4
$\begingroup$

Let $R$ be a commutative rings with unity such that $\langle a, b, c \rangle = 1$. Then show that $\langle a^{15}, b^{16}, c^{17} \rangle = 1$.

If I work in $\mathbb{Z}$, I can argue(due to prime factorisation) that gcd has not changed. But for general ring, I have no clue on how to begin with. One idea which I though of is to write $$xa + yb + zc = 1$$ and multiply it $n$-times so that I could get appropriate $a^n$ terms, but the problem with this approach is that there are cross terms which needed to be taken care.

$\endgroup$
5
$\begingroup$

Your intuition is good. Moreover, if you raise your equation to a high enough power, all the cross terms will end up in the ideal generated by $\{a^{15},b^{16},c^{17}\}$.

Specifically, by the binomial theorem, you can write $$1=(xa+yb+zc)^{n}=\sum_{i+j+k=n}\left(x^iy^jz^k{n\choose k}{i+j\choose i}\right)a^ib^jc^k.$$ Let's ignore the coefficient, and focus on the monomial in $a,\,b$ and $c$. Note that if $n\geq 15+16+17$, then in every summand here, at least one of the following holds: $i\geq 15$ or $j\geq 16$ or $k\geq 17$. Thus, you could factor at least one of $a^{15}$ or $b^{16}$ or $c^{17}$ from each summand and then collect these terms to get some linear combination of them summing to $1$. Thus, just raise your equation to the power of $15+16+17=48$, and you're done!

$\endgroup$
  • $\begingroup$ Sounds good. So essentially the numbers 15, 16, 17 have no specific significance. I was wrecking my brain into thinking that this specific power could somehow manage the crossterm. $\endgroup$ – ManishKumar Singh May 11 '18 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.