Does there exist some subring $k \subset \mathbb{C}$ such that the following assertion holds?


* **($k$-Hodge conjecture)** For each nonsingular algebraic variety $X$ over $\mathbb{C}$, and each $q = 0, 1, \ldots, \dim\_{\mathbb{C}}(X)$, each class $\mathfrak{z} \in H^{2q}(X; k) \cap H^{q,q}(X)$ is a $k$-linear combination of classes of algebraic cycles.


Atiyah-Hirzebruch [proved in 1961](http://hirzebruch.mpim-bonn.mpg.de/151/1/29_Analytic%20cycles%20on%20complex%20manifolds.pdf) that $k \neq \mathbb{Z}$. As commented by Ben Wieland below, the argument of Atiyah-Hirzebruch also shows that $k \neq \mathbb{Z}\_{(p)}$ for any prime $p$. 


**Remark:** If the millenium problem is true, then we may take $k = \mathbb{Q}$.


(The above post has been slightly edited from the original).



Status of Hodge conjecture over subrings of $\mathbb{C}$
The $k$-Hodge conjecture is false for any $k$ containing an irrational real $\alpha$.


Indeed we may clearly assume $\alpha$ is negative. Consider the elliptic curve $E = \mathbb C / \langle 1, \sqrt{\alpha} \rangle$. Let $e\_1$ and $e\_2$ be a basis of $H^1(E,\mathbb Z)$ such that the map to $H^{1,0}(E)$ sends $e\_1$ to $1$ and $e\_2$ to $\sqrt{\alpha}$. Then in $H^1(E,k) \otimes H^1(E,k) \subset H^2(E \times E, k)$, consider the class $$\alpha( e\_1 \otimes e\_1) - (e\_2 \otimes e\_2)$$


In the natural projection to $H^{2,0}$, it is sent to $\alpha(1) - (\sqrt{\alpha})^2=0$. In the natural projection to $H^{2,0}$, which is the complex conjugate of that, it is sent to $\alpha (1) - (-\sqrt{\alpha})^2=0$. So it lies in $H^{1,1}(E\times E) $.


But because $E$ is not CM (because $\alpha$ is irrational $\sqrt{\alpha}$ is not a quadratic irrational), the only algebraic cycle that contributes to $H^1(E) \otimes H^1(E)$ is the diagonal, which has class $e\_1 \otimes e\_2 - e\_2 \otimes e\_1$, and this class is not a multiple of it.




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On the other hand, assuming the $\mathbb Q$-Hodge conjecture, we obtain the $k$-Hodge conjecture for any imaginary quadratic field $k$. Indeed, for $k= \mathbb Q(\sqrt{-D})$, $H^{2q}(X,\mathbb Q(\sqrt{-D}))= H^{2q}(X,\mathbb Q) + \sqrt{-D}H^{2q}(X,\mathbb Q)$. Because complex conjugation exchanges $H^{q,p}$ and $H^{p,q}$, it fixes $H^{q,q}$, so if a class in $H^{2q}(X,\mathbb Q(\sqrt{-D}))$ lands in $H^{q,q}$, both its real and imaginary parts do as well, so under the Hodge conjecture they are $\mathbb Q$-linear combinations of algebraic cycles, and the original is clearly a $\mathbb Q(\sqrt{-D})$-linear combination of algebraic cycles.


I'm not sure about number fields of degree at least $3$ that contain no real irrationals.


