Let $p(x) \in \mathbb Z_{5}[x]$, given by $p(x) = x^{3}+2x^{2}+1$ and let $I = <p(x)>$ be the ideal in $\mathbb Z_{5}[x]$ constructed by $p(x)$.
Determine the inverse of $2x+3+I$ in $\mathbb F$
(I have already determined that $p(x)$ is irreducible and that $\mathbb F$ is a field)
I think I have to calculate a certain $a(x)+I$ with $a(x)(2x+3) = 1 + I$, correct?
So then, $a(x)+I$ with $a(x)(2x+3) = 1 + b(x)I$.
Ie. $a(x)(2x+3)-b(x)(x^{3}+2x^{2}+1) = 1$
Can I solve this with Euclid's algorithm? How?
Since $x^3+2x^2+1=q(x)(2x+3)+c$, $c$ is the value of $x^3+2x^2+1$ in $x=-\frac{3}{2}=1$ (we are in $\mathbb{F}_5$), so $c=4=-1$ and: $$ 1 = x^3+2x^2+2 = (2x+3)q(x), \tag{1}$$ where $q(x)$ is a quadratic polynomial of the form $$ q(x) = 3x^2+dx-1$$ such that $3d-2=0$, so $d=-1$ and through $(1)$ we have that the inverse of $(2x+3)$ in $\mathbb{F}_5[x]/{(x^3+2x^2+1)}$ is just $q(x)=3x^2-x-1$.