A question about quotient spaces, something I do not fully understand yet, and can use some help.
$A = \mathbb F_3[x]$, $P = x^3-x+2\in A$
1) Show that $P$ is irreducible (I did it, it has no roots in $\mathbb F_3$).
2) Find the inverse of $x^2+P$ in $A/P$.
My problem is that I don't know what are the elements of $A/P$. I don't know what that is.
If I understand right, $x^2+P$ is the equivalence class of $x^2$ in $A/P$, but what now?
Elements of $A/P$ are of the form $f(x) + P$, where $f(x) \in A$. However, there is one important simplification : Euclidean division gives you polynomials $t, r \in A$ such that $$ f = tP + r, \text{ and } \deg(r) < 3 $$ Now $f + P = r + P$, and so you can assume that all elements of $A/P$ are of the form $$ (ax^2 + bx + c) + P $$ So you are now trying to find a polynomial of this form such that $$ (x^2 + P)(ax^2 + bx + c + P) = 1 + P $$ $$ \Leftrightarrow ax^4 + bx^3 +cx^2 - 1 \in P $$ Now use the fact that $x^3 + P = x-2+P$, and so this implies $$ ax(x-2) + b(x-2) + cx^2 - 1 \in P $$ $$ \Leftrightarrow (a+c)x^2 + (b-2a)x + (-1-2b) \in P $$ But this is a polynomial of degree $<3$, and there is no such polynomial in $P$ other than the zero polynomial (why?). Hence $$ a+c = 0, \qquad b-2a = 0, \qquad \text{ and } -1-2b = 0 $$ Now use the fact that $2 \equiv -1$ in $\mathbb F_3$ to solve these equations.