abstract algebra - Two elements of $mathbb F_{q^m}$ that share the same minimal polynomial over $mathbb F_q$ have the same multiplicative order

Let $\mathbb F_{q^m}$ be the field with $q^m$ elements, where $q$ is a prime number (constructed as the quotient field $\mathbb F_q[x]/(p)$, where $p$ is an irreducible, monic polynomial over $\mathbb F_q$). Let $\alpha, \beta \in \mathbb F_{q^m}$, so that $\alpha$ and $\beta$ share the same the same minimal polynomial $m_\alpha$ over $\mathbb F_q$.

I now want to show that $\alpha, \beta$ have the same order in $\mathbb F_{q^m}^\times$ (the multiplicative group of $\mathbb F_{q^m}$). Using that, I'm then supposed to conclude that $m \mid \phi(q^m - 1)$, where $\phi$ is Euler's totient function.

Now my idea was to maybe look at the multiplicative, cyclic subgroup $\langle \alpha \rangle$ of $\mathbb F_{q^m}^\times$ (the subgroup that's generated by $\alpha$). Now we have $\mid \langle \alpha \rangle \mid = \text{ order of } \alpha \text{ in } \mathbb F_{q^m}^\times =: ord_{\mathbb F_{q^m}^\times}(\alpha)$.

Now I suspect that it's true that $\langle \alpha \rangle = \langle \beta \rangle$, from which the first statement would follow right away. But I'm not sure how I could derive that from only the fact that $\alpha, \beta$ are roots of the same minimal polynomial in $\mathbb F_q$.

(I'm also open to any other approaches – this was just the first that came to mind.)

Answer

The multiplicative order of $\alpha$ is the minimum $k$ such that $\alpha$ is a root of $x^{k} - 1$.

Now if $F$ is a field and $f \in F[x]$, the following are equivalent

• $\alpha$ is a root of $f$, and



• the minimal polynomial of $\alpha$ over $F$ divides $f$.

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So the multiplicative order of $\alpha$ is the minimum $k$ such that $m_{\alpha} \mid x^{k} -1$. And the multiplicative order of $\beta$ is the minimum $h$ such that $m_{\beta} \mid x^{h} -1$. But $m_{\alpha} = m_{\beta}$, thus...

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As to the second part, $\varphi(q^m - 1)$ is the number of generators of the cyclic group of non-zero elements of the field with $q^{m}$ elements. Every such generator generates the field with $q^{m}$ elements over the field with $q$ elements, and thus its minimal polynomial over the field with $q$ elements has degree $m$. But by the first part all the $m$ roots of such a minimal polynomial have the same multiplicative order, hence the number of generators of the cyclic group of non-zero elements of the field with $q^{m}$ elements is a multiple of $m$.

complex numbers - Squarring the Euler's formula

$$\begin{align}&e^{\pi i} + 1 = 0 &\text{ (Euler's Formula)}\\
\implies &e^{\pi i} = -1&\\
\implies &e^{2\pi i} = 1& \text{ (Squaring both sides)}\\
\implies &e^{2\pi i} = e^0 (e^0 = 1)&\\
\implies &2\pi i = 0&\end{align}$$

how is this possible?

elementary number theory - A sequence divisible by 9

I was trying to solve this series by mathematical induction for every $n$ from $\Bbb N$ : $u_n=n4^{n+1}-(n+1)4^n+1$ is divisible by $9$.

The initiation was pretty easy, but I only managed to prove $u_{n+1}=3k$ while $k$ is an integer and I don't think if it's divisible by $3$ implies that it is divisible by $9$ ; is it ? if not how can I proceed to prove the divisibility ? by mod maybe? thanks in advance for your answer

calculus - uniformly convergence on compact metric space

Let $K$ be a compact metric space. Let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous functions on $K$ such that $f_n$ converges to a function $f$ pointwise on $K$.

on Walt. Rudin's book Principles of mathematical analysis, 7.13, if we assume

(1). $f$ is continuous;

(2). $f_n(x)\geq f_{n+1}(x)$ for all $x\in K$ and all $n$;

then it is proved that $f_n$ converges to $f$ uniformly on $K$.

Is there counterexample satisfying (1) but not (2)? And satisfying (2) but not (1)?

Answer

Yes to both.

If we assume (2) but not (1), then let
$$
f_n(x) =
\begin{cases}
1 & \text{if } x = 0 \\
1-nx & \text{if } 0 \le x \le \frac{1}{n} \\
0 & \text{if } \frac{1}{n} \le x
\end{cases}
$$

$f_n$ is continuous, $f_n(x) \ge f_{n+1}(x)$, and $f_n$ and converges pointwise to the discontinuous characteristic function $\chi_{\{0\}}$.
However, the convergence is not uniform because if $\epsilon < 1$,
then for all $n$ there is a value of $x > 0$ with $f_n(x) > \epsilon$. Alternatively, you know the convergence cannot be uniform because the uniform limit of continuous functions is continuous.

If we assume (1) but not (2), then for $n \ge 2$ let
$$
f_n(x) =
\begin{cases}
nx & \text{if } 0 \le x \le \frac{1}{n} \\
2 - nx & \text{if } \frac{1}{n} \le x \le \frac{2}{n} \\

0 & \text{if } \frac{2}{n} \le x
\end{cases}
$$
The functions are continuous and converge to the continuous $0$, but the convergence is again not uniform.

Modular exponentiation with operations in the exponent

I’m trying to know how to calculate step by step the next equation applying the module in each operation:

$2^\left(4 \times \frac{6}{8}\right) \pmod{11}$

I know that if I just solve the whole equation and then I apply mod 11, the result is 8.

$4 \times \frac{6}{8}= 3$

$2^3 = 8$

$8 \pmod {11} = 8$

But if I try to do with modular arithmetic is not working for me:

$8^{-1} \pmod{11} = 7$

$6 \times 7 \pmod{11} = 9$

$4 \times 9 \pmod{11} = 3$

$2 ^ 3 \pmod{11} = 9$

Something similar happened when I tried to solve first what is inside the brackets in an expression like this:

$2^\left(4\times 6 \times 8\right) \pmod{11} $

But I changed the way to solve the equation and it works

$\left(\left(2^4\right) ^6\right)^8 \pmod{11} $

Now with the division I have no idea what to do, like in the first example I showed.

Any idea how to solve it?

Answer

Your modular computation with exponents is meaningless since, by Euler's formula,
$$ a^r\equiv a^{r\bmod\varphi(n)}\mod n, $$
so you should compute the exponents modulo $\varphi(11)=10 $. Unfortunately, $8$ is not a unit module $10$.

calculus - Proving $lim_{ntoinfty} left(frac {2}{3}right)^n=0$ using limit definition

Prove $\displaystyle\lim_{n\to\infty} \left(\frac {2}{3}\right)^n=0$ with the definition of limit.

From the definition and since $n\in\mathbb N$ I get that ${\Large\mid} \left(\frac {2}{3}\right)^n{\Large\mid}=\left(\frac {2}{3}\right)^n$ but now I'm not sure what to do, I don't see how taking a $\log$ here would help.

I thought of different approach, proving by induction that $\left(\frac {2}{3}\right)^n < \frac 1 n$: The first steps are trivial, then for $n+1$: $\frac {1}{n+1}>\frac {2}{3}\left(\frac {2}{3}\right)^n$, then from the induction hypothesis: $\frac {1}{n+1}>\frac {2}{3}\frac 1 n$ and we get that this true for $n>2$ so we can choose $N_{\epsilon}=\frac 1 {\epsilon}$.

Is there another way that does not involve induction?

Answer

HINT. By limit definition:
$$|a_n-l|<\epsilon$$
i.e.

$$\left|\left(\frac{2}{3}\right)^n\right|=\left(\frac{2}{3}\right)^n<\epsilon$$
whence
$$-n\ln(3/2)<\ln\epsilon$$
and
$$n>-\frac{\ln\epsilon}{\ln \frac{3}{2}}$$

calculus - summation of series by telescoping series method (feedback needed)

i am stuck i did the first part by cancelling out terms since its a telescoping series. But I do not know how I can proceed any further . Please help. I am not sure of whatever i have done so far. so Please see also for the errors.

Answer

What you have done is correct. Now it is straightforward that $$\lim_{n \to \infty} -\ln 2 + \ln(n+2)= -\ln 2 + \lim_{n \to \infty}\ln(n+2)=+\infty$$ since $\ln$ is a monotone increasing function.

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If you need to prove that $\ln n$ is unbounded you need the following: $$\ln' n=\dfrac{1}{n}>0$$ so that $\ln$ is monotone increasing. Moreover $\ln 2>\ln 1=0$. Now, take $M\in \mathbb R$, arbitrarily large. Then there exists $m \in \mathbb N$ such that $$M0$) or equivalently $M< \ln 2^m$. Therefore for any $n>2^m$ you have that $$M<\ln 2^m <\ln n$$ from which you can conclude that $\ln n$ is unbounded since $M$ was arbitrarily large.