This is the problem 1.1 from the book, A Gentle Course in Local Class Field Theory.  Let $p$ be a prime, let $F$ be a field of characteristic $\neq p$, and let $a \in F^{\times} \backslash F^{\times p}$. We make no assumption on the presence of roots of unity in $F$, and we wish to prove that $[F(\alpha):F]=p$, where $\alpha$ is a root of $X^p-a$. For this, we assume $[F(\alpha):F] \neq p$, and work toward a contradiction:   Consider the composite field $F(\mu_p, \alpha)$, and show that $F(\alpha) \subset F(\mu_p)$.   2. Deduce that $F(\alpha)/F$ is Galois, and then that $F(\alpha)=F(\mu_p)$   From the fact that $[F(\alpha):F]$ does not depend on the choice of root $\alpha$, conclude.   Note that $\mu_p$ stands for a pth primitive root of unity in the algebraic closure of $F$, aka $F^{alg}$. Also, $F^{\times p} = \{f^n|f  \in F\}$.  I've had a hard time proving that $F(\alpha) \subset F(\mu_p)$ and $F(\alpha)/F$ is Galois, and didn't get how the assumption for absurdness would fit in the proof. Could anyone please give some help on this?   I have been aware that the problem is kind of equivalent to the proposition that $x^p-a$ is irreducible iff $x$ has no root in $F$ (there is an answer regrading this). However, I am trying to understand the way from the book.
22569
Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:     Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field  and let $c \in  F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$   has no root  in $F$. [Hint: consider two cases: char$(F) = p$ and   char$(F)$ different of $p$.]   I have attempted this a lot. Anyone has an answer?
59613
I need help to solve the following exercise:   If $K$ is an algebraic extension of $\mathbb{Q}$ (finite or infinite), then we let $O_{K}$ denote the ring of algebraic integers in $K.$ If $\frak{P}$ is a prime ideal of $O_{K}$ then $\frak{P}∩\mathbb{Z}$ is a prime ideal of $\mathbb{Z}.$ Moreover, if $\frak{P}$ is non-zero, then so is $\frak{P}∩\mathbb{Z}$ . Thus $\frak{P}∩\mathbb{Z}=$$ p\mathbb{Z}$ for some prime $p.$ Show that $O_{K}/\frak{P}$  is a field, and is an algebraic extension of $\mathbb{F}_{p}$; then show that it is Galois over $\mathbb{F}_{p}$ with abelian Galois group.  Now suppose that $K/F$ is a (finite or infinite) Galois extension. Let $\frak{P},$  $O_{K}$ be as before and put $\frak{p}=\frak{P} \cap $$O_{F}$, a prime ideal of $O_{F}.$  Show that $\mathrm{Gal}(K/F)$ acts transitively on the set of prime ideals of $O_{K}$ above $\frak{p}.$   Thanks in advance
91654
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.
78866
I have proven that in $\mathbb F_{p^2}^*$ exists an element $\alpha$ with $\alpha^8 = 1$.      Let $f(X) := X^4+1 \in \mathbb F_p[X]$. How can I prove that $f$ is reducible over $\mathbb F_p$?    Has $f$ a zero in $\mathbb F_p$ ?
107461
I have to prove the following theorem :     Let $p$ be a prime number and let $n \ge 1$,be any integer, then there exists a field of order $p^n$.   My attempt  I started off by considering the polynomial $f(x)$=$x^{p^n}-x \in \Bbb Z_p[x]$.  I took $F$ to be the splitting field of $f(x)$ over $\Bbb Z_p$.  Since $F$ is the splitting field of $f(x)$ over $\Bbb Z_p$ therefore $f(x)$ has exactly $p^n$ zeros in $F$ counting multiplicity.  Also since $f'(x)$ = $p^nx^{p^n-1}-1$ , therefore $f(x)$ will not have any multiple zeros and hence it will have $p^n$ distinct zeros in $F$.  Then I took $K =\{k \in F | k^{p^n} = k\}$.  I was trying to prove that $K$ is a sub-field of $F$.  $1 \in K$ thus $K$ is non-empty.  For $a,b \in K$ $a^{p^n}=a$ and  $b^{p^n}=b$  So $(a+b)^{p^n}$ (should be) = $a^{p^n}$+$b^{p^n}$, this is where I am currently stuck!  I know that this is true when  $p$ is the characteristic of a field, but why would this expansion be true here?
86188
Let $E$ be an extension field of $F$ and $\alpha \in E$. Then $\alpha$ is transcendental over $F$ if and only if $F(\alpha)$ is isomorphic to $F(x)$, the field of fractions of $F[x]$.   This was a theorem in an abstract algebra textbook with a very brief proof. Can someone please explain why this theorem holds? I'm having difficulty grasping the concepts at hand.  Thanks!
44179
I'm trying to prove that the characteristic of any field $F$ is either $0$ or a prime number, but I have no idea what to do.  Help?
48811
Given an arbitrary finite field $K$ (not necessarily $\mathbb{F}_p$ with $p \in \mathbb{P}$) with $|K| = q$ and an irreducible polynomial $f$ with $\alpha$ as root and degree of $n$. Is $|K(\alpha)| = q^n$ and why? Its clear to me for $K$ isomorphic to $\mathbb{Z}_p$ with $p$ is a prime number.
28718
Let $A$ be a UFD, $K$ its field of fractions, and $f$ an element of $A[T]$ a monic polynomial.  I'm trying to prove that if $f$ has a root $\alpha \in K$, then in fact $\alpha \in A$.  I'm trying to exploit the fact of something about irreducibility, will it help? I havent done anything with splitting fields, but this is something i can look for.
119238
Let $F$ be a field and $K$ a field extension of $F$. Suppose $a,b\in K$ are algebraic over $F$ with degrees $m$ and $n$, where $m,n$ are relatively prime. Then $F(a) \cap F(b) = F$.  I see that the intersection on the LHS must contain $F$, but I don't see why $F$ contains the LHS.
109312
Let $p$ be a prime. Prove that $\mathbb{F}_p[X]/(X^2+X+1)$ is a field iff $p \equiv 2\bmod3$.    So:   If $p \equiv 2 $ mod $ 3$, I have to show that every element of $\mathbb{F}_p[X]/(X^2+X+1)$ has an inverse.  If $\mathbb{F}_p[X]/(X^2+X+1)$ is a field I have to show $p \equiv 2 $ mod $ 3$.
27428
Let $K$ be a field and $f \in K[X]$ of degree $n$ with Galois group $S_n$. Let $a$ be a root of $f$, $L = K(a)$, and let $E$ be a intermediate field of the extension $K \subset L$. Prove that $E = K$ or $E = L$.   Any suggestions on where to start? Thanks.
48704
This is problem 3.18 from Patrick Morandi's Field and Galois Theory.     Give an example of fields $k \subseteq K \subseteq L$ and $l \subseteq L$ for which $l/k$ and $L/K$ are algebraic, $k$ is algebraically closed in $K$, and $lK = L$, but $l$ is not algebraically closed in $L$.   This question has been asked twice, namely  and , though neither had any answers, and only some inconclusive comments. Normally I wouldn't ask the question again, but after struggling with it for weeks at this point, I emailed Morandi, and he gave me some hints.  He told me that he remembers finding an example, but that he isn't sure. He told me that it would need to be over characteristic $p$ and that the extensions (at least some of them I suppose) could not be separable. Moreover, he told me that $K = k(x)(a)$, where $a$ is a root of an irreducible polynomial $f(x,y) \in k(x,y)$ (I'm not 100% what the root of a two variable polynomial is). Finally, he said that $l$ is an algebraic extension of $k$, where $f$ becomes reducible over $l$. Still however, I can't come up with anything. I was hoping that with these extra hints, someone would be able to figure it out.  He did point out that there is a chance that there is an error in the question, but I haven't managed to prove that such an example cannot exist.
12430
Let $\mathbb{F}_{27}$ denote the finite field of size 27. For each $\alpha\in\mathbb{F}_{27}$ , define $ A_{\alpha}$ $=$ { $ 1, 1+\alpha , 1+\alpha+{\alpha}^2 , 1+\alpha+{\alpha}^2+{\alpha}^3 , .... $} . Then  prove that  the number of  $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|$=26 equals 12.  $ 0\in A_\alpha $ if and only if $\alpha\neq 0$ For the first proof , the multiplicative group of the field $\mathbb{F}_{27}$ is cyclic and its order is 26  , so there are 12 generators in that group.   Now  I need  to  show  that  $ 1+\alpha+{\alpha}^2+.. {\alpha}^{25} $ =0  , where order of $\alpha$=26 . Since   $\alpha ^{26}$ =1 ($\alpha$-1)(1+$\alpha$+${\alpha}^2$+...+${\alpha}^{25}$)=0 , since $\alpha\neq 1$  , So $ 1+\alpha+...+{\alpha}^{25} $ =0 My question is , can I write these last two  lines?  I mean are those two lines alright to write? If those two lines are invalid then I need some help to show  $ 1+\alpha+{\alpha}^2+..+{\alpha}^{25} $=0  .Thanks
4760
I want to argue that argue that $\pi(\alpha)=\alpha^3+3\alpha+3$ is an irreducible polynomial over the finite field with 5 elements $\mathbb{F}_5$. My approach was just to check that $\pi$ has no roots in $\mathbb{F}_5$. Is this right?
19029
Let $\alpha_1,\ldots, \alpha_n$ be roots of unity, and let $a=\frac{1}{n}\sum\alpha_i$. Then if $a$ is an algebraic integer, we have either $a=0$ or $a=\alpha_1=\dots=\alpha_n$. Why?
16690
Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$?  One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge |F|$ then I can construct an irreducible polynomial, namely  $ p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )$  where $a_j$ are all the field elements. It is clear that $p(x)$ has no roots in $F$.  This trick doesn't work for $n &lt; |F|$. A counter-example: Let $F = \mathbb{F}_3$ and $p(x) = 1 + (x-1)(x-2)$, then $p(1) = 1$, $p(2) = 1$, $p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3$.  I know there is a way to count them using the Möbius function $\mu(n)$ but I want a proof without it that just shows existence.