0 & -1\\ 1 & 0 & 3\\ 0 & 1 & 0\\ \end{bmatrix} \implies T(\theta) = 0, \; N(\theta) = det \left[m_{\theta}\right] = -1\end{align*} So far so good! What about the characteristic polynomial $char(\theta)$? As per Eq.  \eqref{char_poly}, \begin{equation}\label{char_poly_ex2} char(\theta)(X) = \begin{vmatrix} X & 0 & 1\\ -1 & X & -3\\ 0 & -1 & X\\ \end{vmatrix} = X(X^2 - 3) + 1\end{equation}

This is not a coincidence that $char(\theta)$ is identical to the minimal polynomial of $\theta$ over $\mathbb{Q}$. In fact,

\textbullet \; Proposition-1: $char_{K/\mathbb{Q}}(\alpha) = min[\alpha, \mathbb{Q}]^{[K:\mathbb{Q}(\alpha)]}$ for some $\alpha \in K$.

Proof sketch. Prove using Caley-Hamilton theorem for $K = \mathbb{Q}(\alpha)$, then apply induction. ◻

$\blacktriangleright$ \; Counting embeddings: There exist alternative equivalent ways to define trace and norm via embeddings of a number field $K$ in an algebraic closure, say $\mathbb{C}$. For our purpose here, an embedding is an injective homomorphism from $K$ to $\mathbb{C}$ that fix $\mathbb{Q}$ pointwise.

Let $\sigma_1, \ldots, \sigma_n$ be $n$ embeddings of $K$ in $\mathbb{C}$, where $n = [K:\mathbb{Q}]$, we can also define trace and norm of $\alpha \in K$ as, \begin{align}\label{trace_norm_1} & T_{K/\mathbb{Q}}(\alpha) = \sum_{i=1}^{n}\sigma_{i}(\alpha)\\ & N_{K/\mathbb{Q}}(\alpha) = \prod_{i=1}^{n}\sigma_{i}(\alpha)\end{align}

Well-definedness of trace and norm: How to see that these two definitions (via $\mathbb{Q}$-linear transformation and through embeddings) are equivalent?

Proof sketch. As per Eq.  \eqref{trace_norm}, $T_{K/\mathbb{Q}}(\alpha)$ is the sum of eigenvalues of $\left[m_{\alpha}\right]$, which is identical to sum of the roots of $char_{K/\mathbb{Q}}(\alpha)$. As per Eq.  \eqref{trace_norm_1}, $T_{K/\mathbb{Q}}(\alpha)$ is the sum of roots of $min[\alpha,\mathbb{Q}]$. The relationship between characteristic and minimal polynomials of $\alpha$ via proposition-1, therefore, completes the proof. Similar argument works for the norm. ◻

$\blacktriangleright$ \; Discriminant: Two (equivalent) definitions of the discriminant of $\alpha_{1}, \ldots, \alpha_{n}$, where $\alpha_{i} \in K$ for all $i$,

  • $disc(\alpha_{1}, \ldots, \alpha_{n}) = |T_{K/\mathbb{Q}}(\alpha_{i}\alpha_{j})|$

  • $disc(\alpha_{1}, \ldots, \alpha_{n}) = |\sigma_{i}(\alpha_j)|^{2}$, if we think embeddings only.

Well-definedness of discriminant: $|\sigma_{i}(\alpha_j)|^{2} = |T_{K/\mathbb{Q}}(\alpha_{i}\alpha_{j})|$.

Proof sketch. $\left[\sigma_{i}(\alpha_j)\right]\left[\sigma_{j}(\alpha_i)\right] = \sigma_{1}(\alpha_{i}\alpha_{j}) + \ldots + \sigma_{n}(\alpha_{i}\alpha_{j}) = T(\alpha_{i}\alpha_{j})$, and then apply properties of determinant. ◻

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