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The norm of an ideal $\mathfrak{a}\unlhd \mathcal{O}_K$ in a number field $K$ is defined by $N(\mathfrak{a}):=(\mathcal{O}_K:\mathfrak{a})$. One justifies this definition by the observation that $$ N\big((\alpha)\big)=|N_{K/\mathbb{Q}}(\alpha)|. $$ (See Neukirch's 'Algebraic Number Theory', page 35.) I'm having some trouble computing a should-be-easy example of this 'justification' for non-integer values of $\alpha$. In what follows, let's take $K=\mathbb{Q}(i).$

Example that I understand: If $\alpha=5$ then we know from looking at the characterization of $N_{K/\mathbb{Q}}$ in terms of embeddings into $\overline{\mathbb{Q}}$ that $N_{K/\mathbb{Q}}(5)=25$. To compute the ideal norm, we know $\mathcal{O}_K=\mathbb{Z}[i]$ has integral bases $1,i$ and so $(5)$ has $\mathbb{Z}$-basis $5,5i$. So $\mathcal{O}_K=\mathbb{Z}+i\mathbb{Z}\cong \mathbb{Z}^2$ and $(5)=5\mathcal{O}_K\cong 5\mathbb{Z}^2$. Since $$ \frac{\mathcal{O}_K}{5\mathcal{O}_K}\cong\frac{\mathbb{Z}\oplus \mathbb{Z}}{5\mathbb{Z}\oplus 5\mathbb{Z}}\cong \mathbb{Z}/5\mathbb{Z}\oplus \mathbb{Z}/5\mathbb{Z}\ $$ we get $N\big((\alpha)\big)=25$ as well.

Example where I need help: Now let's take $\alpha=2+i$. In this case, $$N_{K/\mathbb{Q}}(\alpha)=(2+i)(2-i)=5,$$ but I can't seem to show that $N\big((\alpha)\big)=5$. The issue is that I can't seem to get a nice $\mathbb{Z}$-basis for $\mathbb{Z}[i]$ so that the basis elements for $(\alpha)$ are just integer multiples of the basis for $\mathbb{Z}[i]$, as in the previous example. More clearly, I'd like some integral basis $v_1,v_2\in \mathbb{Z}[i]$ so that $\alpha v_i=d_iv_i$, $i=1,2$. If I can find such a basis then $$ (\alpha)=\alpha\mathcal{O}_K=\mathbb{Z} \alpha_1v_1+\mathbb{Z} \alpha_2v_2=\mathbb{Z} d_1v_1+\mathbb{Z} d_2v_2, $$ in which case it would follow that $$ \frac{\mathcal{O}_K}{\alpha\mathcal{O}_K}\cong \frac{\mathbb{Z}\oplus \mathbb{Z}}{d_1\mathbb{Z}\oplus d_2\mathbb{Z}}, $$ which would allow me to compute the ideal norm easily (it would just be the determinant $d_1d_2$, which I know has to be 5). At this point, I feel like it should be an easy linear algebra problem (Smith normal form??), but I can't get it to work. I would appreciate a refresher of how these types of computations work, particularly the linear algebra details.

Arbutus
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  • Why don’t you just show that every Gaussian integer is congruent to $0,1,2,3$, or $4$ modulo $(2+i)$ ? For instance, $i-3=(-1+i)(2+i)$, so that $i\equiv3\pmod{2+i}$. – Lubin May 22 '18 at 04:36
  • @Lubin, thank you so much for your comment! I agree that your method does seem the easiest route in this case. The reason I was trying the more linear algebraic approach was to understand, with an example, Neukirch's comments following his defining of the ideal norm (page 35 in my edition). Essentially, he says that $N((\alpha))=|\det A|$ where $A$ is the transition matrix from an integral basis of $\mathcal{O}_K$ to one for $(\alpha)$. Again, thanks for your comment and very delighted to make your virtual acquaintance. – Arbutus May 22 '18 at 05:14
  • Well, I guess you take the $\Bbb Z$-basis of $\Bbb Z[i]$, namely ${1,i}$, and multiply both by $2+i$ to get ${2+i,-1+2i}$. I guess that’s what @Quasicoherent did in his/her exhaustive answer. – Lubin May 22 '18 at 14:31

1 Answers1

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First, here's a quicker way to compute the norm of $(2 + i)$. Since $\mathbb{Z}[i] \cong \frac{\mathbb{Z}[x]}{(x^2+1)}$, then the Third Isomorphism Theorem implies \begin{align*} \frac{\mathbb{Z}[i]}{(2+i)} &\cong \frac{\mathbb{Z}[x]/(x^2+1)}{(2 + x, x^2+1)/(x^2+1)} \cong \frac{\mathbb{Z}[x]}{(2+x, x^2 + 1)} \cong \frac{\mathbb{Z}[-2]}{((-2)^2 + 1)} = \frac{\mathbb{Z}}{(5)} \, . \end{align*}

One can certainly compute the norm as you have suggested. As you indicated, $$ (2 + i)\mathbb{Z}[i] = (2+i) (\mathbb{Z} \oplus i \mathbb{Z}) = (2+i) \mathbb{Z} \oplus (-1+2i) \mathbb{Z} $$ as $\mathbb{Z}$-modules. Since $$ 2 + i = \begin{pmatrix} 2 & 1 \end{pmatrix} \begin{pmatrix} 1\\ i \end{pmatrix} \quad \text{and} \quad -1 + 2i = \begin{pmatrix} -1 & 2 \end{pmatrix} \begin{pmatrix} 1\\ i \end{pmatrix} $$ then the ideal $(2+i)$ is represented by the matrix $$ A = \begin{pmatrix} 2 & 1\\ -1 & 2 \end{pmatrix} $$ with respect to the basis $1,i$. You are correct that the Smith normal form $D$ of this matrix $A$ will allows us to see the structure of the quotient. We can compute this by using row and column operations, and if we augment $A$ with identity matrices we can simultaneously obtain the change of basis matrices $P, Q$ such that $PAQ = D$. \begin{align*} \left(\begin{array}{rr|rr} 1 & 0 & 2 & 1 \\ 0 & 1 & -1 & 2 \\ \hline 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) &\leadsto \left(\begin{array}{rr|rr} 1 & 1 & 1 & 3 \\ 0 & 1 & -1 & 2 \\ \hline 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \leadsto \left(\begin{array}{rr|rr} 1 & 1 & 1 & 0 \\ 0 & 1 & -1 & 5 \\ \hline 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 1 \end{array}\right)\\ &\leadsto \left(\begin{array}{rr|rr} 1 & 1 & 1 & 0 \\ 1 & 2 & 0 & 5 \\ \hline 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 1 \end{array}\right) \end{align*} From this we can already see that $$ \frac{\mathbb{Z}[i]}{(2+i)} \cong \frac{\mathbb{Z} \oplus \mathbb{Z}}{\mathbb{Z} \oplus 5 \mathbb{Z}} \cong \frac{\mathbb{Z}}{5\mathbb{Z}} $$ as $\mathbb{Z}$-modules. Moreover, $P^{-1}$ gives us a basis $v_1, v_2$ for $\mathbb{Z}[i]$ such that $(2+i) = \mathbb{Z} v_1 \oplus \mathbb{Z} 5 v_2$. (For more details, see this post.) Since $$ P^{-1} = \left(\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right) $$ we find that $v_1 = 2 - i$ and $v_2 = -1 + i$.

For more on Smith normal form, I recommend this note by Keith Conrad, and this post.

Viktor Vaughn
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