# Multiplicative inverse of polynomial modulus an integer

How do you calculate the multiplicative inverse of a polynomial mod a monomial/integer?The specific questions are: Find the multiplicative inverse of 1) x+1 mod 3 2) x^2+x-1 mod 3 3) x^2+x-1 mod 32

I understand that you need to use the Extended Euclidean algorithm to solve it. For integer mod integer (e.g. 11 mod 26) and polynomial mod polynomial, its clear.But how to solve x+1 mod 3?

Neither $x+1$ nor $x^2+x-1$ are invertible modulo $3$, i.e. in the polynomial ring $\mathbf Z/3\mathbf Z[x]$, which is a polynomial ring over a field, because in such a case, for any polynomials $f,g$ we have $\deg fg=\deg f+\deg g \ge\deg f,\deg g$, which is not compatible with $fg=1$ unless $f$ and $g$ are constants.
If the quotient ring is not a field, this may not be true. However $x^2+x-1$ invertible in $is impossible since its dominant coefficient is$1$, hence$\deg f(x)(x^2+x-1)=\deg f+2$. For an example with coefficients in$\mathbf Z/32\mathbf Z$you can check that$\;(16x^2-4x+1)(4x+1)=1$. • Thank you for the response. – jaynjayn Jul 14 '15 at 16:06 • do you mean that for a polynomial f to have a multiplicative inverse mod g then inlinedeg fg=deg f+deg g ≥deg f,deg g ? If that is the case then x+1 mod 3 meets the criteria because deg ((x+1)3)=1' and deg f=1 and deg g=0` . Kindly explain – jaynjayn Jul 15 '15 at 7:03 • If$fg=1$, we must have$\deg f+\deg g=\deg 1=0$, whence$\deg f=\deg g=0$, i. e.$f$and$g\$ must be non-zero constants. – Bernard Jul 15 '15 at 7:29