When is an Ideal of a PID ring also a vector space?

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I'm reading David R. Finston and Patrick J. Morandi's book Abstract Algebra: Structure and Application and in section 7.1 page 107 Theorem 7.6 it mentions

... if $C$ is an ideal of $\mathbb{Z}_2[x]/(x^n+1)$, then $C$ is a $\mathbb{Z}_2$ vector subspace of $\mathbb{Z}_2[x]/(x^n+1)$ ...

I wonder, in general, if $C$ is an ideal of a PID ring $R$, when $C$ can be seen as a vector space?

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An ideal $I$ of a ring $R$ with identity is always a unital module over a subring $S$ of $R$. This just follows from the definition of "ideal." $I$ is already an additive group, and the way $S$ acts on it via multiplication already satisfies the module axioms (since it satisfies the ring multiplication axioms.)

In particular, if $S$ is a field, you could call $I$ a vector space (it's just a specific term for a module over a field or division ring.)

Of course, there is no sense in asking "can something be seen as a vector space" without a field being mentioned. You'd have to specify what field you were interested in.