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4

Assume that $\sigma$ is an automorphism of order $p$, also $\sigma$ is a permutation of the roots $a_1, \ldots , a_p$ of $f(x)$. Thus $\sigma$ must permute these roots in a cycle. This means that all the roots are in fact conjugate. Thus $f(x)$ is irreducible.


2

Here's a very rough sketch of how to get started: Suppose for contradiction $f$ were reducible. Then $f(x) = g(x)h(x)$, such that $\deg(g) + \deg(h) = p$. Choose any root $\alpha_1$ of $g$. then $\mathbb{Q}[\alpha_1]$ is a field extension of $\deg(g) <p$ degree. Suppose that, in this new extension, $g(x)$ now splits into $g(x) = \displaystyle ...


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When $a$ is not a $p$-th power in $F$, we can always define $F[\sqrt[p]{a}] = F[X]/(X^p-a)$. We can also specify that we are working within an algebraic closure. In characteristic $p$, $p$-th roots are unique, so this is even somewhat canonical.


2

Let $Fix(\sigma):=\{x\in F(\alpha_1,\ldots,\alpha_n)|\sigma(x)=x\}$. It follows straight from the definition that $Fix(\sigma)$ is a field. By assumption, $F\subset Fix(\sigma)$ and for all $i$ $\alpha_i\in Fix(\sigma)$, hence $Fix(\sigma)=F(\alpha_1,\ldots,\alpha_n)$.


2

The notation $F / K$ is just a traditional notation that indicates that $F$ is an extension of $K$, i.e. that $F$ contains $K$. So, for now, you can regard $F / K$ as the same thing as $F$, with the notation reminding you that $F$ contains the subfield $K$. When you start to study maps between fields, then the distinction takes on more significance. An ...


1

Just look at $f$ and $h$ in some splitting field. They have $x-\alpha$ as a common factor. Now the gcd can be calculated algebraically from $f$ and $g$ by Euclid's algorithm. Thus this gcd will be in any field that contains the coefficients of $f$ and $g$.


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Hint for approach #1: Use the fact in a finite field the non-zero squares form a cyclic subgroup of the multiplicative group, so the sum (well, half of it) is a segment of a geometric series such that... Hint for approach #2: Let $$ S=\sum_{\alpha\in K}\alpha^2=\sum_{\alpha\in K^*}\alpha^2. $$ Because $p^8>3$ (this is enough) there exists an element ...


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What you must do is check if the subset satisfies the subfield axioms. So for example for the first subset, if you want to check the closedness under addition, you take $x,x' \in A$, $y,y' \in B$. Then $x+y$ and $x'+y'$ are in the subset $X = \{ b+a \mid b \in B, a \in A \}$, so their sum must be in $X$ again. But their sum is $(x+y)+(x'+y') = ...


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An element of that field looks as $\;f(\alpha_1,...,\alpha_n)\;$ , with $\;f(x_1,...,x_n)\in F(x_1,...,x_n)\;$ Thus, we have that $$f(\alpha_1,...,\alpha_n)=\sum_{i_1,...,i_n}a_{i_1,...,i_n}\alpha_1^{i_1}\cdot\ldots\cdot\alpha_n^{i_n}\implies$$ $$\sigma f(\alpha_1,...,\alpha_n)=\sum_{i_1,...,i_n}\sigma ...


1

Associativity involves three elements, but they do not have to be distinct. So with a two element field $a+(b+c)=(a+b)+c$ means that whatever values $a,b,c$ take in the field, the equation is true. There are eight possibilities to check, which is not too bad. Associativity can be a pain to prove if you don't have a short cut. The associativity of the two ...


1

Define $F_0 = \mathbb{Q} \cup \{a\}$, which is clearly countable, as the union of a countable set and a finite set. Having defined $F_n$, define $F_{n+1}$ by $$F_{n+1} = \{x+y: x,y \in F_n\} \cup \{-x: x \in F_n\} \cup \{x\cdot y: x,y \in F_n\} \cup \{\frac{1}{x}: x \in F_n, x \neq 0\}$$ This defines a sequence of subsets of $\mathbb{R}$, and they are all ...



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