202306021206

Type :Note Tags : Number Theory

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Trace and Norm

Note

Let $K$ be a number field of degree $n$, we define two function $T^K$ and $N^K$ on $K$ as follows: Let ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n$ be the $n$ embeddings of $K$ into $\mathbb{C}$. For each $\alpha \in K$, set \begin{align} T^K(\alpha) = \sum_{i=1}^{n}\sigma_{i}(\alpha) \\ N^K(\alpha) = \prod_{i=1}^{n}\sigma_{i}(\alpha) \end{align}

• Write $T,N$ instead of $T^K,N^K$ if context is clear.
• $T(\alpha +\beta )=T(\alpha )+T(\beta )$
• $N(\alpha \beta )=N(\alpha )N(\beta )$
• For $r\in \mathbb{Q}$, $T(r)=nr$, $N(r)=r^n$.
• For $r\in \mathbb{Q},\alpha \in K$ $T(r\alpha )=rT(\alpha )$, $N(r\alpha )=r^{n}N(\alpha )$

Theorem 1:

Note

Let $\alpha$ have degree $d$ over $\mathbb{Q}$. Let $t(\alpha ),n(\alpha )$ denote the sum and product of its conjugates respectively. Then \begin{align} T(\alpha) = \frac{n}{d}t(\alpha) \\ N(\alpha) = (n(\alpha))^{n/d} \end{align}

where $n=[K:\mathbb{Q}]$.

Proof:

Every embedding of $\mathbb{Q}(\alpha )$ into $\mathbb{C}$ extends to $\frac{n}{d}$ embeddings of $K$ into $\mathbb{C}$. (refer to Number Field)

Corollary 1:

Note

$T(\alpha ),N(\alpha )\in \mathbb{Q}$

Proof:

$t(\alpha ),n(\alpha )\in \mathbb{Q}$ since they are coefficients of ${\alpha }^{\prime }s$ minimal polynomial, and the result follows.

Corollary 2:

Note

If $\alpha$ is an algebraic integer, $T(\alpha ),N(\alpha )\in \mathbb{Z}$.

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Generalisations

Note

Let $K,L$ be number fields with $K\subset L$. Denote by ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n$ the $n$ embeddings of $L$ in $\mathbb{C}$ which fix $K$ pointwise. Here $n=[K:L]$. For $\alpha \in L$, the relative trace and relative norm are defined by: \begin{align} T ^{L}_{K}(\alpha) = \sum_{i=1}^{n}\sigma_{i}(\alpha) \\ N ^{L}_{K}(\alpha) = \prod_{i=1}^{n}\sigma_{i}(\alpha) \end{align}

• $T_K^L(\alpha +\beta )=T_K^L(\alpha )+T_K^L(\beta )$
• $N_K^L(\alpha \beta )=N_K^L(\alpha )N_K^L(\beta )$
• For $r\in K$, $T_K^L(r)=nr$, $N_K^L(r)=r^n$.
• For $r\in K,\alpha \in L$ $T_K^L(r\alpha )=r{T}_K^L(\alpha )$, $N_K^L(r\alpha )=r^n{N}_K^L(\alpha )$

Theorem 2:

Note

Let $\alpha$ have degree $d$ over $K$. Let $t(\alpha ),n(\alpha )$ denote the sum and product of its conjugates over $K$ respectively. Then \begin{align} T^L_K(\alpha) = \frac{n}{d}t(\alpha) \\ N^L_K(\alpha) = (n(\alpha))^{n/d} \end{align}

where $n=[K:\mathbb{Q}]$.

Proof:

Every embedding of $\mathbb{Q}(\alpha )$ into $\mathbb{C}$ extends to $\frac{n}{d}$ embeddings of $K$ into $\mathbb{C}$. (refer to Number Field)

Corollary:

Note

$T_K^L(\alpha )$ and $N_K^L(\alpha )$ are in $K$. If $\alpha \in \overline{\mathbb{Z}}\cap L$ then they are in $\overline{\mathbb{Z}}\cap K$.

Theorem 3:

Note

Let $K,L,M$ be number fields with $K\subset L\subset M$. Then for all $\alpha \in M$, we have \begin{align} T_{K}^{L}(T_{L}^{M}(\alpha)) = T_{K}^{M}(\alpha)\\ N_{K}^{L}(N_{L}^{M}(\alpha)) = N_{K}^{M}(\alpha) \end{align}

Proof:

Let ${\sigma }_1,\dots ,{\sigma }_n$ and ${\tau }_1,\dots ,{\tau }_m$ be the embeddings of $L$ and $M$ which fix $K$ and $L$ pointwise, respectively. Now we want to compose the ${\sigma }^{\prime }$s with the ${\tau }^{\prime }$s but for that we need to extend them to a normal extension $N$ of $M$. So choose any extension of each ${\sigma }_i$ and ${\tau }_j$, and relabel those automorphisms of $N$ as ${\sigma }_i$ and ${\tau }_j$.

Now, $$ \begin{align} T_{K}^{L}(T_{L}^{M}(\alpha)) = T_{K}^{L}\left( \sum_{i=1}^{m}\tau_{i}(\alpha) \right) &= \sum_{j=1}^{n}\sigma_{j}\left( \sum_{i=1}^{m}\tau_{i}(\alpha) \right) \ &= \sum_{i,j} \sigma_{j}(\tau_{i}(\alpha)) \ \end{align}

Now note that each $\sigma_{j}\circ \tau_{i}$ is an embedding of $M$ which fixes $K$. We know that there are $mn$ embeddings of $M$ fixing $K$, if we can show that the $\sigma_{i} \circ \tau_{j}'s$ are distinct, we will be done. If $\sigma_{i}\circ\tau_{j} = \sigma_{k}\circ\tau_{\ell}$, then applying both sides to an element of $L$, we get $\sigma_{i}=\sigma_{k}$ are equal on the image of $L$ under the embeddings $\tau_{j},\tau_{\ell}$. This gives $i = k$. Since $\sigma_{i}=\sigma_{k}$ is an automorphism, we can multiply by its inverse to get $\tau_{j} = \tau_{\ell}$, and hence $j = \ell$. --- # References [[Algebraic Integers]] [[Number Field]]