Ideals and order of a polynomial ring

Question

Consider the ideal $I=(X^3+\hat2X+\hat1)$ of polynomial ring $R=\mathbb Z_3[X]$.

1. Is $R/I$ an integral domain?
2. How many elements does $R/I$ have?
3. Find the inverse of $X^3+\hat1$ in $R/I$.

(1) $(X^3+\hat2X+\hat1)$ has no roots in $\mathbb Z_3$ and it's irreducible, so it can't be written as a product of polynomials. Thus, $R/I$ is an integral domain.

(2) I don't know if this means the same as the order in quotient groups, probably not.

(3) I think I am supposed to find a polynomial $h\in R/I$ so that $(X^3+\hat1)h=\hat0=X^3+\hat2X+\hat1$.

Answer 1

Your first answer is correct. If $f=x^3+2x+1$ would be reducible, it would have a root in ${\Bbb Z}_3$, but it hasn't.

The quotient field ${\Bbb Z}_3[x]/\langle f\rangle$ consists of the elements $ax^2+bx+c$ with $a,b,c\in{\Bbb Z}_3$. Thus it has $3^3=27$ elements.

In order to find the inverse an element $g\in{\Bbb Z}_3[x]/\langle f\rangle$ with degree $\leq 2$, use the extended Euclidean algorithm to find the repesentation $rf+sg=1$ in the quotient field. Then $s$ is inverse to $g$.

Here $f$ has a root in the quotient field and so $x^3=-2x-1=x+2$. Then $x^3+1$ is congruent to $(x+2)+1 = x$ mod $f$.

Answer 2

For $2$ since $P=X^3+2X+1$ is irreducible, $F_3[X]/P$ is a $3$-dimensional vector field over $F_3$ so it has $3^3=27$ elements.

$\hat X^3=-2\hat X-\hat 1$ we deduce that $\hat X^3+\hat1=-2\hat X-1+1=-2\hat X=\hat X$, $\hat X^3+2\hat X=-\hat 1$ implies that $2\hat X^3+\hat X=\hat 1$ and $\hat X(2\hat X+\hat 1)=\hat 1$.