Are there any homological checks I can use to check if a projective morphism is flat? For example, I would expect the following projective morphism to be flat $$ \textbf{Proj}\left( \frac{\mathbb{C}[s][x,y,z]}{x^3 + y^3  xyz + sz^3} \right) \to \textbf{Spec}(\mathbb{C}[s]) $$ while $$ \textbf{Proj}\left( \frac{\mathbb{C}[s][x,y,z]}{s(x^4 + y^4 + z^4)} \right) \to \textbf{Spec}(\mathbb{C}[s]) $$ would not be flat.
For a scheme to be flat over $A^1_s = Spec(C[s])$ is equivalent to not having irreducible (or embedded) components supported over (closed) points of $A^1_s$. Since your schemes are hypersurfaces in $A^1_s \times A^3_{x,y,z}$, having such a component is equivalent to containing $\{a\} \times A^3_{x,y,z}$ as a subvariety for some point $a \in A^1_s$. The later is equivalent to the equation of the hypersurface being nonzero after substitution $s = a$. So, it easily follows that the first hypersurface is flat, while the second is not.

$\begingroup$ Isn't $\textbf{Spec}(\mathbb{C}[t][x,y]/(xy  t) \to \textbf{Spec}(\mathbb{C}[t])$ a flat morphism? If you try and resolve $\mathbb{C}$ as $\mathbb{C}[t] \xrightarrow{\cdot t} \mathbb{C}[t]$, you get vanishing $Tor_1$ $\endgroup$ – 54321user Aug 17 '16 at 22:20

2$\begingroup$ That doesn't contradict Sasha's answer, the source of your map is irreducible (and smooth). $\endgroup$ – abx Aug 18 '16 at 6:05

$\begingroup$ @abx but the closed fiber over $0$ is reducible. $\endgroup$ – 54321user Aug 18 '16 at 16:32

1$\begingroup$ Yes, but what counts is the irreducible components of the total space. If it is reduced and irreducible, it is automatically flat. See Hartshorne, Proposition 9.7. $\endgroup$ – abx Aug 18 '16 at 16:54