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If $X$ is an integral, projective scheme over a ring $A$, $Z\subset X$ a nonempty, reduced, closed subscheme, and $U = X\setminus Z$, can $U$ detect whether the $Z$ is reducible? What about $Z$ smooth or nonsingular?

What about the special cases: (1) $Z$ is pure codimension 1 (2) $X = \mathbb{P}^n_k$

To elaborate, if we assume (1), then I believe $U$ will be affine, in which case we now have two rings $\Gamma(Z)$ and $\Gamma(U)$. Given $X$, each of $Z$ and $U$ (with its embedding) determines the other, so we might hope that certain information in $\Gamma(Z)$ is reflected in $\Gamma(U)$. I am interested in any kind of information that is reflected, but reducibility was the first specific property I was considering.

The presence of zero-divisors in $\Gamma(Z)$ determines its reducibility. Is there a corresponding property in $\Gamma(U)$?

I guess I am vaguely thinking of $X$ as some kind of dualizing space, which is why I figured it would need to be sufficiently nice - certainly projective, and possibly even projective space itself.

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    What kind of information are you allowed to put in? As it stands, the question is vague. For example, if you assume both (1) and (2) above, what would be a satisfactory answer for you? – 2017-01-21
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    I tried to make the question a little more precise. – 2017-01-22
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    If you only assume (1), there is no reason for $U$ to be affine. For example, let $X$ be the blow up of a point on a smooth variety and let $Z$ be the exceptional divisor. I think, the question is still too vague for a decent answer. Also, while it is true that zero divisors on $\Gamma(Z)$ may show reducibility, $Z$ can be reducible with no zero divisors in $\Gamma(Z)$. – 2017-01-22
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    Ok, I guess I should assume both (1) and (2) in hopes of an interesting answer. How can $Z$ be reducible with no zero divisors? – 2017-01-22
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    Take $Z$ to be a hypersurface in projective space of dimension at least 2. Whether $Z$ is reducible or not, $\Gamma(Z)=k$, the base field. – 2017-01-22
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    Yes, you are right. I was thinking $Z$ itself was affine. I guess this question has a lot more issues than I thought. – 2017-01-22

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