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?
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).