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Let $X$ be a smooth quasi-projective variety over a field $k$, with pure dimension $d$, $K/k$ an arbitrary field extension.

  • For any algebraic cycle $\eta$ of codimension $1$ on $X_K$ ($\eta\in Z^1(X_K)$), does there exist an algebraic cycle $\xi\in Z^1(X)$ (ie. defined over $k$) such that $\xi_K-\eta$ is effective?

  • Fix a collection $\{Y_1,\ldots, Y_n\}$ of $k$-subvarieties of $X$, and further assume each component of $\eta$ intersects each of the $(Y_i)_K$'s in prescribed fixed codimension $p$. Does there exist $\xi\in Z^1(X)$ such that each component of $\xi$ intersects each of the $Y_i$'s in codimension $p$, and $\xi_K-\eta$ is effective?

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The answer to the both questions is no. Consider $X=\mathbb{A}^1$, $k=\mathbb{Q}$, and $K=\mathbb{C}$ (any transcendental extension will do here). Let $\eta=\{\pi\}$. The cycles on $X_K$ of the form $\xi_K$ are precisely the finite Galois-stable linear combinations of elements of $\bar{\mathbb{Q}}\subset\mathbb{C}$, so $\xi_K-\eta$ can never be effective.

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  • $\begingroup$ I’m confused. If you define $Z^1(X)$ to be formal locally finite $\mathbf{Z}$-linear combinations of symbols $[Z]$ and define effective cycles as those whose total degree is positive, what’s wrong in taking $[\{\pi\}]+[\{0\}]$, or $[\varnothing]$ already effective? The OP doesn’t ask for an effective $\xi$. $\endgroup$
    – user97068
    Commented Jan 13, 2018 at 17:27
  • $\begingroup$ “symbols $[Z]$”: $Z$ over all closed sub varieties of $X$ of codimension one. $\endgroup$
    – user97068
    Commented Jan 13, 2018 at 17:37
  • $\begingroup$ @P.S. I don't understand the examples in your first comment. $[\{\pi\}]+[\{0\}]$ is not a cycle on $X$, but only on $X_K$. It we take $\xi=[\varnothing]$, then $\xi_K-\eta=-[\{\pi\}]$ is not effective. $\endgroup$
    – Julian Rosen
    Commented Jan 13, 2018 at 17:49
  • $\begingroup$ @P.S. An algebraic cycle is effective if all of its coefficients are non-negative. $\endgroup$
    – anon
    Commented Jan 13, 2018 at 17:54
  • $\begingroup$ Ah. Misremembered! $\endgroup$
    – user97068
    Commented Jan 13, 2018 at 17:55

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