This statement was given in my number theory textbook when analyzing quadratic fields, and I am not seeing how to prove it. $m$ is a squarefree (not divisible by the square of any number) integer and $d$ and $b$ can be any integers which satisfy $d^2 \lvert mb^2$. I can't invoke Euclid's lemma since $m$ and $d^2$ are not necessarily coprime. Since $m$ is squarefree it is clear that $d^2 \not \lvert m$, but I can't figure out how to make use of that.
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1try with the fundamental theorem of arithmetic. – L F May 11 '15 at 03:16
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1See here – Bill Dubuque May 11 '15 at 03:17
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More generally, $d^n\mid mb^n,\Rightarrow, d\mid b$ for $n\ge 2$ by analoguous proof to robjohn's or egreg's. – user26486 May 11 '15 at 16:49
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In above comment $m$ is square-free of course. – user26486 May 12 '15 at 15:12
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For any prime $p$, suppose the exponent of $p$ in the factorization of $d$ is $e_d$ and the exponent of $p$ in the factorization of $b$ is $e_b$. We know the exponent of $p$ in the factorization of $m$ is $0$ or $1$. Since $d^2\mid mb^2$, we have $$ 2e_d\le1+2e_b $$ Since both are integers, we have $$ e_d\le e_b $$ Since this is true for any prime $p$, we get $$ d\mid b $$
robjohn
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Your $e_d$ is by convention written as $\upsilon_p(d)$. It is popular notation, seen, e.g., in LTE paper. – user26486 May 11 '15 at 15:04
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Let $k=\gcd(d,b)$, so $d=kD$ and $b=kB$, where $D$ and $B$ have no common divisors. If $d^2\mid mb^2$, we have $$ mb^2=nd^2 $$ that gives $mB^2=nD^2$ or $D^2\mid mB^2$. Suppose $D>1$ and take a prime divisor $p$ of $D$: then $p^2\mid mB^2$, but $p\nmid B$, so $p\mid m$. Thus $p\mid\frac{m}{p}B^2$, but this is impossible, because neither $p\mid B$ nor $p\mid \frac{m}{p}$ (the latter because $m$ is squarefree).
Hence no such prime can exist, so $D=1$, which is the same as saying that $k=d$, so $d\mid b$.
egreg
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