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Suppose $n=ab=cd$ in an non-UFD where $a,b,c,d$ are irreducibles.

Do $c^{-1}\bmod a$, $c^{-1}\bmod b$, $d^{-1}\bmod a$, $d^{-1}\bmod b$ and $a^{-1}\bmod c$, $a^{-1}\bmod d$, $b^{-1}\bmod c$, $b^{-1}\bmod d$ exist?

If so can we find using any modification of Extended Euclidean Algorithm?

If not how to find them?

If we know $c^{-1}\bmod a$, $c^{-1}\bmod b$ can we get $c^{-1}\bmod ab$?

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I think the answer is negative.

For example in $\Bbb Z[i\sqrt{5}]$ it is negative. Since $(1-i\sqrt{5})^{-1}=\frac{1+i\sqrt5}6$ and $6$ is not invertible $\bmod 2$ or $\bmod 3$ yet $2\cdot3=(1+i\sqrt 5)(1-i\sqrt 5)$. I was not sure whether $2$ or $3$ were invertible $\bmod (1+i\sqrt{5})$ or $\bmod (1-i\sqrt{5})$.

But could there be examples where inverses exist (in quadratic case it generally seems negative because of the norm in denominator)?

May be there are cubic and higher order cases where we can expect the unexpected?

Turbo
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  • @EricWofsey updated I just want to know if we can do $c^{-1}\bmod b$? – Turbo Dec 25 '16 at 03:14
  • @EricWofsey sorry clarified the query .. is it still no? – Turbo Dec 25 '16 at 03:17
  • Can you write what your assumptions are more clearly? You say you are in a "non-UFD", but what does that mean? Are you assuming you are in an arbitrary integral domain? Are $a$ and $b$ both supposed to not be associates of $c$ or $d$? – Eric Wofsey Dec 25 '16 at 03:19
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    Euclidean domains are PIDs so UFDs, so what do you mean by the Euclidean algorithm in a non-UFD? – Bill Dubuque Dec 25 '16 at 03:22
  • @EricWofsey I think the answer is negative. For example in $\Bbb Z[i\sqrt{5}]$ it is negative. Since $(1-i\sqrt{5})^{-1}=\frac{1+i\sqrt5}6$ and $6$ is not invertible $\bmod 2$ or $\bmod 3$ yet $2\cdot3=(1+i\sqrt 5)(1-i\sqrt 5)$. I was not sure whether $2$ or $3$ were invertible $\bmod (1+i\sqrt{5})$ or $\bmod (1-i\sqrt{5})$. But could there be examples where inverses exist (in quadratic case it generally seems negative because of the norm in denominator)? May be there are cubic and higher order cases where we can expect the unexpected? – Turbo Dec 25 '16 at 03:24
  • @BillDubuque you missed 'modification' and 'If not how to find them?'. – Turbo Dec 25 '16 at 03:33
  • If $,c^{-1}!\pmod a$ exists then $,(c,a) = 1,$ so $,c\mid ab,\Rightarrow,c\mid b,$ by Euclid's Lemma (in Bezout form).. Therefore $,c^{-1}\pmod b\ $ does not exist. – Bill Dubuque Dec 25 '16 at 03:43

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Assuming that $a$ and $b$ are not associates of $c$ or $d$ so that we really do have two distinct factorizations of $n$, these inverses can never exist. For instance if $c$ had an inverse mod $a$, then we would have $$d\equiv c^{-1}cd\equiv c^{-1}ab\equiv 0\pmod{a}$$ so $d$ is divisible by (and hence associate to) $a$, contrary to our assumption. In the language of ring theory, what's going on is that $c$ and $d$ are zero-divisors in the quotient ring mod $a$, and zero-divisors can never be units (in a nonzero ring).

Eric Wofsey
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  • How about if $a=cu$ and $b=vd$ holds where $uv=1$? – Turbo Dec 25 '16 at 03:46
  • Then $c^{-1}$ mod $a$ certainly doesn't exist, since $c$ is $0$ mod $a$, and similarly for many of the others. Inverses like $c^{-1}$ mod $b$ may or may not exist. – Eric Wofsey Dec 25 '16 at 03:48