Marcus,problem 12, chapter 3 I'm trying to solve this exercise from Marcus' book "Number fields". Following Marcus notation, let me call $S=\mathbb{Z}[\alpha]$, where $\alpha=\sqrt[3]{2}$, and $R=\mathbb{Z}$.


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*Firstly I proved that $5S=P_1P_2$ where $P_1=(5,\alpha+2)$ and $P_2=(5,\alpha^2+3\alpha-1)$. Indeed $P_1P_2=5(5,\alpha^2+4\alpha+1,\alpha^2+\alpha)$. Now, remember $1+\alpha+\alpha^2$ is a unit in $S$. It easy to see that $3 \in (5,\alpha^2+4\alpha+1,\alpha^2+\alpha)$, and so $(\alpha^2+4\alpha+1)-(3\alpha) =1+\alpha+\alpha^2 \in (5,\alpha^2+4\alpha+1,\alpha^2+\alpha) $.Thus, $(5,\alpha^2+4\alpha+1,\alpha^2+\alpha) =S$.

*There is an isomorphism between $\mathbb{Z}_5[x]/(x^3+3x-1) $ and $ \mathbb{Z}[x]/(5,x^3+3x-1)$ which rises up from the homomorphism



$\phi :\mathbb{Z}_5[x] \to \mathbb{Z}[x]/(5,x^3+3x-1) \qquad f(x) \mapsto f(x) + (5,x^3+3x-1)$



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*Moreover, there is an homomorphism $\psi$ from $\mathbb{Z}[x]/(5,x^3+3x-1)$ to $S/(5,\alpha^2+3\alpha-1)$.

*Now, troubles start. Using the previous points I should be able to conclude that either   $S/(5,\alpha^2+3\alpha-1)$ is a field of order 25, or $S=(5,\alpha^2+3\alpha-1)$. I have no ideas, any hints?

*The last point is showing that $S \neq (5,\alpha^2+3\alpha-1)$. I think this is because, otherwise, $P_1P_2=P_1S=5S$, and $\alpha+2 \in P_1S$ but $\alpha+2$ doesn't lie in $5S$. I'm not sure about this last my statement, if you have other ideas, they will be kindly accepted. 


Here you can read the exercise

 A: What (b) says is that
$$\mathbb Z[x]/(5,x^2+3x-1) \cong (\mathbb Z/5\mathbb Z)[x]/(x^2+3x-1).$$
Since $x^2 + 3x - 1$ is irreducible mod $5$, that means the latter is a field extension of degree $2$ of $\mathbb F_5$ i.e. the field with 25 elements.
That means in (c) we have shown that there is a surjection from the field $\mathbb F_{25}$ to $S/(5,\alpha^2+3\alpha-1)$. The only possible homomorphic images of a field are itself or the 0 ring, and so either $S/(5,\alpha^3+3\alpha -1)$ is $0$ or all of $\mathbb F_{25}$. The first would say that the ideal you are modding out by is all of $S$.
Your argument for (e) seems to be on the right track. From (a), we have $P_1 P_2 = 5S$. If $P_2 = S$, then that implies $P_1 = 5S$ but as you observed, $\alpha + 2$ is not in $5S$. I am not sure why you are unsure about this. Note that $5S$ is the ideal of $\mathbb Z$-linear combinations of $1,\alpha,\alpha^2$ whose coefficients are all divisible by $5$, and evidently the coefficients of $\alpha+2$ are not divisible by $5$.
Surjection in part (c). First of all, there is clearly a map
$$\mathbb Z[x] \rightarrow S/(5,\alpha^2+3\alpha - 1)$$
which is induced by evaluating $x$ at $\alpha$ to go to $S$ and then applying the quotient map. This is a surjection because both maps are surjections. The quotient is always a surjection, and for the first recall that $S = \mathbb Z [\alpha]$, so all its elements are polynomials in $\alpha$ with $\mathbb Z$ coefficients. All we have to do now is check that the kernel of this map contains $(5,x^2 +3x - 1)$ and then the above surjection factors through the quotient to give us the map described above.
But this is easy; $5$ and $x^2 + 3x - 1$ are sent to the classes $\bar 5, \overline{\alpha^2 + 3\alpha - 1}$ in $S/(5,\alpha^2+3\alpha - 1)$ and those are certainly $0$ in that quotient.
