Even more VW handout

by Wolstenholme, Nov 27, 2014, 7:36 PM

$1.4.a)$ It is clear that $\alpha$ is algebraic so let its minimal polynomial have degree $d.$ Letting $\alpha_2, \alpha_3, \dots, \alpha_d$ be the conjugates of $\alpha$ in $\mathbb{Q}[\alpha]$ we can define $f_0(x) = c(x - \alpha)(x - \alpha_2)\dots(x - \alpha_d)$ where $c$ is the smallest positive integer such that $f \in \mathbb{Z}[x]$ (in other words, such that $f$ is primitive).

Now letting $f_1(x^2) = (-1)^df(x)f(-x) = c^2(x^2 - \alpha^2)(x^2 - \alpha_2^2)\dots(x^2 - \alpha_d^2)$ we have by Gauss's Lemma that $f_1(x)$ is primitive.

Defining $f_2, f_3, \dots$ similarly and continuing in this fashion we have for all $m \in \mathbb{N}$ that $f_m$ is primitive where $f_m(x) = c^{2^m}\left(x - \alpha^{2^m}\right)\left(x - \alpha_2^{2^m}\right)\dots\left(x - \alpha_d^{2^m}\right).$

Now since $n\alpha^k$ is an algebraic integer for all $k \in \mathbb{N}$ we have for all $k \in \mathbb{N}$ that $g_k(x) = (nx - n\alpha^k)\dots(nx - n\alpha_d^k) = n^d(x - \alpha^k)\dots(x - \alpha_d^k)$ is an integer polynomial.

This implies that $\frac{g_{2^m}(x)}{f_m(x)} = \frac{n^d}{c^{2^m}}$ is an integer for all $m \in \mathbb{N}$ (since the $f$ 's are primitive). But this is clearly false unless $|c| = 1$ which implies that $\alpha$ is an algebraic integer as desired.

$1.4.b)$ Hmm is it me or is the previous problem much harder? First, assume WLOG that $f$ is primitive. Basically define $g_k(x) = \frac{f^{(k)}(x)}{f(x)}.$ Assume that $g_1(x) \in \mathbb{Z}[[x]].$ Then usng product rule we have that $f^{(k + 1)}(x) = g_k'(x)f(x) + g_k(x)f'(x)$ and upon dividing by $f(x)$ we find that $g_{k + 1}(x) = g_k'(x) + g_k(x)g_1(x)$ so by an easy induction we get that $g_k(x) \in \mathbb{Z}[[x]]$ for all $k \in \mathbb{N}.$

Now we get that $g_k(x)f(x) = f^{(k)}(x) = k!\sum_{i = 0}^{\infty}\binom{k + i}{k}a_{i + k}x^i$ and since $f$ is primitive this implies that $k! \vert g_k(x).$ Writing the product as $\frac{g_k(x)}{k!} \cdot f(x) = \sum_{i = 0}^{\infty}\binom{k + i}{k}a_{i + k}x^i$ and letting $x = 0$ implies that $f(0) \vert a_k$ for all $k \in \mathbb{N}$ as desired so the answer is yes.

$1.4.c)$ Well they both use Gauss's Lemma so yeah

$1.5)$ By the Frobenius endomorphism we have that $\left((1 + \omega)^p - (1 + \omega^p)\right)^2 \equiv 0 \pmod{p^2}$ and upon expanding and rearranging and dividing by $(1 + \omega)^p(1 + \omega^p)$ we immediately obtain the desired result.

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For problem 1.5, how you do you know the expression is actually an integer?

by Naysh, Nov 27, 2014, 10:32 PM

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Hmm you're right

by Wolstenholme, Nov 28, 2014, 2:06 AM

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
the random chinese tst problem is the only thing I read, but I'll assume your blog is nice and give you a shout even though you probably never use aops anymoer
by fukano_2, Jun 14, 2020, 6:24 AM
wolstenholme - op
by AopsUser101, Jan 29, 2020, 8:27 PM
this blog is so hot
by mathleticguyyy, Jun 5, 2019, 8:26 PM
Hi. Nice Blog!
by User360702, Jan 10, 2019, 6:03 PM
helloooooo
by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
You were just featured on AoPS's facebook page.
by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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Wolstenholme's blog

Even more VW handout

by Wolstenholme, Nov 27, 2014, 7:36 PM

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