Help understanding prime decomposition.

Question

I've been thinking about prime decomposition a lot the last couple of days, and I think I mostly understand it. I just have, I think, only one question.

If we have some function

$$p_{1}^{q} \times f(x) = p_{2}^{r} \times g(x)$$

For some prime numbers, $p_{1}$, $p_{2}$ and some integer powers of $q$, and $r$, and where $f(x)$ and $g(x)$ are both integers for all x. Then we can say that both sides of the equation will be of the form $$2^{a_{1}} \times 3^{a_{2}} \times 5^{a_{3}} \times \dots \times p_{1}^{a_{m}} \times \dots \times p_{2}^{a_{n}} \times \dots$$ and on and on for all $a$ values where $a_{s} \geq 0$ and are all integers for all values of $s$.

That's a bit of a mouthful, and I think in most cases, all those a values will just be 0, but the point is, does the original equation mean that

$$f(x) = p_{2}^r$$ $$g(x) = p_{1}^{q}$$ Why couldn't they both have a common factor such that

$$f(x) = p_{2}^r \times k$$ $$g(x) = p_{1}^{q} \times k$$ For some integer $k$? It would make the original equation given still true, but it might change the values of f(x) and g(x) depending on what they are. Is there a way to prove that in all cases, $k = 1$? It would certainly be convenient if $k = 1$.

I have a sneaking suspicion that this has something to do with the fundamental theorem of algebra but I'm not terribly familiar with it yet. I'm currently studying up on it.

Thanks for your help!

By unique fractionization in the linked dupe: if $,p_1,p_2,$ are coprime then $$ \dfrac{f}g = \dfrac{p_2^r}{p_1^q}\iff \begin{align}f = np_2^r\ g = np_1^q\end{align},\text{ for some } n\in\Bbb Z\backslash 0$$ — Bill Dubuque, Aug 27 '22 at 08:23
Likely you mean the fundamental theorem of arithmetic (not algebra), i.e. the existence and uniqueness of prime factorizations of integers. This (or closely related results like Euclid's Lemma) are indeed used in proofs of this. — Bill Dubuque, Aug 27 '22 at 08:29

score 0 · Accepted Answer · answered Aug 27 '22 at 07:48

0

Yes, the general solution (assuming $p_1,p_2$ are distinct primes) is $f(x)=p_2^rk$ and $g(x)=p_1^qk$ for some integer $k$. If you also know that $f(x)$ and $g(x)$ are coprime then you can deduce that $k=1$, so $f(x)=p_2^r$ and $g(x)=p_1^q$.

answered Aug 27 '22 at 07:48

Thomas Bloom

2,221

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Aug 27 '22 at 08:25

Help understanding prime decomposition.

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