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Corrected error pointed out by Charles Siegel
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inkspot
inkspot
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A scheme $Z$ is connected if and only if the only idempotent $f\in H^0(Z,\mathcal O_Z)$ are $f=0,1$. Since $H^0(S,\mathcal O_S)=H^0(S\times Y,\mathcal O_{S\times Y})$ for any proper $k$-variety $Y$ such that $k$ is algebraically closed in $H^0(Y,\mathcal O_Y)$, it follows that $S$ is connected if and only if $S\times Y$ is so.

A scheme $Z$ is connected if and only if the only idempotent $f\in H^0(Z,\mathcal O_Z)$ are $f=0,1$. Since $H^0(S,\mathcal O_S)=H^0(S\times Y,\mathcal O_{S\times Y})$ for any proper $k$-variety $Y$, it follows that $S$ is connected if and only if $S\times Y$ is so.

A scheme $Z$ is connected if and only if the only idempotent $f\in H^0(Z,\mathcal O_Z)$ are $f=0,1$. Since $H^0(S,\mathcal O_S)=H^0(S\times Y,\mathcal O_{S\times Y})$ for any proper $k$-variety $Y$ such that $k$ is algebraically closed in $H^0(Y,\mathcal O_Y)$, it follows that $S$ is connected if and only if $S\times Y$ is so.

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inkspot
inkspot
  • 3.1k
  • 20
  • 15

A scheme $Z$ is connected if and only if the only idempotent $f\in H^0(Z,\mathcal O_Z)$ are $f=0,1$. Since $H^0(S,\mathcal O_S)=H^0(S\times Y,\mathcal O_{S\times Y})$ for any proper $k$-variety $Y$, it follows that $S$ is connected if and only if $S\times Y$ is so.