Let $m \in \mathbb{N}^*$. Denote the $m$-th cyclotomic polynomial by $\Phi(x)$ and a complex primitive $m$-th root of unity by $\omega$.
Let $K = \mathbb{Q}[x]/\langle \Phi(x) \rangle$ (which is isomorphic to $\mathbb{Q}(\omega)$).
So, I know that the ring of integers $\mathcal{O}_K$ is $\mathbb{Z}[\omega]$ (which is isomorphic to $\mathbb{Z}[x]/\langle \Phi(x) \rangle$) and therefore can be seen as a $\mathbb{Z}$-module of dimension $\varphi(m)$.
I also know that $\mathbb{R}$ is a vector space over $\mathbb{Q}$, however with bases of infinity cardinality.
But the tensor product is computed by tensoring vectors in each basis. Then, how could I define the set $\mathcal{O}_K \otimes_\mathbb{Q} \mathbb{R}$? Are the elements therein polynomials?
More context
I have found such tensor product in Lemma 3 of this article.
First of all, $\mathcal O_K$ is not a $\mathbb Q$-vector space, so I presume it's a type which should say $\mathcal O_K \otimes_{\mathbb Z}\mathbb R$. This tensor product is a change of base rings / extension of scalars, so:
$$\mathcal O_K\otimes_{\mathbb Z}\mathbb R = \mathbb R[x]/(\Phi(x))$$ Yes, the elements are equivalence classes of polynomials. However, you don't need to think about $\mathbb R$ being an infinite dimensional vector space over $\mathbb Q$ here, that's what you would do if you considered this as a $\mathbb Q$-vector space or abelian group. It's more appropriate to view it just as a finite dimensional real vector space.