In my abstract algebra class we discussed the idea that for a finite field $F$, that the characteristic of $F$ is a prime number. The proof would go more or less like : Suppose that $char(F) = nm$ for $n,m \in \Bbb Z$ and $n,m \geq 2$. Then $nm(1_F)=n(1_F)m(1_F)=0$ implying that either $n(1_f)=0$ or $m(1_f)=0$ since $F$ is an integral domain, but this contradicts the minimality of $char(F)$. This lead me to two questions:  $1$)What effect does $char(F)$ have on the size of $F$  $2$)Are there finite fields of cardinality $k$ for any $k\in \Bbb Z, k \geq 2$
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?
74055
In the integral domain every prime is irreducible. But the converse is not true, for example, $1+\sqrt{-3}$ is an irreducible but not a prime in ${\Bbb Z}[\sqrt{-3}]$. In a UFD, "prime" and "irreducible" are equivalent.   Here is my question:     Is there a non-UFD integral domain such that prime is equivalent to irreducible?
3605
Blue was correct, I need to fix my understanding of this:     Finite fields have cardinality of a prime order because they have a prime subfield that has finite characteristic.    I do not completely understand why finite field of each characteristic is unique. Perhaps we look at a polynomial and express the finite field as a splitting field? And we use uniqueness of splitting fields?  Then, why is it that  $\mathbb F_{p^m} \subseteq \mathbb F_{p^n}$ if and only if $m$ divides $n$?
54881
I am told that $K \leq L$ is a field extension of degree 2 and that $K$ has characteristic not 2. I am then asked to show that $\exists \: \alpha \in L$ such that $L = K(\alpha), \; \alpha^2\in K $  So far I have that:  $\exists \; \alpha_1, \alpha_2 \in L$ such that $L = K( \alpha_1, \alpha_2)$  $\Rightarrow |L:K| = |L:K(\alpha_1)| |K(\alpha):K| $ so $|K(\alpha):K| = 1$ or $2$  $|K(\alpha_1):K| = 2 \Rightarrow L = K(\alpha_1)$ and $deg(f_{\alpha_1}(t)) = 2  $ where $f_{\alpha_1}(t)$ is the minimal polynomial of $\alpha_1$ over $K$   So $f_{\alpha_1}(t) = t^2 + at + b$ for some $a,b \in K$  Also, $|K(\alpha_1):K|=1 \Rightarrow K(\alpha_1) = K$ and $L = K(\alpha_2)$ which means $degf_{\alpha_2}(t) = 2$ where $f_{\alpha_2}(t)$ is the minimal polynomial of $\alpha_2$ over $K$.   So $f_{\alpha_2}(t) = t^2 + at + b$ for some $a,b \in K$  Given that we consider the same polynomial each time, relabel to consider:   $f_{\alpha}(t) = t^2 + at+ b $ for some $a,b \in K$  We have already demonstrated that $L=K(\alpha)$ for some $\alpha \in L$ so now we much show that $\alpha^2 \in K$  $\alpha^2 \in K \Leftrightarrow a = 0_K$  I now have no idea how to show that this coefficient of $f_{\alpha}(t)$ is necessarily $0$.   I know that $f_{\alpha}(t)$ has two roots in $L$ and that because $K$ and thus $L$ has characteristic not 2, they cannot be the same root, but I don't know how to then conclude that they must sum to $0$  Any help would be greatly appreciated.
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.
113998
I'm having a hard time showing this:     If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.   I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.  Any help would be appreciated.
51676
If $k&gt;0$ is a positive integer and $p$ is any prime, when is $\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in\mathbb Z_p\}$ a field? Find necessary and sufficient condition.   Attempt: Since we know that a finite integral domain is a field and since $\mathbb Z_p[\sqrt{k}]$ is finite, it suffices to find the condition when $\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in \mathbb Z_p\}$ forms an integral domain.  The given set forms a commutative ring with unity. Hence, the only condition that needs to be satisfied is the absence of any zero divisors.  $(a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k) = (a_1a_2+b_1b_2k)+(a_1b_2+a_2b_1)\sqrt k$  Case $1$: When $k$ is not a perfect square  $(a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k)=0 \implies (a_1a_2+b_1b_2k) \bmod p=0~~ \&amp; ~~ (a_1b_2+a_2b_1) \bmod p =0$  Case $2$: When $k$ is a perfect square $ = u^2$  $(a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k) = 0 \implies (a_1a_2+b_1b_2 u^2 + (a_1b_2+a_2b_1)u ) \bmod p=0 $  Both the above cases can have many cases and seems a bit complicated. Am I missing out on something? The book which I am reading (Gallian) hasn't introduced Quadratic Residues as of yet. Thank you for the help.
74924
I know that $1$ is not a prime number because $1\cdot\mathbb Z=\mathbb Z$ is, by convention, not a prime ideal in the ring $\mathbb Z$.  However, since $\mathbb Z$ is a domain, $0\cdot\mathbb Z=0$ is a prime ideal in $\mathbb Z$. Isn't $(p)$ being a prime ideal the very definition of $p$ being a prime element?  (I know that this would violate the .)    Edit: Apparently the answer is that a prime element in a ring is, by convention a non-zero non-unit (see ).  This is strange because a prime ideal of a ring is, by convention, a proper ideal but not necessarily non-zero (see ).  So, my question is now: Why do we make this awkward convention?
77003
Is there any field of characteristic two that is not $\mathbb{Z}/2\mathbb{Z}$?  That is, if a field is of characteristic 2, then does this field have to be $\{0,1\}$?
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.
104963
Reading Chapter 2 of Koblitz's Introduction to Elliptic Curves and Modular Forms, I got stuck on the following question. I would like to proceed my reading, so I would appreciate any hint to this. I have no background in Algebraic Number Theory, but if someone explains in that language, I would like to see it and learn it as much as I can at this moment. References will also be appreciated!  Let $q$ be a power of prime $p$ such that $q \equiv 1 \mod 4$ and $p \nmid 2n$.  Question. Let $\chi_{4} : \mathbb{F}_{q}^{*} \rightarrow S^{1}$ be the character of order $4$, show that $\chi_{4}(-4) = 1$.  The hint in the book says we may show that $\chi_{4}(4) = \chi_{4}(-1) = 1$ for $q \equiv 1 \mod 8$ and $\chi_{4}(4) = \chi_{4}(-1) = -1$ for $q \equiv 5 \mod 8$.  If $q \equiv 1 \mod 8$, then $\chi_{4}(4) = \chi_{4}(2)^{2} = \chi_{2}(2) = \pm 1$, where $\chi_{j}$ denotes the character of order $j$. So I think $2$ must be quadratic residue in this case, but I cannot figure out why.
98034
I'm trying to understand the following explicit construction of a finite field with $p^n$ elements where $p$ is prime, $n \geq 2$:  Take any irreducible polynomial $f(X) \in \mathbb{F}_p$ of degree $n$. Then $\mathbb{F}_p[X] / (f(X))$ is a field with $p^n$ elements.  Now, I understand that if $f$ is irreducible, the ideal generated by $f$ is maximal, so $\mathbb{F}_p[X] / (f(X))$ is indeed a field.  I don't understand the following points:   Why does for every $n \geq 2$ an irreducible $f(X) \in \mathbb{F}_p[X]$ of degree $n$ exist? Why does the field $\mathbb{F}_p[X] / (f(X))$ have $p^n$ elements?   Thanks in advance for any help!
100951
I'm trying to show that the minimal polynomial of a linear transformation $T:V \to V$ over some field $k$ has the same irreducible factors as the characteristic polynomial of $T$. So if $m = {f_1}^{m_1} ... {f_n}^{m_n}$ then $\chi = {f_1}^{d_1} ... {f_n}^{d_n}$ with $f_i$ irreducible and $m_i \le d_i$.  Now I've managed to prove this using the primary decomposition theorem and then restricting $T$ to $ker({f_i}^{m_i})$ and then using the fact that the minimal polynomial must divide the characteristic polynomial (Cayley-Hamilton) and then the irreducibility of $f_i$ gives us the result.  However I would like to be able to prove this directly using facts about polynomials/fields without relying on the primary decomposition theorem for vector spaces. Is this fact about polynomials true in general?  We know that $m$ divides $\chi$  and so certainly $\chi = {f_1}^{m_1} ... {f_n}^{m_n} \times g$ but then how do we show that $g$ must have only $f_i$ as it's factors? I'm guessing I need to use the fact that they share the same roots. And I'm also guessing that it depends on $k$, i.e. if $k$ is algebraically closed then it is easy because the polynomials split completely into linear factors.  Help is much appreciated, Thanks
90126
There is something I do not understand in my lecture notes, mainly because I do not understand quotients very well. We consider a number field $K = \mathbb{Q}(\sqrt{d})$ where $d$ is squarefree.  If $\mathcal{O}_K$ denotes the ring of integers of $K$, it has a basis $\langle 1, \omega \rangle$ where $\omega$ is either $\sqrt d$ or $\frac{1+\sqrt{d}}{2}$ depending on $d \bmod 4$. We let $f(x) \in \mathbb{Z}[x]$ be the minimal polynomial of $\omega$ (so $f$ is monic of degree 2).  What I do not understand is first of all that we can write  $ \mathcal{O}_K = \mathbb{Z}[x] / f(x) $. Also, if $p \in \mathbb{Z}$ is prime, the notes then say $$ \frac{\mathcal{O}_K}{(p)} = \frac{\mathbb{Z}[x]}{(p,f(x))} = \frac{\mathbb{F}_p[x]}{f(x) \bmod p} $$  I do understand how to go from the first expression to the second(trivial when you know the above) but not how to go from the second to the third.  Also, the lecturer then gives the expression $\frac{\mathcal{O}_K}{(p)}$ when $p$ is ramified, split or inert. I do not really see why, in the sense how does it help to know that when $p$ is inert then  $$ \frac{\mathcal{O}_K}{(p)} = \frac{\mathbf{F}_p}{f(x)} = \mathbf{F}_{p^2} $$ for example ?  Thank you very much for your help !
46247
I saw it's been answered before but I could not understand the answers as it delved into material that I did not study yet.  My question is as follows:  Suppose $p \in F[x]$ is the minimal polynomial of some matrix $A \in M_n(F)$ and $F \subseteq K$  I need to show that $p$ is also the minimal polynomial of $A$ over the field $K$. In other words: Show that the minimal polynomial remains the same when we move to a bigger field.  What I thought of:  Obviously $p$ is irreducible in $F$, suppose it is reducible in $K$, which means there are $q,t \in K, q,t \notin F$ such that $p=qt$, now we just need to show that $q(A) \neq 0$ and $t(A) \neq 0$...But why is that so?  Edit: Another way of presenting this question is "Show that if $p$ is the minimal polynomial of $A$ over the field $K$ while the entries of$~A$ are contained in a subfield $F \subseteq K$, then all the coefficients of $p$ are elements of $F$".
123230
I am looking at   and trying understand the proof for the theorem that states:  Given a nonzero ideal I in $P(F)$, there is a monic polynomial $p(t)$ such that   $$I=\left\{q(t)p(t) | q(t) \in P(F) \right\}$$  Proof goes like  Since $I$ contains at least one nonzero element, we can let $p(t)$ be a nonzero element of $I$ of minimum degree. We can show that every element of $I$ is a multiple of $p(t)$. Let $f(t) \in I$. Then there exist $q(t),r(t) \in P(F)$ such that  $$f(t) = q(t)p(t) + r(t)$$ and deg $r(t)$ &lt; deg $p(t)$. Since $p(t) \in I$, so is $q(t)p(t)$, and therefore so is $$r(t) = f(t) -q(t)p(t)$$.  By the choice of $p(t),r(t) = 0$, so $f(t)$ is a multiple of $p(t)$.  What I am having trouble understanding is why $r(t)$ must equal $0$. If someone can explain why it is, I'd appreciate it.
32407
I want to prove the following theorem: Theorem. The minimal polynomial of $\alpha$ has integer coefficients if and only if $\alpha$ is an algebraic integer. Here is what I have done and I want to ask whether it has any problematic parts or not: ($\implies$) part is trivial. For the converse, let $q(X) \in \mathbb{Q}[X]$ be the minimal polynomial of $\alpha$ and $f(X) \in \mathbb{Z}[X]$ such that $f(\alpha)=0$. Then, since $q(X)$ is the minimal polynomial of $\alpha$, $f(X)=q(X)r(X)$ for some $r(X) \in \mathbb{Q}[X]$. Assume that $q$ has at least one non-integer coefficient. Define $ V = lcm\{$denominators of coefficients of $q(X) \}$. $ U = lcm\{$denominators of coefficients of $r(X) \}$. Then, $UVf(X) \equiv 0$ modulo ($UV$) but $Vq(X)Ur(X) \equiv UVq(X)r(X) \not \equiv 0$ . So, I think we are done.