I'm currently reading through Ireland and Rosen's A Classical Introduction to Modern Number Theory, and I'm working on proving that a later definition of trace and norm of arbitrary finite algebraic extensions generalizes an earlier definition for the case of finite fields.
The early definition from Chapter 11.2 goes as follows. Let $F$ have $q$ elements and suppose $E \supset F$ has $q^n$ elements (that is, $[E: F] = n$). Let $a \in E$. The trace of $a$ is $\text{tr}(a) = a + a^q + a^{q^2} + \dots + a^{q^{s-1}}$. The norm of $a$ is $N(a) = a \cdot a^q \cdot a^{q^2} \dots a^{q^{s-1}}$.
The later definition in Chapter 12.1 goes as follows. Now just suppose $E \supseteq F$ is just an algebraic extension with degree $n$ with no further assumption on the base field $F$. Let $\{\beta_1, \beta_2, \dots, \beta_n \}$ be a basis for $E$ over $F$. Then $a \beta_i = \sum_j a_{ij} \beta_j$ for some constants $a_{ij} \in F$ determined by just $i$ and $j$. Thus, we can put the $a_{ij}'s$ into a matrix, $[a_{ij}]$. The norm of $a$ is defined as determinant of $[a_{ij}]$ and the trace of $a$ is just the trace of $[a_{ij}]$, which is $a_{11} + a_{22} + \dots + a_{nn}$.
How can I show the second definition generalizes the first? Going back to the assumption that $F$ and $E$ are finite fields, I can write a basis for $E$ over $F$ as $\{1, \beta, \beta^2, \dots, \beta^{n-1} \}$ where $\beta$ is the root of some monic irreducible polynomial of degree $n$ with coefficients in $F$. Write $a = \sum\limits_{j = 0}^{n-1} a_j \beta^j$ with $a_j \in F$. To determine the trace, we would have to find the coefficient of $\beta^i$ in $a \beta^i = \sum\limits_{j = 0}^{n-1} a_j\beta^{i+j}$, for all $i$. We know one of the terms in the coefficient is $a_0$ since when $j = 0$, the term of the sum is $a_0 \beta^i$. However, when the $i + j > n-1$, we would have to rewrite $\beta^{i+j}$ in terms of lower powers via the minimal polynomial of $\beta$, which is where things can get really messy. This is where I'm stuck. I haven't begun working on the determinant.
Answer
If $F\subseteq E$, then you defined the map $E\to End_F(E)\cong M_n(F)$ by sending $\alpha\mapsto T_\alpha$ such that $T_\alpha(\beta)=\alpha \beta$.
This is a homomorphism, and since $E$ is a field, it is also injective. Consider the characteristic polynomial $p_\alpha$ of the matrix $T_\alpha$ - then the trace and determinant (up to a sign) are the coefficients of $x^{n-1}$ and $1$.
You now have that $p_\alpha \in F[x]$ has $T_\alpha$ as a root, and therefore has $\alpha$ as a root (using the injective homomorphism $\alpha \mapsto T_\alpha$), namely $p_\alpha (\alpha)=0$. If $\sigma$ is any Galois automorphism of $E$ over $F$ (or in general non finite field, $\sigma :E\to F^{alg}$ which fixes $F$), then
$$0=\sigma (0)=\sigma (p_\alpha(\alpha))=\sigma(p_\alpha)(\sigma(\alpha))=p_\alpha(\sigma(\alpha))$$ so you get that $\sigma(\alpha)$ is also a root of $p_\alpha$ (this argument is just the usual argument that if a complex number is a root of a real polynomial, then its complex conjugate is also a root). If you assume that all the $\sigma(\alpha)$ are distinct, then you get all the $n$ roots so that $p_\alpha(x)=\prod_\sigma(x-\sigma(\alpha))$, so that the trace and determinant are $\sum \sigma(\alpha)$ and $ \prod\sigma(\alpha)$ (EDIT: removed the $\pm 1$). For finite fields, the Galois group is generated by the Frobenius automorphism $\alpha\mapsto \alpha^p$ which proves the equality you wanted.
If the $\sigma(\alpha)$ are not distinct, you first find the trace and determinant for $F[\alpha]:F$ and then the determinant for $E:F[\alpha]$ and the composition will give you what you need. In this second case, multiplying by $\alpha$ is just multiplication by scalar from the base field so that trace is just $[E:F[\alpha]]\cdot \alpha$ and the determinant is $\alpha^{[E:F[\alpha]]}$.
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