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Let $K$ be a number field over $\mathbb{Q}$ and let $n$ be $[K:\mathbb{Q}]$

Then it is clear for an ideal $I \in I(K)$ that $I$ is a vector space over $K$ of degree $\leq n$.

Now it is clear that a principal ideal $P \in P(K)$ is a vector space over $\mathbb{Q}$ of degree $n$.

It follows therefore that an ideal $I$ is a vector space over $\mathbb{Q}$ of degree $\leq n^n$.

Can you do better? It's wishful thinking to suppose $I$ is a vector space over $\mathbb{Q}$ of degree $\leq n$ or isn't it?

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The ideal $I$ is not a vector space over either $K$ or $\mathbb Q$ (or any field), so it does not have a $K$- or $\mathbb Q$-dimension.

An ideal $I$ of a number field $K/\mathbb Q$ is by definition an ideal of the ring of integers $\mathcal O_K$. In particular, it's an $\mathcal O_K$-module. The ring $\mathcal O_K$ is a free $\mathbb Z$-module of rank $n = [K\,\colon\mathbb Q]$, so if $I$ is a principal ideal, then $I$ is also a free $\mathbb Z$-module of rank $n$. On the other hand, it is true that any $I$ can be generated as an ideal by two elements (this is a consequence of the fact that $\mathcal O_K$ is a Dedekind domain). Note however that if $I$ is not principal (so you really need at least two generators), then it cannot be free as a $\mathbb Z$-module, so it is not isomorphic to any $\mathbb Z^k$. The best you can say is that $I$ can be generated by $2n$ elements over $\mathbb Z$.

Explicitly, if $\alpha_1,\dots ,\alpha_n$ forms an integral basis for $\mathcal O_K$, so that the $\alpha_i$ generate $\mathcal O_K$ as a $\mathbb Z$-module, and if $I$ is generated by $\theta, \phi\in\mathcal O_K$, then $\{ \theta\alpha_1, \dots , \theta\alpha_n, \phi\alpha_1, \dots , \phi\alpha_n \}$ is a set of generators for $I$ over $\mathbb Z$.

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