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

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 $e d = [L : K]$.

**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

**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

(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

**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

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

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

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.

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