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

Question

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

Attempt

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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?

Answer 1

Hint:

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

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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 $x=1$.

Answer 2

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.

Answer 3

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.