I understand how Galois fields of prime powers (e.g. $GF(2^8)$, $GF(17)$, $GF(257)$) behave. What I could never figure out is how they behave with sizes other than simple powers of primes. For example, $GF(15)$, $GF(100)$, or $GF(2^43^5)$.    Is it even possible?    Searching for this yielded no results.
2216
Let $F$ be a finite field. How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?
78390
Is $3^{2015} - 2^{2015}$ a prime. If not, why?
42691
Exactly what the name entails. The function $2^n-1$ I see largely tends to generate primes when $n$ is prime. However, a week ago I heard that this was horribly false. Please show me a disproof.
102517
$F$ is a finite field of size $s$. Prove that if $s=2^m$ for some $m&gt;0$, then all elements of $F$ have a square root.   (Hint: For some integer $i$, ($1\leq i \leq s-1$), $i$ is odd if and only if $(s-1)+i$ is even.)  My attempt:  If $s=2^m$, then $s$ is even. This means $s-1$ is odd. So the multiplicative group has an odd number of elements.  Let $x$ be the primitive element of $F$ such that: $F/\{0\}=\{x, x^2, x^3, ... , x^{s-1}\}$.  I tried to arrive at a contradiction assuming there existed an element without a square root, but I couldn't think of anything. Also I'm not quite sure how to use the hint.
43064
This is from A Course in Arithmetic by JP Serre          Theorem 1           ii) Let $p$ be a prime number and let $q = p^f(f \geq 1)$ be a power of $p$. Let      be an algebraically closed field $\Omega$ of characteristic $p$. There exists a  unique subfield $F_q$ of $\Omega$ which has $q$ elements. It is the set of roots of the polynomial$X^q-X$.           iii) All finite fields with $p^f$ elements are isomorphic to $F_q$.      The proof of the last part says          Assertion iii) follows from ii) and from the fact that all fields with $p^f$     elements can be embedded in $\Omega$, since $\Omega$ is algebraically closed.      Can someone explain me how iii) follows fom ii)?
81563
I have finite Galois extensions: E/K and E/L. $$M:=K \cap L$$ I am trying to prove that if the extension E/M is finite then it is also Galois. Any suggestions? Thanks
105119
How do I find the Galois group of $x^4-5$ over $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt5)$ and $\mathbb{Q}(\sqrt{-5})$? I've managed to do so over $\mathbb{Q}$ but I don't know how to find the others.  I'd appreciate any help. Thanks!
9606
We need to construct a field of 25 elements.  By using a result : For a prime p and a monic irreducible polynomial $p(x)$ in $\Bbb F_p [x]$ of degree $n$ , then the ring $(\Bbb F_p [x] / &lt;p(x)&gt;)$ is a field of order $p^{n}$.  So , for the question problem above , we need a irreducible polynomial of degree $2$ in $\Bbb Z_5 [x]$.  Solution says , $x^{2} + x +1$ does our job , since it's irreducible in $\Bbb Z_5 [x]$.  My question : Can't we take $ x^{2} + 2$ , it has no zeroes in $\Bbb Z_5$ , and is irreducible. In every solution I've gone through , the polynomial $x^{2} + x+1$ is taken. Can't we consider $ x^{2} + 2$ ?
64493
I need to describe all numbers of the form $x^2 + 2y^2$.  So far I've reduced the problem to primes, and showed p=2 satisfies it. I've also shown that any primes mod 5 or 7 can't be a written in this form. How do I proceed to show that it holds for all primes mod 1 or 3? I think I'll need to use quadratic reciprocity somehow (only since it is the topic of the homework). I have a hunch the supplementary quadradic recprocity laws will be important.  Any help would be appreciated. Thanks!
106169
I got the following task:  Let $F = GF(2^6 )$ be K[x] modulo the primitive polynomial $h(x) = 1 +x ^2 +x ^3 +x ^5 +x ^6$ , and let $\alpha$ be the class of x. I have a table with the binary representation of the elements of $GF(2^6 ) = \{0; 1; \alpha^1; \alpha^2 ; ...; \alpha^{62}\}$.  Is there a quicker way to check $\alpha^{52}$ than performing a long division of $x^{52}$ by $h(x)$ ?  It has $2^6$ elements so after $\alpha^{62}$ it will be back to 1 again, but maybe there is some trick to find it faster rather than performing such a long division?  I have also $\alpha^{13}$ so it's $\alpha^{13^4}$, but no idea how to use it
113514
How large should the prime $p$ and generator $g$ values be in a Diffie-Hellman handshake if the messages are encrypted.   If the key that encrypted the Diffie-Hellman messages becomes compromised, then does that compromise the key generated using Diffie-Hellman?
121335
GF(q) is a finite field of order q, where q is odd. Prove that $a\in GF(q), a\neq0$ has a root in $GF(q)$ iff $a^{(q-1)/2}=1$.  I tried to prove it this way:  Suppose a has a root in $GF(q)$, so there is some $b\in GF(q)$ so that $b^2=a$. There is a primitive element $g$ in $GF(q)$, so that $g^x=a, g^y=b$. Therefore $(g^y)^2=g^x\to 2y=x$. Furthermore $a^{(q-1)/2}=(g^x)^{(q-1)/2}=g^{(x/2)*(q-1)}=g^{y(q-1)}$ and since $y$ is an integer, $g^{y(q-1)}=1^y=1$. Now suppose $a^{(q-1)/2}=1$, then again $(g^x)^{(q-1)/2}=1\to g^{(x/2)*(q-1)}=1$. That means that $x/2$ is an integer (or else $g^{(x/2)*(q-1)}\neq 1$), and therefore $0\le x/2\le q-1$ being an integer, $g^{x/2}$ creates a certain element $b\in GF(q)$ so that $b^2=a$.  I am a bit unsure of this proof - first of all since I don't use the fact that $q$ is odd, which means I am wrong somewhere. Second of all, I feel extremely 'weak' with my argument regarding $x/2$ being an integer, yet I have no idea how to strengthen it. Any hints/directions will be extremely appreicated!
52950
I have seen many textbooks state this result without proof.   $``$ If $E$ is the splitting field for the polynomial $f=x^p-1 \in \mathbb{Q}[X]$ where $p$ is prime, then the Galois group $Gal(E:Q)\simeq \mathbb{F}_{p}^{\times} \simeq\mathbb{Z}_{p-1} . "$  I would like to know how to prove this.  The things that I know from reading are that $E=Q(\zeta)$, where $\zeta$ is a root of the irreducible polynomial $x^{p-1} + x^{p-2} \dots + 1$. So $E$ a cyclotomic extension of Q of degree $p-1$ and that the roots of $x^p - 1$ are the $p^{th}$ roots of unity. Also that the Galois group is comprised of the automorphisms of $E$ that fix $\mathbb{Q}$ and permute the roots of $f$.  As a second question, if I am asked to prove that the Galois group $Gal(E/Q)$ is cyclic, when $E$ is the splitting field for some specific $f$ and $p$ is small, say $5$ or $7$, is there an easier way to prove this than first proving the more general statement above and then drawing the conclusion from that?   Any help will be much appreciated! Thanks.
7115
I have some problems with this exercise. I don't know if it can be done. Consider the polynomial $ x^n - a \in \mathbb{Q} $ Can I compute the Galois group of this over $\mathbb{Q}$? Maybe having a nice "basis.  The splitting field is given by $\mathbb{Q}(\zeta_n,\alpha)$ , where $\zeta_n$ is a primitive root of unity , and $\alpha$ is some number such that $\alpha^n = a $. Well first of all, I want a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different (in this case the computation of the galois group is very simple).  For example one easy case, it's when $a&gt;0$, then $(a)^{\frac{1}{n}} \in \mathbb{R}$ , so clearly the minimal polynomial of $(a)^{\frac{1}{n}} , \zeta_n$ are distincs, and I'm done. If $n$ is odd then , it's also easy, since one root it's also real, for example $x^3-3 $, the real root is $ \root 3 \of { - 3}  =  - \root 3 \of 3  $ , so I can consider the splitting field as $\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )$. The difficult case is when $n$ is even and $a&lt;0$ , for example  $x^8+20$ or $x^4+20$ in some cases as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general. It can be done?  $\underline{Remark}$ I'm searching a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different since in this case the computation of the galois group is very simple.
493
The following is a well known problem   If $a|b^2; b^2|a^3; a^3|b^4; ....$ then $a=b$.  I started with $a$ is a prime $p$ and the conclusion immediately holds.  So I was comparing powers of primes but can't get anything useful.  Please help.
28798
I am reading a book about algebraic number theory and I'm wondering whether the following statement is true:  Let $K$ be a number field and $\mathcal{O}_K$ its algebraic integers. Let $\beta \in \mathcal{O}_K$, is it true that $N(\beta)/\beta \in \mathcal{O}_K$? ($N$ is the norm of K over $\mathbb{Q}$)  If $K$ is a Galois extension of $\mathbb{Q}$ then we have that the norm is just the product of all Galois conjugates of a given element. Thus $N(\beta)/\beta$ would be the product of all Galois conjugates of $\beta$ except the one corresponding to the identity. Since Galois conjugates of an algebraic integer are also algebraic integers, the above follows.   Now I'm struggling with what happens when $K$ is only a separable extension. (separable always holds since we are in characteristic 0)
91247
Let $p$ be a prime number, and let $\mathbb{F}_p$ be the field with $p$ elements. How many elements of $\mathbb{F}_p$ have cube roots in $\mathbb{F}_p$?   I had this question on an exam and after reviewing I am still not sure. Any help would be appreciated.