Let \(i=\sqrt{-1}\). Lets look at \(K=\mathbb {Q}(i)\). Using Proposition 2.2.19 we see that \(\Delta (\mathcal{O}_K)=-4\). Moreover, \(K\) has no real embeddings, so \(r_1=0\) and \(r_2=1\). Putting this together we get

Theorem 4.0.3 tells us that each ideal class contains an ideal of norm \(\leq 1\). But the only ideal of norm \(1\) is the trivial ideal. So \(\operatorname{Cl}_K\) is trivial. If follows that \(\mathcal{O}_K=\mathbb {Z}[i]\) is a PID.

Let \(K=\mathbb {Q}(\sqrt{-5})\). From Theorem 2.1.11 we know that \(\mathcal{O}_K=\mathbb {Z}[\sqrt{-5}]\). Here again we have \(r_1=0\), \(r_2=1\) and \(\Delta (\mathcal{O}_K)=-20\). Therefore

This means every class contains an ideal of norm \(1\) or \(2\). We know the trivial ideal has norm \(1\), so lets look at what ideal has norm \(2\). As we saw in the proof of Corollary 3.4.10, its enough to look at what prime ideals divide \(2\). By Theorem 3.5.17 we have

and therefore \(\mathfrak {p}_2:=(2,1+\sqrt{-5})\) is the unique ideal of norm \(2\).

So, this means the class group \(\operatorname{Cl}_K\) has either \(1\) or \(2\) elements, depending of whether or not these two ideals are in the same class, i.e., if \([1]=[\mathfrak {p}_2]\). In other words, is \(\mathfrak {p}_2\) a principal ideal (since principal ideals are the things which \([1]\) consists of). If it where principal, then Proposition 3.4.5 would mean that there is some element in \(\mathcal{O}_K\) of norm \(\pm 2\). Now

and its easy to see that this is never equal to \(\pm 2\). So \(\mathfrak {p}_2\) is not principal and therefore

so \(\operatorname{Cl}_K\) is cyclic of order \(2\).

Let \(K=\mathbb {Q}(\sqrt{-31})\). In this case by Theorem 2.1.11 we have \(\mathcal{O}_K=\mathbb {Z}[\alpha ]\) where \(\alpha =\frac{1+\sqrt{-31}}{2}\), \(n=[K:\mathbb {Q}]=2\) and \(r_1=0,r_2=1\). Lastly, Exercise 2.2.28 gives \(\Delta (\mathcal{O}_K)=-31\). This means \(M_K=\frac{2}{\pi }\sqrt{31} {\lt} 4\).

So we need to find the ideals of norm less than \(4\). Using Theorem 3.5.17 gives

with \(\mathfrak {p}_2 \neq \mathfrak {p}_2'\), both of which have norm \(2\). Moreover, we see that \((3)\) is inert in \(K\), so has norm \(9\).

So the ideals of norm \({\lt}4\) are \((1),\mathfrak {p}_2,\mathfrak {p}_{2}'\). We know \(\mathfrak {p}_2 \neq \mathfrak {p}_2'\), but this does NOT mean we have \([\mathfrak {p}_2] \neq [\mathfrak {p}_2']\). We need to see whether or not this is the case. We know that \(\mathfrak {p}_2\mathfrak {p}_2'=(2)\) which means we have \([\mathfrak {p}_2][\mathfrak {p}_2']=[(2)]=[1]\) so we have \([\mathfrak {p}_2]=[\mathfrak {p}_2']^{-1}\). Therefore if we had \([\mathfrak {p}_2] = [\mathfrak {p}_2']\) this would mean that \([\mathfrak {p}_2]^2=[1]\) which means that \(\mathfrak {p}_2\) would have to be principal. Can this happen? Assume this is the case, then we have \(\mathfrak {p}_2^2=(\beta )\) for some \(\beta \in \mathcal{O}_K\) (its in \(\mathcal{O}_K\) and not \(K\) since \(\mathfrak {p}_2\) is a ideal in \(\mathcal{O}_K\) not a fractional ideal). Now, we know that \(N(\mathfrak {p}_2)=2\) (this follows from Corollary 3.5.11). This would mean \(\beta \) has norm \(4\). Note that the norm of an arbitrary element in \(\mathcal{O}_K\) is

Therefore we need to check if this can be equal to \(4\). Clearly this can happen, and moreover it can only happen if \(x=\pm 2\) and \(y=0\). This would mean that we have \(\beta =2\) and therefore

but we know that

therefore, by uniqueness of factorization and the fact that \(\mathfrak {p}_2 \neq \mathfrak {p}_2'\) we see that \(\mathfrak {p}_2^2\) cant be principal and therefore \([\mathfrak {p}_2] \neq [\mathfrak {p}_2']\).

It remains to check if \([\mathfrak {p}_2]=[1]\) or \([\mathfrak {p}_2']=[1]\). Similar to what we did above, we now need to see if there is some element of norm \(\pm 2\).

So the question is, can we find \(x,y\) which make \(x^2+xy+8y^2\) this equal to \(2\). The first thing to note is that \(N_{K/\mathbb {Q}}(x+y\alpha ) \geq 7y^2\), (to see this, note that \(x^2+xy+y^2 \geq 0\)) so we are going to need \(y=0\). This means we would need \(x^2=2\) which cant happen, therefore \([\mathfrak {p}_2] \neq [1] \neq [\mathfrak {p}_2']\). Thus we have

so its cyclic of order \(3\). From this we can deduce that \([\mathfrak {p}_2^2]=[\mathfrak {p}_2']\) and \([\mathfrak {p}_2^3]=[1]\) (i.e. \(\mathfrak {p}_2^3\) is principal). Lets finish off by writing down the multiplication table for the group:

Lets look at \(K=\mathbb {Q}(\sqrt[3]{7})\) and let \(\alpha =\sqrt[3]{7}\). The minimal polynomial is clearly \(x^3-7\) which has discriminant \(-1323=-3^3 \cdot 7^2\). Next, lets check that \(\mathcal{O}_K=\mathbb {Z}[\sqrt[3]{7}]\). For this we need to check if any of

can be algebraic integers for \(0 \leq x_i \leq 2\), \(0\leq y_i \leq 6\). Since the minimal polynomial is clearly Eisenstein at \(7\) we know from Lemma 2.2.21 that we only need to worry about \(3\). Here is a trick: note that

is Eisenstein at \(3\) and at \(7\) so if \(\beta \) is one of its roots, then \(\mathbb {Z}[\beta ]=\mathcal{O}_{K'}\) where \(K'=\mathbb {Q}(\beta )\), but \(\mathbb {Q}(\alpha )=\mathbb {Q}(\beta )\) and \(\mathbb {Z}[\alpha ]=\mathbb {Z}[\beta ]\). Therefore \(\mathcal{O}_K=\mathbb {Z}[\sqrt[3]{7}]\).

Knowing this means we can freely use Theorem 3.5.10. Now, lets calculate the Minkowski bound. In this case \(n=3\), \(\Delta (\mathcal{O}_K)=-1323\), and \(r_1=r_2=1\). Putting this together we get

So we need to find all ideals of norm at most \(10\). So lets factor the ideals given by \(2,3,5,7\) using Theorem 3.5.10.

• \(x^3-7 \equiv (x-1)(x^2+x+1) \mod 2\)

• \(x^3-7 \equiv (x-1)^3 \mod 3\)

• \(x^3-7 \equiv (x-3)(x^2+3x-1) \mod 5\)

• \(x^3-7 \equiv x^3 \mod 7\)

From this we deduce

• \((2)=\mathfrak {p}_2\mathfrak {p}_2'=(2,\alpha -1)(2,\alpha ^2+\alpha +1)\)

• \((3)=\mathfrak {p}_3^3=(3,\alpha -1)^3\)

• \((5)=\mathfrak {p}_5\mathfrak {p}_5'=(5,\alpha -3)(5,\alpha ^2+3\alpha -1)\)

• \((7)=\mathfrak {p}_7^3=(7,\alpha )^3=(\alpha )^3\)

By Theorem 3.5.7 we know that \(N({\mathfrak {p}_i})=p^{f_{\mathfrak {p}_i|p}}\) so if we calculate the residue degrees, we will know the norms of these ideals. But again Theorem 3.5.10 tells us that this is exactly the degrees of the polynomials appearing on the right hand side of each factorization modulo \(p\). So combining this we have:

• \(N(\mathfrak {p}_2)=2\) and \(N(\mathfrak {p}_2')=4\)

• \(N(\mathfrak {p}_3)=3\)

• \(N(\mathfrak {p}_5)=5\) and \(N(\mathfrak {p}_5')=25\)

• \(N(\mathfrak {p}_7)=7\)

So by combining these ideals we can construct any ideal of norm at most \(10\). The next question is are these ideals distinct in \(\operatorname{Cl}_K\)? Lets see if any of them are principal:

Clearly, \((7)=(\alpha )^3\) is principal so this gives the trivial class in the class group. Note that since \((2)=\mathfrak {p}_2\mathfrak {p}_2'\) we have \([\mathfrak {p}_2]^{-1}=[\mathfrak {p}_2']\).

So lets look at the other ideals.

Note that

So, do we have elements of norm \(2,3\) or \(5\)? We’ll from the above we see that the norm of an element must be a cube modulo \(7\). But the only cubes modulo \(7\) are \(\pm 1\). So there are no elements of norm \(2,3,5\). So these ideals are not principal. So \(\operatorname{Cl}_K\) is generated by \(\mathfrak {p}_{2},\mathfrak {p}_3,\mathfrak {p}_5\).

The next question is, do these ideals give distinct classes in \(\operatorname{Cl}_K\)? First note that \(\mathfrak {p}_3^3=(3)\) therefore \([\mathfrak {p}_3]\) has order \(3\) in \(\operatorname{Cl}_K\) (since we already know its not trivial). Note that \(N_{K/\mathbb {Q}}(2+\sqrt[3]{7})=15\) and \(N_{K/\mathbb {Q}}(-1+\sqrt[3]{7})=6\).

Therefore, since we only have one prime ideal of norm \(2,3\) and \(5\) we must have

thus \([\mathfrak {p}_5][\mathfrak {p}_3]=[1]\) and \([\mathfrak {p}_{2}][\mathfrak {p}_3]=[1]\) giving \([\mathfrak {p}_5]=[\mathfrak {p}_3]^{-1}=[\mathfrak {p}_2]\). Thus \(\operatorname{Cl}_K\) is generated by \([\mathfrak {p}_3]\) and is therefore cyclic of order \(3\). The multiplication table is then