3

I want to prove: $$\text{If }a^2|z^2,\text{ then }a|z.$$ Assume everyone is a positive integer, etc. Unless I'm deluding myself, this is pretty easy to show using unique prime factorization.

But I want to do it without using primes or the (usual statement of the) FTA. That is, using coprime is fine, using the so-called Bezout identity (XGCD algorithm), etc. is fine. Is this even possible without essentially defining at least irreducibles, if not primes and prime factorization, along the way?

(See here for a more vague question I asked a while ago on this.)

Community
  • 1
kcrisman
  • 2,195
  • In $\mathbb Z/4\mathbb Z$ we have $0^2|2^2$ but $0 \nmid 2$. So any proof that works for $\mathbb Z$ must use some property that $\mathbb Z/4\mathbb Z$ does not have. – marlu Jan 24 '13 at 21:10

3 Answers3

3

Hint $\ $ If $\: (z/a)^2 = n\:$ then $\,z/a\,$ is a root of $\:x^2 - n\:$ so $\:z/a\in \Bbb Z\:$ by the rational root test (RRT). Domains satisfying the monic case of the RRT are called integrally-closed. They are a much wider class of domains than UFDs. For example, the usual proof of RRT uses only gcds, so any gcd domain is integrally-closed.

Or by $\,(a^2,z^2)=(a,z)^2$ [see here or here], cancelling the gcd $\, (a,z)\,$ twice from $\,z^2/a^2\,$ reduces to the coprime case $(a^2,z^2)\! =\! 1,\,$ so $\,a^2\,|\,z^2\Rightarrow a^2 = \pm 1\Rightarrow a = \pm 1,\,$ so $\,a\mid z$.

If you desire to learn more about this specific property then try searching on root-closed domains. A domain D is called $n$ root-closed if for every fraction $x$ over D we have $\,x^n\in{\rm D}\Rightarrow x\in\rm D.\,$ If this holds for all $\,n\in \Bbb N\,$ then one calls D root-closed. One interesting result using this property is the following (which applies to rings of algebraic integers).

D is a Dedekind domain with torsion class-group $\iff$ D is root-closed and for every $\,\{a,b,\ldots\}\subset\rm D\,$ there is an $\,n\in\Bbb N\,$ such that the ideal $(a^n,b^n,\ldots)\,$ is principal. Domains that satisfy the latter ideal-theoretic property are sometimes called almost-PIDs or API domains.

Bill Dubuque
  • 272,048
Math Gems
  • 19,574
  • I like the great generality, though I have to say that I am glad I didn't encounter all these different types of domains when they could have distracted me... cool theorem about the class group. – kcrisman Jan 25 '13 at 02:48
1

Since Bezout is fair game, one can use that if $gcd(a,b)=1$ and $a|bc$ then $a|c$.

Now if $a^2|b^2$ we may first remove the gcd from $a,b$ and have then $gcd(a,b)=1.$ Pick $k$ with $a^2k=b^2$. Note that $gcd(b,a^2)=1$ and we have $b|a^2k$ [since $b \cdot b=a^2k$]

Therefore we have $b|k$ and can write $k=bk'$. Put this into $a^2k=b^2$ to get $$a^2\cdot bk'=b^2,$$ then we have $a^2k'=b$, so that $a \cdot (ak')=b$, i.e. $a|b$.

Not sure if the use of Bezout needs to be flushed out some, but I think that's all I used.

EDIT: At one point I used that $gcd(a,b)=1$ implies $gcd(a^2,b)=1$ This is OK since Bezout works both ways, that is $gcd(a,b)=1$ if and only if there are integers $x,y$ with $ax+by=1$. So from $gcd(a,b)=1$ there are $x,y$ with $ax+by=1$, so $a(ax+by)x+by=1$, and multiplying out we have $a^2x+abxy+by=1$ and then factoring $b$ out of the second two terms gives $a^2x+b(axy+y)=1.$ So by reverse Bezout we arrive at $gcd(a^2,b)=1.$ [Thanks to kcrisman for pointing out I originally had shown the converse of the desired implication!]

coffeemath
  • 29,884
  • 2
  • 31
  • 52
  • Yes, that's fine, it's essentially the same as the alternative proof I gave. – Math Gems Jan 24 '13 at 23:12
  • Thanks, Math Gems; I was a bit worried I didn't stick to the OP's requirements. +1 on your answer, which gave the right name Euclid's Lemma to the essential step (I had forgotten what that was called). – coffeemath Jan 24 '13 at 23:25
  • This is more the level of detail I'm hoping to leave for posterity. Fix the edit so the logical implication makes sense and I'll accept it (am I right that currently it looks like you proved $gcd(a^2,b)=1$ implies $gcd(a,b)=1$, not the converse which you need? Or am I again missing something? $ax+by=1$ doesn't universally mean that $x$ can be a multiple of $a$...). – kcrisman Jan 25 '13 at 02:50
  • 1
    I.e. I guess we need that if $gcd(a,b)=1$ then $ax+by=1=a\cdot 1 \cdot x +by=1$ so that $a(ax+by)x+by=1$ or $a^2 x^2+b(ayx+y)=1$, which as you point out is $\iff gcd(a^2,b)=1$. – kcrisman Jan 25 '13 at 03:09
  • Oops. I did go the wrong way and showed $\gcd(a^2,b)=1$ implies $gcd(a,b)=1$. It looks like your last comment is the proof, same one I obtained working on a solution set for a number theory class I taught a few years back! Will fix. Thanks. – coffeemath Jan 25 '13 at 08:53
  • Sweet; in fact, I am also teaching NT right now and just hate the "usual" way of using the sledgehammer of FTA for everything, but wasn't clever enough late at night to come up with the right argument in time. I'll accept and duly acknowledge both of you in the course notes - thanks! – kcrisman Jan 25 '13 at 16:03
0

Here's one more presentation of the proof (which I don't claim to have invented):

Lemma: $\gcd(x^2,y^2) = \gcd(x,y)^2$. $\ \ $ Proof: see this answer.

Now,

Claim: $a^2 \mid z^2$ if and only if $a \mid z$.

Proof: Just the case $a>0$: $a^2 \mid z^2$ if and only if $\gcd(a^2,z^2)=a^2$, if and only if $\gcd(a,z)^2 = a^2$, if and only if $\gcd(a,z)=a$, if and only if $a \mid z$. $\square$

Bill Dubuque
  • 272,048
Zach Teitler
  • 3,020
  • (I guess the point is just that $\gcd(x,y)^2$ can be written as $\gcd(x^2,y^2)$, you don't need $\gcd(x^2,y^2,xy)$.) – Zach Teitler Apr 03 '23 at 18:36
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. The lemma is proved here in many places (so I replaced your proof by a link to the same), as is this application (e.g. in the $10$ year-old answer by Math Gems in this thread). – Bill Dubuque Apr 04 '23 at 18:44