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Allen Knutson
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Let $B={\rm Spec}\\, k[x,y]/(xy)$, i.e., the intersectionunion of two lines. There is an obvious flat morphism to the line $p:B\to A={\rm Spec}\\, k[x]$. Now let $X$ be a reduced scheme. $Z=X\times B$, and $f:Z\to A$ the composition of the projection to $B$ with $p$. The projection is flat, and hence so is $f$.

Now the fiber over $(x)\in A$ is "$2X$", a non-reduced scheme with support equal to $X$.

As I mentioned in my comment above, one can definitely not prescribe the scheme structure on $X$.

Let $B={\rm Spec}\\, k[x,y]/(xy)$, i.e., the intersection of two lines. There is an obvious flat morphism to the line $p:B\to A={\rm Spec}\\, k[x]$. Now let $X$ be a reduced scheme. $Z=X\times B$, and $f:Z\to A$ the composition of the projection to $B$ with $p$. The projection is flat, and hence so is $f$.

Now the fiber over $(x)\in A$ is "$2X$", a non-reduced scheme with support equal to $X$.

As I mentioned in my comment above, one can definitely not prescribe the scheme structure on $X$.

Let $B={\rm Spec}\\, k[x,y]/(xy)$, i.e., the union of two lines. There is an obvious flat morphism to the line $p:B\to A={\rm Spec}\\, k[x]$. Now let $X$ be a reduced scheme. $Z=X\times B$, and $f:Z\to A$ the composition of the projection to $B$ with $p$. The projection is flat, and hence so is $f$.

Now the fiber over $(x)\in A$ is "$2X$", a non-reduced scheme with support equal to $X$.

As I mentioned in my comment above, one can definitely not prescribe the scheme structure on $X$.

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Sándor Kovács
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Let $B={\rm Spec}\\, k[x,y]/(xy)$, i.e., the intersection of two lines. There is an obvious flat morphism to the line $p:B\to A={\rm Spec}\\, k[x]$. Now let $X$ be a reduced scheme. $Z=X\times B$, and $f:Z\to A$ the composition of the projection to $B$ with $p$. The projection is flat, and hence so is $f$.

Now the fiber over $(x)\in A$ is "$2X$", a non-reduced scheme with support equal to $X$.

As I mentioned in my comment above, one can definitely not prescribe the scheme structure on $X$.