Definition 9.20.1. Let $L/K$ be a finite extension of fields. For $\alpha \in L$ we define the trace $\text{Trace}_{L/K}(\alpha ) = \text{Trace}_ K(\alpha : L \to L)$ and the norm $\text{Norm}_{L/K}(\alpha ) = \det _ K(\alpha : L \to L)$.
9.20 Trace and norm
Let $L/K$ be a finite extension of fields. By Lemma 9.4.1 we can choose an isomorphism $L \cong K^{\oplus n}$ of $K$-modules. Of course $n = [L : K]$ is the degree of the field extension. Using this isomorphism we get for a $K$-algebra map
\[ L \longrightarrow \text{Mat}(n \times n, K),\quad \alpha \longmapsto \text{matrix of multiplication by }\alpha \]
Thus given $\alpha \in L$ we can take the trace and the determinant of the corresponding matrix. Of course these quantities are independent of the choice of the basis chosen above. More canonically, simply thinking of $L$ as a finite dimensional $K$-vector space we have $\text{Trace}_ K(\alpha : L \to L)$ and the determinant $\det _ K(\alpha : L \to L)$.
It is clear from the definition that $\text{Trace}_{L/K}$ is $K$-linear and satisfies $\text{Trace}_{L/K}(\alpha ) = [L : K]\alpha $ for $\alpha \in K$. Similarly $\text{Norm}_{L/K}$ is multiplicative and $\text{Norm}_{L/K}(\alpha ) = \alpha ^{[L : K]}$ for $\alpha \in K$. This is a special case of the more general construction discussed in Exercises, Exercises 111.22.6 and 111.22.7.
Lemma 9.20.2. Let $L/K$ be a finite extension of fields. Let $\alpha \in L$ and let $P$ be the minimal polynomial of $\alpha $ over $K$. Then the characteristic polynomial of the $K$-linear map $\alpha : L \to L$ is equal to $P^ e$ with $e \deg (P) = [L : K]$.
Proof. Choose a basis $\beta _1, \ldots , \beta _ e$ of $L$ over $K(\alpha )$. Then $e$ satisfies $e \deg (P) = [L : K]$ by Lemmas 9.9.2 and 9.7.7. Then we see that $L = \bigoplus K(\alpha ) \beta _ i$ is a direct sum decomposition into $\alpha $-invariant subspaces hence the characteristic polynomial of $\alpha : L \to L$ is equal to the characteristic polynomial of $\alpha : K(\alpha ) \to K(\alpha )$ to the power $e$.
To finish the proof we may assume that $L = K(\alpha )$. In this case by Cayley-Hamilton we see that $\alpha $ is a root of the characteristic polynomial. And since the characteristic polynomial has the same degree as the minimal polynomial, we find that equality holds. $\square$
Lemma 9.20.3. Let $L/K$ be a finite extension of fields. Let $\alpha \in L$ and let $P = x^ d + a_1 x^{d - 1} + \ldots + a_ d$ be the minimal polynomial of $\alpha $ over $K$. Then where $e d = [L : K]$.
\[ \text{Norm}_{L/K}(\alpha ) = (-1)^{[L : K]} a_ d^ e \quad \text{and}\quad \text{Trace}_{L/K}(\alpha ) = - e a_1 \]
Proof. Follows immediately from Lemma 9.20.2 and the definitions. $\square$
Lemma 9.20.4. Let $L/K$ be a finite extension of fields. Let $V$ be a finite dimensional vector space over $L$. Let $\varphi : V \to V$ be an $L$-linear map. Then and
\[ \text{Trace}_ K(\varphi : V \to V) = \text{Trace}_{L/K}(\text{Trace}_ L(\varphi : V \to V)) \]
\[ \det \nolimits _ K(\varphi : V \to V) = \text{Norm}_{L/K}(\det \nolimits _ L(\varphi : V \to V)) \]
Proof. Choose an isomorphism $V = L^{\oplus n}$ so that $\varphi $ corresponds to an $n \times n$ matrix. In the case of traces, both sides of the formula are additive in $\varphi $. Hence we can assume that $\varphi $ corresponds to the matrix with exactly one nonzero entry in the $(i, j)$ spot. In this case a direct computation shows both sides are equal.
In the case of norms both sides are zero if $\varphi $ has a nonzero kernel. Hence we may assume $\varphi $ corresponds to an element of $\text{GL}_ n(L)$. Both sides of the formula are multiplicative in $\varphi $. Since every element of $\text{GL}_ n(L)$ is a product of elementary matrices we may assume that $\varphi $ either looks like
\[ E_{12}(\lambda ) = \left( \begin{matrix} 1
& \lambda
& \ldots
\\ 0
& 1
& \ldots
\\ \ldots
& \ldots
& \ldots
\end{matrix} \right) \quad \text{or}\quad E_1(a) = \left( \begin{matrix} a
& 0
& \ldots
\\ 0
& 1
& \ldots
\\ \ldots
& \ldots
& \ldots
\end{matrix} \right) \]
(because we may also permute the basis elements if we like). In both cases the formula is easy to verify by direct computation. $\square$
Lemma 9.20.5. Let $M/L/K$ be a tower of finite extensions of fields. Then
\[ \text{Trace}_{M/K} = \text{Trace}_{L/K} \circ \text{Trace}_{M/L} \quad \text{and}\quad \text{Norm}_{M/K} = \text{Norm}_{L/K} \circ \text{Norm}_{M/L} \]
Proof. Think of $M$ as a vector space over $L$ and apply Lemma 9.20.4. $\square$
The trace pairing is defined using the trace.
Definition 9.20.6. Let $L/K$ be a finite extension of fields. The trace pairing for $L/K$ is the symmetric $K$-bilinear form
\[ Q_{L/K} : L \times L \longrightarrow K,\quad (\alpha , \beta ) \longmapsto \text{Trace}_{L/K}(\alpha \beta ) \]
It turns out that a finite extension of fields is separable if and only if the trace pairing is nondegenerate.
Lemma 9.20.7. Let $L/K$ be a finite extension of fields. The following are equivalent:
$L/K$ is separable,
$\text{Trace}_{L/K}$ is not identically zero, and
the trace pairing $Q_{L/K}$ is nondegenerate.
Proof. It is clear that (3) implies (2). If (2) holds, then pick $\gamma \in L$ with $\text{Trace}_{L/K}(\gamma ) \not= 0$. Then if $\alpha \in L$ is nonzero, we see that $Q_{L/K}(\alpha , \gamma /\alpha ) \not= 0$. Hence $Q_{L/K}$ is nondegenerate. This proves the equivalence of (2) and (3).
Suppose that $K$ has characteristic $p$ and $L = K(\alpha )$ with $\alpha \not\in K$ and $\alpha ^ p \in K$. Then $\text{Trace}_{L/K}(1) = p = 0$. For $i = 1, \ldots , p - 1$ we see that $x^ p - \alpha ^{pi}$ is the minimal polynomial for $\alpha ^ i$ over $K$ and we find $\text{Trace}_{L/K}(\alpha ^ i) = 0$ by Lemma 9.20.3. Hence for this kind of purely inseparable degree $p$ extension we see that $\text{Trace}_{L/K}$ is identically zero.
Assume that $L/K$ is not separable. Then there exists a subfield $L/K'/K$ such that $L/K'$ is a purely inseparable degree $p$ extension as in the previous paragraph, see Lemmas 9.14.6 and 9.14.5. Hence by Lemma 9.20.5 we see that $\text{Trace}_{L/K}$ is identically zero.
Assume on the other hand that $L/K$ is separable. By induction on the degree we will show that $\text{Trace}_{L/K}$ is not identically zero. Thus by Lemma 9.20.5 we may assume that $L/K$ is generated by a single element $\alpha $ (use that if the trace is nonzero then it is surjective). We have to show that $\text{Trace}_{L/K}(\alpha ^ e)$ is nonzero for some $e \geq 0$. Let $P = x^ d + a_1 x^{d - 1} + \ldots + a_ d$ be the minimal polynomial of $\alpha $ over $K$. Then $P$ is also the characteristic polynomial of the linear maps $\alpha : L \to L$, see Lemma 9.20.2. Since $L/k$ is separable we see from Lemma 9.12.4 that $P$ has $d$ pairwise distinct roots $\alpha _1, \ldots , \alpha _ d$ in an algebraic closure $\overline{K}$ of $K$. Thus these are the eigenvalues of $\alpha : L \to L$. By linear algebra, the trace of $\alpha ^ e$ is equal to $\alpha _1^ e + \ldots + \alpha _ d^ e$. Thus we conclude by Lemma 9.13.2. $\square$
Let $K$ be a field and let $Q : V \times V \to K$ be a bilinear form on a finite dimensional vector space over $K$. Say $\dim _ K(V) = n$. Then $Q$ defines a linear map $Q : V \to V^*$, $v \mapsto Q(v, -)$ where $V^* = \mathop{\mathrm{Hom}}\nolimits _ K(V, K)$ is the dual vector space. Hence a linear map
\[ \det (Q) : \wedge ^ n(V) \longrightarrow \wedge ^ n(V)^* \]
If we pick a basis element $\omega \in \wedge ^ n(V)$, then we can write $\det (Q)(\omega ) = \lambda \omega ^*$, where $\omega ^*$ is the dual basis element in $\wedge ^ n(V)^*$. If we change our choice of $\omega $ into $c \omega $ for some $c \in K^*$, then $\omega ^*$ changes into $c^{-1} \omega ^*$ and therefore $\lambda $ changes into $c^2 \lambda $. Thus the class of $\lambda $ in $K/(K^*)^2$ is well defined and is called the discriminant of $Q$. Unwinding the definitions we see that
\[ \lambda = \det (Q(v_ i, v_ j)_{1 \leq i, j \leq n}) \]
if $\{ v_1, \ldots , v_ n\} $ is a basis for $V$ over $K$. Observe that the discriminant is nonzero if and only if $Q$ is nondegenerate.
Definition 9.20.8. Let $L/K$ be a finite extension of fields. The discriminant of $L/K$ is the discriminant of the trace pairing $Q_{L/K}$.
By the discussion above and Lemma 9.20.7 we see that the discriminant is nonzero if and only if $L/K$ is separable. For $a \in K$ we often say “the discriminant is $a$” when it would be more correct to say the discriminant is the class of $a$ in $K/(K^*)^2$.
Exercise 9.20.9. Let $L/K$ be an extension of degree $2$. Show that exactly one of the following happens
the discriminant is $0$, the characteristic of $K$ is $2$, and $L/K$ is purely inseparable obtained by taking a square root of an element of $K$,
the discriminant is $1$, the characteristic of $K$ is $2$, and $L/K$ is separable of degree $2$,
the discriminant is not a square, the characteristic of $K$ is not $2$, and $L$ is obtained from $K$ by taking the square root of the discriminant.
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #1770 by Carl on