Problem 12

Question

Suppose that $L: K$ is a Galois extension of degree $n$ with Galois group $G .$ If $x \in L$, let $$ \operatorname{tr}(x)=\sum_{\sigma \in G} \sigma(x), \quad N(x)=\prod_{\sigma \in G} \sigma(x) $$ The mapping $N$ is the norm. Suppose that $\alpha \in L$ has minimal polynomial $$ x^{r}-a_{1} x^{r-1}+\cdots+(-1)^{r} a_{r} $$ Show that $\operatorname{tr}(\alpha)=(n / r) a_{1}$ and $N(\alpha)=a_{r}^{n / r}$.

Step-by-Step Solution

Verified

Answer

The trace is $(n/r) a_1$ and the norm is $a_r^{n/r}$.

1Step 1: Understand the norm and trace definitions

Given the norm and trace, we have $ \operatorname{tr}(x) = \sum_{\sigma \in G} \sigma(x) $ and $ N(x) = \prod_{\sigma \in G} \sigma(x) $. Here, $L:K$ is a Galois extension with the Galois group $G$ and degree $n$.

2Step 2: Analyze the polynomial

The minimal polynomial of $\alpha$ in $L$ is given by $x^{r} - a_1 x^{r-1} + \cdots + (-1)^{r} a_r$, which implies that $\alpha$ has $r$ Galois conjugates.

3Step 3: Relate degree and conjugates

The degree $r$ of the minimal polynomial reflects the number of Galois conjugates, so $r$ divides the degree $n$ of the extension. This means if the roots are permuted by $G$, they form $n/r$ orbits, each of size $r$.

4Step 4: Calculate the trace using the conjugates

The trace $\operatorname{tr}(\alpha)$ is the sum of all conjugates of $\alpha$. As there are $n/r$ complete sets of $r$ conjugates forming the roots, $\operatorname{tr}(\alpha) = (n/r) a_1$ since $a_1$ is the coefficient that represents the trace of roots when considering the Galois orbit repetition.

5Step 5: Calculate the norm through polynomial constant term

The norm $N(\alpha)$ is equivalent to the product of $\alpha$'s conjugates. Since this is the constant term in the polynomial, $N(\alpha) = a_r^{n/r}$ because there are $n/r$ sets of these conjugates.

Key Concepts

Galois extensionnorm and trace in field theoryminimal polynomial

Galois extension

In the study of Galois Theory, a **Galois extension** is a crucial concept. It refers to a field extension $L:K$ that is both normal and separable. This ensures the extension is well-suited to applying the properties of field automorphisms. For a Galois extension, every element in the field $L$ has all its conjugates in $L$, making the roots of any polynomial fully reside within $L$. This is one of Galois' great insights: matching field extensions with groups of symmetries.

*Normality* implies that if a polynomial with coefficients in $K$ has one root in $L$, all roots of that polynomial are in $L$.
*Separability* ensures polynomials have distinct roots, avoiding any repeated root complexities.

The degree $n$ of the extension $L:K$ is the number of elements in the Galois group $G$, which maps elements of $L$ back onto themselves while preserving arithmetic operations. Knowing this group and how it acts helps us understand algebraic structures deeply and solve polynomial equations.

norm and trace in field theory

Within Galois extensions, we encounter useful tools called the **norm** and **trace** when analyzing elements. These provide ways to measure combined "size" and "aggregate" of field elements under the action of the Galois group.The **trace** $\operatorname{tr}(x)$ is calculated as the sum of the images of $x$ under all automorphisms in the Galois group $G$, given by $\operatorname{tr}(x) = \sum_{\sigma \in G} \sigma(x)$. It gives a sense of how an element behaves when exposed to all symmetries of the field.The **norm** $N(x)$ is the product of these images: $N(x) = \prod_{\sigma \in G} \sigma(x)$. This formula assesses the cumulative influence of the element over these transformations.

In the exercise, the trace $\operatorname{tr}(\alpha)$ links to the coefficient $a_1$, reflecting the sum of $\alpha$'s conjugates.
The norm $N(\alpha)$ connects to the last coefficient $a_r$, embodying the product of all conjugates.

These functions show deep algebraic connections within the Galois framework, playing essential roles in proving fundamental theorems.

minimal polynomial

The concept of a **minimal polynomial** is indispensable when discussing elements of algebraic field extensions. For an element $\alpha$ in the extension field $L$, its minimal polynomial is the monic polynomial of smallest degree with coefficients in the base field $K$ for which $\alpha$ is a root. The minimal polynomial is essential due to its role in characterizing $\alpha$ in $K[x]$.Each root of the minimal polynomial correspondsto Galois conjugates of $\alpha$, confirming its structure and evidencing how $\alpha$ interacts algebraically with $K$.In the exercise example, the polynomial is represented as:\[ x^{r} - a_1 x^{r-1} + \cdots + (-1)^{r} a_r \]

The degree $r$ gives insight into $\alpha$'s number of conjugates, suggesting $\alpha$ and its conjugates are deeply connected by their shared minimal polynomial.
This structure signifies why $\operatorname{tr}(\alpha)$ and $N(\alpha)$ tie directly to the coefficients of this polynomial $a_1$ and $a_r$ respectively.

Understanding the minimal polynomial makes it easier to navigate algebraic expressions and deduce meaningful results in Galois extensions.

Problem 11

Problem 13

Other exercises in this chapter

Problem 10

Suppose that $L: K$ is a Galois extension with Galois group $G$. If $x \in L$, let $$ \operatorname{tr}(x)=\sum_{\sigma \in G} \sigma(x) $$ Show that \(\m

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Problem 11

Suppose that char $K=p$, that $L: K$ is a cyclic extension of degree $p$ and that $\tau$ generates $\Gamma(L: K)$. Let $z$ be an element of $L$ wi

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Problem 13

(Hilbert's theorem 90) Suppose that $L: K$ is a Galois extension of degree $n$ with cyclic Galois group generated by $\tau$, say. (i) Suppose that \(\alph

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Problem 14

Suppose that $L: K$ is a Galois extension of degree $n$ with cyclic Galois group generated by $\tau$, say, and that $\alpha \in L$. Show that \(\operato

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