<s>[INST] Quotient $\mathbf{F}_3[X]/(X^5+1)$ 
Factor $X^5+1\in\mathbf{F}_3[X]$ into irreducibles. What does the quotient $\mathbf{F}_3[X]/(X^5+1)$ look like?

Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $\mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.
I am hesitating about the following part:
$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$. 
By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$\frac{\mathbf{F}_3[X]}{(X^5+1)}\cong\frac{\mathbf{F}_3[X]}{(X+1,2)}\cong\frac{\mathbf{Z}[X]/(3)}{(X+1,2)}\cong \frac{\mathbf{Z}[X]/(X+1)}{(3,2)}\cong\mathbf{Z}.$$
But $\mathbf{Z}$ is not a field, and $\frac{\mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?
 [/INST] I think the mistake you made is that the ideal $(X^5 + 1) \neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) \neq (X + 1,5)$. Long division only gives us "$\subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) \cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.
If you want to compute the quotient $\mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$\mathbb{F}[X] / (X^5 + 1) \cong \mathbb{F}[X] / (X+1) \times \mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.
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