$\mathbb Z$-bases for ideals I am asked to find a $\Bbb{Z}$-basis for the ideal of $\Bbb{Z}[\sqrt{-5}]$ given by $\left< 2, 1 +\sqrt{-5} \right>$. I must be missing something because these are both integers on $\Bbb{Z}[\sqrt{-5}]$, they are generators so don't they already comprise an integer basis of the ideal?
 A: It is true that $2, 1 + \sqrt{-5}$ are a $\Bbb{Z}$-basis of the ideal  $$I = \left< 2, 1 + \sqrt{-5} \right>,$$ but this requires a proof. 
I am giving one here, but the proof of Thomas Andrews in his comment is simpler.
Consider the set
$$
J = \{ 2 a + (1 + \sqrt{-5}) b : a, b \in \Bbb{Z} \} \subseteq I.
$$ 
You want to prove that $J$ is an ideal of $\Bbb{Z}[\sqrt{-5}]$, and so $J = I$. For this, compute
$$
\begin{align}
(2 a + (1 + \sqrt{-5}) b) \cdot \sqrt{-5}
&=
2 a \sqrt{-5} + (\sqrt{-5} - 5) b
\\&=
- 5 b + (2 a + b) \cdot \sqrt{-5}
\\&=
2 (- a - 3 b)
+
(1 + \sqrt{-5} )(2 a + b) \in J.
\end{align}$$
Then maybe it is safe noting that $0 = 2 a + (1 + \sqrt{-5}) b = (2 a + b) + b \sqrt{-5}$ if and only if $b = a = 0$.
PS Just to give you a different example, which indicates that a proof is indeed needed, if you consider the ideal
$$
I' = \left\langle 2, \sqrt{-5} \right\rangle,
$$
we have $- 5 = \sqrt{-5} \cdot \sqrt{-5} \in I'$, but $-5$ is not a $\Bbb{Z}$-linear combination of $2$ and $\sqrt{-5}$.
A: There is a simple criterion using norms to test if such modules are ideals. If $\rm\:M = [a,b\!+\!\sqrt{d}]\:$ is an ideal then it contains the norm $\rm\: N(b\!+\!\sqrt{d}) = (b\!+\!\sqrt{d})(b\!-\!\sqrt{d}) = b^2\!-\!d,\ $ so $\rm\,\ a\mid b^2\!-\!d.\:$  This necessary condition is also sufficient. The proof is easy:
The module $\rm\:M = [a,b\!+\!\sqrt{d}]\:$ is an ideal of $\rm\,R = \Bbb Z[\sqrt{d}]\iff$  M is closed under multiplcation by elements of $\rm\,R\iff$ $\rm\sqrt{d}\ M \subseteq M\iff a\sqrt{d},\, (b\!+\!\sqrt{d})\sqrt{d} \in M.\:$ The first inclusion is clear since $\rm\:a\sqrt{d}\, =\, a(b\!+\!\sqrt{d})-ab\in M.\:$ For the second inclusion
$$\rm\begin{eqnarray} -(b\!+\!\sqrt{d})\sqrt{d} &=\,&\rm (b\!+\!\sqrt{d})(b\!-\!\sqrt{d})-b(b\!+\!\sqrt{d})\\
&=\,&\rm b^2\!-d -b(b\!+\!\sqrt{d})\end{eqnarray}$$
The prior is $\rm\,\in\, M = [a,b\!+\!\sqrt{d}]\iff a\mid b^2\!-\!d = N(b\!+\!\sqrt{d})\:$ where $\rm\:N = $ the norm.
This is true for $\rm\,[2,1\!+\!\sqrt{-5}],\ $ i.e. $\rm\:2\mid N(1\!+\!\sqrt{-5}) = (1\!+\!\sqrt{-5})(1\!-\!\sqrt{-5}) = 1\!+\!5 = 6.$
The criterion generalizes to test idealness of the module $\rm\,[a,b\!+\!c\,\omega]\,$ in the ring of integers of a quadratic number field, e.g. see section 2.3 in  Franz Lemmermeyer's notes.
This is a special case of module normal forms that generalize to higher degree number fields, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's $ $ A Course in Computational Number Theory.
