3
$\begingroup$

If $a^k|b^{k+100} $ for every $k$, then show $a|b$.

Attempt


If $k=1$ it shows that if $p | a$ then $p|b$ too, so $b$ has every prime divisors which $a$ contains. But I don't Know what to do now. Could anyone help?

$\endgroup$
6
$\begingroup$

Hint:

  • Use the fact if $d=\gcd(a,b)$, then $a=dx$ and $b=dy$ where $x,y$ are relatively prime.
  • Use Gauss lemma.

So you got $$d^kx^k\mid d^{k+100}y^{k+100}\implies x^k\mid d^{100}y^{k+100}$$ Now since $x,y$ are relatively prime we have: $$x^k\mid d^{100}$$ which is true for all $k$. So if $x>1$ then $x^k$ is bigger then $d^{100}$ for big enough $k$ which is impossible. So $x=1$.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

Let $p$ be a prime that divides $a$, let $p^m$ be the highest power of $p$ that divides $a$ and let $p^n$ be the highest power of $p$ that divides $b$. We show that $m\leqslant n$.

By hypothesis, we have that $mk\leqslant nk + 100n$ for every $k\geqslant 1$. Divide by $k$ to obtain that $m\leqslant n + \frac{100n}k$ for every $k\geqslant 1$. Letting $k\to \infty$, we obtain our result.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Notice of course that, in the argument above, the $100$ in $k + 100$ could be substituted by any expression that is $o(k)$ as $k\to\infty$. $\endgroup$ – Fimpellizieri Nov 1 '19 at 18:20
  • $\begingroup$ There's no need to use primes here - see my answer. $\endgroup$ – Gone Nov 1 '19 at 21:02
2
$\begingroup$

Hint $\ \forall k\!:\ \color{#c00}{b^{100}} (b/a)^k\in \Bbb Z\,\Rightarrow\, b/a\in \Bbb Z,\,$ since unbounded powers of proper fractions have unbounded denominators, so they can't all be written with $\,\color{#c00}{b^{100}}$ as a denominator. $\ $ QED

Remark $\ $ This is true far more generally. Suppose $ \:D\:$ is any Noetherian integrally closed domain, e.g. any PID. Suppose that $ \:w\:$ is a fraction over $ \:D\:$ such that some unbounded sequence of powers of $ \:w\:$ has a common denominator $ \:0 \ne d\in D,\:$ i.e. $ \:d\!\:w^{n_i}\in D\:$ for all $ \:n_i.\:$ Then $ \:w\in D.$

Proof $\ $ By ACC the sequence of ideals $ (d, dw^{n_1}, dw^{n_2},\ldots)$ eventually stabilizes, which implies that for some $ \:k\:$ we have $ \: dw^{\large n_k}\in (dw^{\large n_{k-1}},\ldots, dw^{\large n_1}, d),\:$ which implies

$$ d\: w^{\large n_k} + c_{\large n_{k-1}} d\: w^{\large n_{k-1}} +\:\! \cdots +\: c_{n_1} d\: w^{\large n_1} + d\: =\: 0$$

Cancelling $ \:d\:$ yields $ \:w\:$ is integral over $ \:D,\:$ hence $ \:w\in D,\:$ since $ \:D\:$ is integrally closed. $\ $ QED

Elements whose powers have such a common denominator are called almost integral. Clearly integral elements are almost integral. By above the converse holds true in Noetherian domains.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.