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

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?

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 $$x=1$$.

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.

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

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.