I'm reading David R. Finston and Patrick J. Morandi's book  and in section 7.2 page 109 it mentions     It’s true that every finite field has a primitive element. However, the   proof of this fact involves more complicated ideas that we will   consider, so we do not give the proof.   I understand some proof might be hard, based on more complex concepts, but leaving this dangling it’s a bit itchy .. May I ask where can I find a relative simple proof?
25850
Why is the multiplicative group $(K\smallsetminus\{0\},\cdot)$ of a finite field $(K,+,\cdot)$ always cyclic?
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I'm trying to understand a proof of  which says that in a integral domain every prime element is irreducible:    I didn't understand why this implication $p=ab\implies p|a$ or $p|b$, is not the contrary $p=ab\implies a|p$ and $b|p$ ?  I'm a little confused  Thanks in advance
55925
I'm reading Fraliegh's text on abstract algebra, and his statement for the factor theorem is, "An element $a\in F$ ($F$ is a field) is a zero of $f(x)\in F[x]\iff$ $(x-a)$ is factor of $f(x)$ in $F[x]$." The proof is straightforward, and seems to be the most common one, but I don't see where exactly you need to use the fact that $F$ is a field; it seems like this result would hold for any ring $R$. Is this true?
70031
My problem is to build, for every prime $p$, a field of characteristic $p$ in which every polynomial of degree $\leq n$ ($n$ a fixed natural number) has a root, but such that the field is not algebraically closed.  If I'm not wrong (please correct me if I am) such a field cannot be finite, by counting arguments. But on the other hand, the union of all finite fields (or of any ascending chain of finite fields) of characteristic $p$, which is what I get if I start with $F_p$ and add a root to each polynomial of degree $\leq n$ in each step, is the algebraic closure of $F_p$, hence algebraically closed. I don't see how I can control this process so that in the end I get a field that is not algebraically closed.  Any hint will be welcome. Thanks in advance.
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I'm currently reading S. Lang's &quot;Undergraduate Algebra&quot;. After the primitive root element theorem (Field theory chapter), there are a bunch of exercises to find one primitive element of extensions and, then, their degrees. However, I don't even know how I should start. They are as below:   In each case find an element $\gamma$ such that $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\gamma)$. Prove every  statement you make.  a) $\alpha = \sqrt{-5}, \beta = \sqrt{2}$ b) $\alpha = \sqrt[3]{2} , \beta = \sqrt{2}$ c) $\alpha = $ root of $t^3 -t + 1$ , $\beta = $ root of $t^2-t-1$ d) $\alpha = $ root of $t^3 -2t + 3$, $\beta = $ root of $t^2 + t + 2$ $\quad$2. Find the degrees of the fields $\mathbb{Q}(\alpha, \beta)$ over $\mathbb{Q}$ in each case of exercise 1.  I think that exercises a) and b) proceed pretty much the same way, but I'm not sure about c) and d).
60751
Here is a proof I saw somewhere of the fact $R/I$ is a field if and only if $I$ is maximal:  $\implies$ Suppose that $R/I$ is a field and $B$ is an ideal of $R$ that properly contains $I$. Let $b \in B$ but $b \notin I$. Then $b + I$ is a nonzero element of $R/I$ and therefore there exists an element $c + I$ such that $(c + I)(b + I) = 1 + I$. Since $b \in B$ we have $bc \in B$. Because $1 + I = (c + I)(b + I) = bc + I$ we have $1 - bc \in I \subset B$. So $1 = (1-bc) + bc \in B$. Hence $B = R$.  $\Longleftarrow$ Now suppose $I$ is maximal and let $b \in R$ but $b \notin I$. Consider $B = \{br + a \mid r \in R, a \in I \}$. This is an ideal properly containing $I$. Since $I$ is maximal, $B = R$. Thus $1 = bc + a^\prime$ for some $a^\prime \in I$. Then $1 + I = bc + a^\prime + I = bc + I = (b + I)(c + I)$.  I thought this was fairly long so I tried to come up with a shorter proof. Can you tell me if this is right:  $\implies$ Assume that $R/I$ is a field and $I$ is not maximal. Then there exists an $x \in R - I = I^c$ that is not a unit (otherwise $I$ would be maximal). Then $x + I$ does not have an inverse hence $R/I$ is not a field.  $\Longleftarrow$ Assume $I$ is maximal and $R/I$ is not a field. Then there is an $x$ such that $x + I \neq 0 + I$ does not have an inverse. This $x$ is not in $I$ and $x$ is not a unit. Hence $I \subsetneq I + (x) \subsetneq R$. Which contradicts $I$ being maximal.
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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)?
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I'm looking for a simple proof that up to isomorphism every group of order 2p (p prime) is either $\mathbb{Z}_{2p}$ or $D_{p}$ (The Dihedral group of order 2p).  I should note that by simple I mean short and elegant and not necessarily elementary. So feel free to use tools like Sylow Theorems, Cauchy Theorem and similar stuff.  Thanks a lot!
71313
the definition of a prime subfield that I have been provided with is the following:      It can be shown that for any subset $S$ of $F$ there exists a minimal subfield $K$ of $F$ which contains $S$. One very important case is when the set $S$ is empty, i.e., every field $F$ contains a unique minimal subfield. This subfield is called the prime subfield.   I further know that any subfield $K$ of a field $F$ contains the elements $0$ and $1$ and is closed under addition, multiplication, and taking negatives and inverses of non-zero elements.   And since the prime subfield is the intersection of all possible subfields across $F$, why is it not the case that the prime subfield is always equal to $0$ and $1$?   Thank you for the help!
122655
I'm just wondering if the following theorem is reasonable and whether the proof is makes sense or not? Also I have an application of it which I was trying to do earlier but I made a lot of mistakes so I would be very grateful if anyone would give some comments on this in case I have some very strong misunderstandings about Galois theory.    Theorem Let $E_1$ and $E_2$ be Galois field extensions of $F$ with trivial intersection ($E_1 \cap E_2 = F$), then $E_1 E_2$ has Galois group $\text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.  Proof   $E_1$ and $E_2$ are splitting fields (over $F$) of some polynomials $f_1$ and $f_2$ so $E_1 E_2$ is the splitting field of $f_1 f_2$ over $F$ and therefore it is Galois and we are justified using the "Gal" notation. Let $\sigma_1 \in \text{Gal}(E_1/F)$ and $\sigma_2 \in \text{Gal}(E_2/F)$ then by the fundamental theorem of Galois theory we can lift these both into elements of $\text{Gal}(E_1 E_2/F)$. The claim is that $\sigma_1 \sigma_2 = \sigma_2 \sigma_1$, to see this consider an element $\alpha \in E_1$, we have both $\sigma_2 \alpha = \alpha$ and $\sigma_2 \sigma_1 \alpha = \sigma_1 \alpha$ because it is invariant with respect to $\sigma_2$, but we can combine these equalities to get $\sigma_1 \sigma_2 \alpha = \sigma_2 \sigma_1 \alpha$. Both groups $\text{Gal}(E_i/F)$ ($i = 1,2$) are included in $\text{Gal}(E_1 E_2/F)$ and the degrees say there are no more elements, furthermore since they commute we can conclude $\text{Gal}(E_1 E_2/F) = \text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.     So hopefully that theorem is, if not correct, fixable.. and can be used to prove that $\mathbb Q (\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})$ has Galois group $C_2^m$ ($m \le n$) over $\mathbb Q$?
111869
The problem is to show that a polynomial $f(x) \in F[x]$ (F is a field) has no repeated roots if and only if f(x) and f'(x) (the derivative of f(x)) are relatively prime. I've managed to prove one direction of this equivalence (if there are no repeated roots, f(x) and f'(x) are relatively prime), but the other one gives me headaches... can anyone help out?
3848
I read ampersand's question , where he asked why Shamir's Secret Sharing Scheme uses arithmetic in a finite field of prime order.  The answers to that question explain that a prime field is not necessary for Shamir's scheme, but, rather, that any finite field can be used.  However, they don't really address the other part of the question, namely why do we need a finite field at all?  Couldn't we just use, say, ordinary integer arithmetic instead?  Can someone please explain (in the simplest possible manner) the reason that Shamir's Secret Sharing Scheme uses finite field arithmetic?
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!
48145
The question states:     Let $g$ by a primitive root of the odd prime $p$. Show that $-g$ is a primitive root , or not, according as $p \equiv 1 \pmod 4$ or not.   For me, I cannot see any connection between the type of primes and the primitive root. Any Hint is highly appreciated.  Thanks
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.
81578
One can see on every text (book, lesson, comments) that a direct sum/coproduct of abelian groups is the same as a finite product but in the infinite case, the direct sum/coproduct is only a subgroup of the product.  Why does an element in an infinite direct sum only have finitely many components different from the neutral elements?  I don't see where the "finite" comes from when I look at the universal property.  Also, in another situation, a condition of "finiteness" appears and I think that is because of the same structure: a freely generated group, vectors space, algebra etc... has elements that are finite product/sum of generating elements. Because, if one really wants an infinite "sum" (or sequence) one may just take a formal serie. That is a ring, hence also an abelian group.
7186
I am trying to understand the proof for the following theorem:     Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible.   Here is the proof:    I typed this a while ago and I don't understand the part where if $x | bc$, then $x=bc$? Is something wrong at this step?  Also, is the definition of prime elements where $p$ is prime if whenever $p|ab$, then either $p|a$ or $p|b$? Now I don't feel so sure. This could be the reason why I am not understanding the proof...