# Calculating this class number

Let $f = x^5+2x^4-2$ and $\alpha \in \mathbb C$ with $f(\alpha) = 0$. Show that $\mathbb Z [\alpha ]$ is a principal ideal ring.

What I have done so far:

My idea was to first prove that $\mathcal O_K = \mathbb Z[\alpha ]$ (with $K := \mathbb Q[\alpha]$) and after that show that the class number $h_K = 1$. This would imply the claim. So I calculated the discriminant:

$$d\left(\mathbb Z[\alpha ]\right) = D(f) = -15536 = (-1)\cdot 2^4 \cdot 971$$ From this I can derive $\mathcal O_K = \mathbb Z[\alpha ]$ (Since $f$ is an Eisenstein-polynomial for $p=2$ we know that $2 \nmid \left[ \mathcal O_K : \mathbb Z [\alpha] \right]$, but $[ \mathcal O_K : \mathbb Z [\alpha] ]^2 \cdot d(\mathcal O_K)= d(\mathbb Z[\alpha ]) = (-1)\cdot 2^4 \cdot 971$. Therefore $[ \mathcal O_K : \mathbb Z [\alpha] ]= 1$ and $\mathcal O_K = \mathbb Z[\alpha ]$)

However I am having problems showing $h_K = 1$. I have calculated the Minkowski bound $M$. It is $M<7$. So I only need to look at the prime ideals above $2,3$ and $5$ and show that they are principal already (right?). But how do I do that exactly, I am not sure how to tell whether they are split, inert or ramified. Any help would be much appreciated.

Use $f$ to find the factorization of a prime $p$ in the given ring. We have that:
$$f = x^5 \in \mathbb{F}_2[x] \implies 2\mathcal{O}_K = (2,\alpha)^5 = \mathfrak{p}_2^5$$ $$f = x^5+2x^4+1 \in \mathbb{F}_3[x] \implies 3\mathcal{O}_K = (3)$$ $$f = x^5+2x^4+3 \in \mathbb{F}_5[x] \implies 5\mathcal{O}_K = (5)$$
So from here we have that the problem reduces to proving that $\mathfrak{p}_2 = (2,\alpha)$ is a principal ideal. But this is true, as $2=\alpha(\alpha^4 + 2\alpha^3)$ and so $2 \in (\alpha)$. Hence we have that $\mathfrak{p}_2 = (\alpha)$. Hence as the class group of the ring of integers is generated by principal ideals we have that $h_K =1$
• I get it now, thx! but is there a smart way to see directly that this polynomial is irreducible (in $\mathbb F_3[x]$ and $\mathbb F_5[x]$)? The only thing i can come up with is equating coefficients. – user9620780 Jun 19 '18 at 10:01
• @user9620780 It's easy to check that the polynomial doesn't have a root modulo $3$. So if it's reducible it has to be a product of a quadratic and a cubic term. All quadratic irreducible factors modulo $3$ are $x^2+1,(x+1)^2+1$ and $(x+2)^2+1$. (They are all quadratic factors of $x^9-x$ modulo $3$). You can check that neither of them is divisor of $f$. – Stefan4024 Jun 19 '18 at 10:54
• @user9620780 As expected modulo $5$ we have more such polynomials and they are $x^2+2,(x+1)^2+2,(x+2)^2+2,(x+3)^2+2,(x+4)^2+2$ $x^2+3,(x+1)^2+3,(x+2)^2+3,(x+3)^2+3,(x+4)^2+3$. You can check that neither of them divides $f$. I don't know whether this method is faster, but it's another way you can prove the irreducibility. – Stefan4024 Jun 19 '18 at 10:56