Let \(\mathscr{P}\) be a property of morphisms of schemes such that
a closed immersion has \(\mathscr{P}\);
a composition of two morphisms having \(\mathscr{P}\) has \(\mathscr{P}\);
\(\mathscr{P}\) is stable under base extension.
Then show that
a product of morphisms having \(\mathscr{P}\) has \(\mathscr{P}\);
if \(f: X \to Y\) and \(g: Y \to Z\) are two morphisms, and if \(g \circ f\) has \(\mathscr{P}\) and \(g\) is separated, then \(f\) has \(\mathscr{P}\);
if \(f: X \to Y\) has \(\mathscr{P}\), then \(f_{\text{red}}: X_{\text{red}} \to Y_{\text{red}}\) has \(\mathscr{P}\).
Hint: for (e), consider the graph morphism \(\Gamma_f: X \to \fiberprod{X}{Z}{Y}\) and note that it is obtained by base extension from the diagonal morphism \(\Delta: Y \to \fiberprod{Y}{Z}{Y}\).
Solution
Assume throughout that the property \(\mathscr P\) satisfies (a), (b), and (c).
1. Let\[ f:X\to Y, \qquad f':X'\to Y' \]have \(\mathscr P\).Then the morphism\[ f\times\id_{X'}:X\times X'\longrightarrow Y\times X' \]has \(\mathscr P\).
The displayed morphism is the base change of \[ f:X\to Y \] along the projection \[ Y\times X'\longrightarrow Y. \] Property (c) says that \(\mathscr P\) is stable under arbitrary base extension, so the base-changed morphism has \(\mathscr P\).
2. The morphism\[ \id_Y\times f':Y\times X'\longrightarrow Y\times Y' \]has \(\mathscr P\).
This is the base change of \[ f':X'\to Y' \] along the projection \[ Y\times Y'\longrightarrow Y'. \] Apply property (c).
3. The product morphism\[ \boxed{ f\times f':X\times X'\longrightarrow Y\times Y' } \]has \(\mathscr P\).
It factors as \[ X\times X' \xrightarrow{f\times\id_{X'}} Y\times X' \xrightarrow{\id_Y\times f'} Y\times Y'. \] Both factors have \(\mathscr P\) by <1>1–<1>2, so their composition has \(\mathscr P\) by property (b). This proves part (d).
4. Let\[ X\xrightarrow{f}Y\xrightarrow{g}Z \]with \(g\circ f\) having \(\mathscr P\) and \(g\) separated.The graph\[ \Gamma_f:X\longrightarrow X\times_ZY, \qquad x\longmapsto(x,f(x)), \]is a closed immersion.
Consider the Cartesian square \[ \begin{array}{ccc} X&\xrightarrow{\Gamma_f}&X\times_ZY\\ \downarrow{\scriptstyle f}&&\downarrow{\scriptstyle f\times\id_Y}\\ Y&\xrightarrow{\Delta_g}&Y\times_ZY. \end{array} \] The graph is the base change of the diagonal \[ \Delta_g:Y\to Y\times_ZY. \]
Since \(g\) is separated, \(\Delta_g\) is a closed immersion. Closed immersions are stable under base change, so \(\Gamma_f\) is a closed immersion.
5. The graph morphism \(\Gamma_f\) has \(\mathscr P\).
By 4 it is a closed immersion. Property (a) says every closed immersion has \(\mathscr P\).
6. The projection\[ p_2:X\times_ZY\longrightarrow Y \]has \(\mathscr P\).
The projection \(p_2\) is the base change of \[ g\circ f:X\longrightarrow Z \] along \[ g:Y\longrightarrow Z. \] Since \(g\circ f\) has \(\mathscr P\), property (c) gives \(\mathscr P\) for \(p_2\).
7. The morphism\[ \boxed{f:X\to Y} \]has \(\mathscr P\).
The factorization \[ f=p_2\circ\Gamma_f \] has both factors satisfying \(\mathscr P\) by <1>5–<1>6. Property (b) gives \(\mathscr P\) for their composition. This proves part (e).
8. Let\[ f:X\longrightarrow Y \]have \(\mathscr P\). The composite\[ X_{\mathrm{red}} \hookrightarrow X \xrightarrow{f} Y \]has \(\mathscr P\).
The reduction morphism \[ i_X:X_{\mathrm{red}}\hookrightarrow X \] is a closed immersion, hence has \(\mathscr P\) by (a). The morphism \(f\) has \(\mathscr P\) by hypothesis. Therefore the composite \[ f\circ i_X:X_{\mathrm{red}}\to Y \] has \(\mathscr P\) by (b).
9. The composite in 8 factors uniquely as\[ X_{\mathrm{red}} \xrightarrow{f_{\mathrm{red}}} Y_{\mathrm{red}} \xrightarrow{i_Y} Y. \]
The scheme \(X_{\mathrm{red}}\) is reduced. By the universal property of the reduction of \(Y\), every morphism from a reduced scheme to \(Y\) factors uniquely through \[ i_Y:Y_{\mathrm{red}}\hookrightarrow Y. \] The resulting morphism is exactly \(f_{\mathrm{red}}\).
10. The morphism\[ i_Y:Y_{\mathrm{red}}\hookrightarrow Y \]is separated.
It is a closed immersion. Every closed immersion is separated.
11. The reduced morphism\[ \boxed{ f_{\mathrm{red}}:X_{\mathrm{red}}\longrightarrow Y_{\mathrm{red}} } \]has \(\mathscr P\).
By <1>8–<1>9, the composite \[ i_Y\circ f_{\mathrm{red}} \] has \(\mathscr P\). By 10, \(i_Y\) is separated. Applying part (e), proved in <1>4–<1>7, to \[ X_{\mathrm{red}} \xrightarrow{f_{\mathrm{red}}} Y_{\mathrm{red}} \xrightarrow{i_Y} Y \] shows that \(f_{\mathrm{red}}\) has \(\mathscr P\). This proves part (f).
12. Q.E.D.
Step 3 proves part (d), 7 proves part (e), and 11 proves part (f).