# Methods to show an ideal in the ring of integers $\mathcal{O}_K$ is a proper ideal

I was wondering if there were general "tactics" to show if an ideal in the ring of integers $\mathcal{O}_K$, where $K/\mathbb{Q}$ is a number field of degree $n$.

For example, consider $\mathcal{O}_K = \mathbb{Z}[\sqrt{-5}]$ and $P = (2,1+\sqrt{5})$. One method to show this is a proper idea is to argue that if $a + b\sqrt{-5} \in P,$ then $a+b\sqrt{-5} = 2x + y(1+\sqrt{-5})$ for integers $x,y$ and comparing coefficients, we get $a - b \equiv 0 \bmod 2,$ so $P$ cannot be the whole ring. However, this method seems quite laborious, and this may just be a special case when modding out by a prime actually works.

For me, the following isomorphism seems quite intuitive: as $\mathbb{Z}[\sqrt{-5}] \cong \mathbb{Z}[x]/(x^2+5),$ we should have that \begin{align*} \mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{5}) &\cong \mathbb{Z}[x]/(x^2+5,2,1+x) \\ &\cong \mathbb{Z}/(6,2) \text{ by mapping } x \mapsto -1 \\ &\cong \mathbb{Z}/2\mathbb{Z} \end{align*} so the ideal $P$ is indeed proper. But I can't seem to set up an explicit isomorphism at the moment, so I was wondering if this works generally?

Furthermore, are there any useful techniques one might use to show an ideal is indeed proper?

• Could you please explain what you mean when you say that you can't set up an explicit isomorphism? Doesn't your isomorphism send a coset $a + b\sqrt{-5} + (2, 1+\sqrt{-5})$ in $\mathbb Q [\sqrt{-5}]/(2,1+\sqrt{5})$ to the element $a - b \in \mathbb Z / 2 \mathbb Z$ (i.e. we think of $\sqrt{5}$ as $x$, then map $x \mapsto -1$, like in your second method)? And doesn't this agree perfectly with your first method? – Kenny Wong May 31 '17 at 23:43
• There are other ways to see that $(2, 1+\sqrt{-5})$ is a proper ideal. For example, we can stare at the ideal-theoretical equation $(2)(3) = (1+\sqrt{-5})(1 - \sqrt{-5})$ and deduce that any prime ideal that divides $(2)$ must also divide $(1+\sqrt{-5})$ or $(1-\sqrt{-5})$, and hence, any such prime ideal divides $(2, 1+\sqrt{-5}) = (2, 1- \sqrt{-5})$. Or we can apply Dedekind's factorization theorem to show that $(2) = (2, 1+\sqrt{-5})^2$. But admittedly, these methods don't work for every ideal in every ring of integers. – Kenny Wong May 31 '17 at 23:47
• your isomorphism isn't intuitive at all considering that if you map $x$ to $2$ instead you get something completely different. – mercio Jun 1 '17 at 9:20
• @user1952009 I'm sure you know much more about this than me! For the second method, we're essentially testing whether or not $2$, $1+x$ and $x^2+1$ have g.c.d. $=1$, and it would be natural to use the Euclidean algorithm. I wonder if this is essentially equivalent to using Grobner bases? (I can't remember how the Grobner basis algorithm works off-hand...) – Kenny Wong Jun 1 '17 at 9:23
• @mercio Could you please explain why we would choose to map $x \mapsto 2$ here? The reason we map $x \mapsto -1$ is that $x+ 1$ is in the ideal, so we're quotienting by $x + 1$. (We could just as well map $x \mapsto 85$, because $x + 85 = (x+1)+42\times 2$ is in the ideal, and ultimately we'll get the same answer. But I don't think we can map $x \mapsto 2$...) – Kenny Wong Jun 1 '17 at 9:25

I'm sure there is a much better way, but yes it works generally, assuming you know the multiplication law in $\mathcal{O}_K$ seen as a free $\mathbb{Z}$ module.

Find $r \in I \cap \mathbb{Z}$, list all the elements of the finite ring $$R_0 = \mathcal{O}_K/(r)$$ then write $I = (u_1,\ldots,u_n)$ and list all the elements of the finite quotient rings $$R_1= R_0/(u_1), \quad R_2= R_1/(u_2), \quad R_{m+1} = R_m/(u_{m+1})$$ You'll get $$I = \mathcal{O}_K \qquad \Longleftrightarrow \qquad \mathcal{O}_K / I = R_n = \{0\}$$

The more general setup is $$\mathcal{O}_K = \mathbb{Z}[X_1,\ldots,X_k]/J$$ for some ideal $J$, and we want to know if $$\mathcal{O}_K= I \qquad \Longleftrightarrow \qquad (I,J) = \mathbb{Z}[X_1,\ldots,X_k]$$ there is an algorithm for that using Gröbner basis

A "proper" ideal is any ideal "which is strictly smaller than the whole ring," right?

Then it's enough to show that the given ideal does not contain 1. If the ideal contains a unit, it must also contain 1 and must therefore be the whole ring.

In your example, we have $\mathfrak P = \langle 2, 1 + \sqrt{-5} \rangle$, which is simply the set of all numbers in this domain of the form $2x + y(1 + \sqrt{-5}) = a + b \sqrt{-5}$. The contribution of $2x$ to $a$ and $b$ is even. We can certainly choose $y$ so that $a = 1$, but then $b \neq 0$ like we want, since in fact $b$ must be odd for $a$ to be odd.

I can't set up an isomorphism either, that's a deficiency of mine, but it's kind of overkill if you just want to show that an ideal is proper.

To know is an ideal $\mathfrak{a}$ is proper or not isn't easier if we just compute the size of $\mathcal{O}_K/ \mathfrak{a}$ thinking of $\mathcal{O}_K$ as a lattice?.

As long as you know a $\mathbb{Z}$-basis for $\mathcal{O}_K$ say $\{ \gamma_1, \ldots, \gamma_n \}$ and you're given $\mathfrak{a}$ is terms of $\mathcal{O}_K$-generators say $(\beta_1,\ldots,\beta_m)$ you can esaily find a $\mathbb{Z}$-basis $\{ \alpha_1, \ldots, \alpha_n \}$ for $\mathfrak{a}$ ( reducing the $\mathbb{Z}$-generators of $\mathfrak{a}$ $\{\beta_i \gamma_j \}$ to a $\mathbb{Z}$-basis, for example by hermite form). Then you simply write $\alpha_i=\sum a_{ij} \cdot \gamma_j$ and $| \mathcal{O}_K/ \mathfrak{a} |=|\text{det}(a_{ij})|$ would be $1$ iff $\mathcal{O}_K=\mathfrak{a}$.

In your example $\{1, \sqrt{-5} \}$ is a $\mathbb{Z}$-basis for $\mathcal{O}_K$, $\{2,2\sqrt{-5},1+\sqrt{-5}, (1+\sqrt{-5})\sqrt{-5} \}$ reduces to the $\mathbb{Z}$-basis $\{1+\sqrt{-5},2 \sqrt{-5}\}$ for $P$, because the hermite form of $$\begin{bmatrix} 2 & 0 \\ 0 & 2 \\ 1 & 1\\ -5 & 1 \end{bmatrix}$$ is the matrix

$$\begin{bmatrix} 1 & 1 \\ 0 & 2 \\ 0 & 0\\ 0 & 0 \end{bmatrix}$$

and so $| \mathcal{O}_K/ P |=\text{det}\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}$$=2\neq 1$.

• this is what does sagemath (see for example the function __cmp__ where self.pari_rhnf is the Hermite normal form of the ideal) – reuns Jun 2 '17 at 17:22