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How is it true that:

If $a_1, a_2,\ldots,a_n$ are pairwise relatively prime positive integers,

then $M_i = \dfrac{(a_1a_2\cdots a_n)}{a_i} $ is relatively prime to $a_i$ ?

This is supposed to be an obvious step in the proof of Chinese remainder theorem, but to me it is not obvious.

Community
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Chan
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  • Note that $M_i$ is the product of all the $a_j$ except for $a_i$. Now use the fact that the product of numbers relatively prime to $a$ is relatively prime to $a$. – André Nicolas Feb 07 '12 at 19:34
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    Note: "primal" is actually a technical term in ring theory, distinct from "prime". "Relative primality" can mean something different in general context. – Arturo Magidin Feb 07 '12 at 19:51
  • @Arturo If you refer to Cohn's notion of primal, then I don't think there is any danger of confusion, since "relatively primal" is not in use (what could it mean?) – Math Gems Feb 07 '12 at 20:53
  • @MathGems: In a commutative ring, an ideal $Q$ is primal if all elements that are not relatively prime to $Q$ form an ideal. An element $a$ is relatively prime to $Q$ if $ab\in Q$ implies $b\in Q$. Since two elements are "relatively prime" if there is no prime ideal that contains both, they would be "relatively primal" if there is no primal ideal that contains both. – Arturo Magidin Feb 07 '12 at 21:00
  • @Arturo That meaning of "relatively primal" does not appear to be in use, so, again, there does not appear to be any danger of confusion. – Math Gems Feb 07 '12 at 21:13
  • @MathGems: "Relative primality" is not in standard use either, though I agree it is more grammatical. However, my ring theory colleagues objected strenously when I noted, in a thesis, that grammatically one should use "primality" in some instances when refering to issues relating to prime (E.g., prove "relative primality" instead of "relative primeness", "prove primality" instead of "prove prime"); the objection was precisely that since "primal" has a specific meaning, and the terms such as "relative primality" are not standard, it is bound to cause confusion. – Arturo Magidin Feb 07 '12 at 21:26
  • @Arturo The OP's "relative primality" is widely used, as a google search quickly confirms, e.g. it is used by your advisor G. Bergman in "Right orderable groups that are not locally indicable", by K. Ribet in "Hodge classes on certain types of abelian varieties", by K. Conrad in "Selmer's example", by H. Davenport in "Notes on a result of Chalk", etc. – Math Gems Feb 07 '12 at 22:00
  • @MathGems: Perhaps my ring theory colleagues misinformed me, then. – Arturo Magidin Feb 07 '12 at 22:04

3 Answers3

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Since $\gcd(a_j,a_i) = 1$ when $i \neq j$, this means that each $a_j$ has no common prime factors with $a_i$. Thus neither does $M_i$.

What is really going on is that if $\gcd(a_j, N) =1$ for $j = 1, \dots, n$, then $\gcd(a_1a_2\dots a_n,N) = 1$ as well.

JavaMan
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Hint: $\:(a, m) = 1\ \iff\ a$ is a unit (invertible) mod $m$. But units are closed under product since $(xy)^{-1} = y^{-1} x^{-1}$. So all $a_i$ units implies their product is a unit mod $m$, so it is coprime to $m$.

Math Gems
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I always like to use that $(a,b)=1$ if and only if $ax+by=1$ has integer solutions $x,y$.

We'll prove the lemma for $i=1$.

Then for $j>1$, there are $x_j,y_j$ such that $a_1x_j + a_jy_j = 1$, since the numbers are pairwise relatively prime.

Now multiply these together to get: $1 = \prod_{j>1} (a_1x_j + a_jy_j)$.

But you can expand that product out, and every term has either at least one multiple of $a_1$ in it, or consists of $\prod_{j>1} a_jy_j$. That means we have a solution to:

$$a_1X + \left(\prod_{j>1} a_j\right)Y=1$$

Thomas Andrews
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