Hartshorne Exercise II.5.12.
Let $\varphi: \PP{k}{n}\ra\PP{k}{m}$ be a morphism. Then:
(a) Let $X$ be a scheme over a scheme $Y$, and let $\mathcal L,\mathcal M$ be two very ample invertible sheaves on $X$. Show that $\mathcal L\otimes\mathcal M$ is also very ample. [Hint: Use a Segre embedding.]
(b) Let $f:X\ra Y$ and $g:Y\ra Z$ be two morphisms of schemes. Let $\mathcal L$ be a very ample invertible sheaf on $X$ relative to $Y$, and let $\mathcal M$ be a very ample invertible sheaf on $Y$ relative to $Z$. Show that $\mathcal L\otimes f^*\mathcal M$ is a very ample invertible sheaf on $X$ relative to $Z$.
(a) We have immersions $\iota_1:X\ra \PP{Y}{n}$ and $\iota_2:X\ra\PP{Y}{m}$ such that $\iota_1^*\OO(1)=\mathcal L$ and $\iota_2^*\OO(1)=\mathcal{M}$ . We note that these need not be closed immersions, according to Hartshorn. When they are closed immersions, then they are projective morphisms. Apparently Hartshorn’s definition of very ample is slightly stronger than the one given in EGA. In EGA, a line bundle is very ample if it gives a locally closed immersion in some projective space bundle $\Bbb P(\mathcal E)$. Although weaker, a composition of locally closed immersions remains a locally closed immersion. Hartshorn’s definition of immersion, however, is not closed under composition. I feel as though to do this problem, we must have that a composition of two immersions is an immersion. Hence I will assume that an “immersion” is a locally closed one.
Consider the Segre embedding $\iota:\PP{Y}{n}\times\PP{Y}{m}\ra \PP{Y}{N}$ with $N=nm+m+n$. We see that $\iota^*\OO_{\PP{Y}{N}}(1)=\OO_{{\PP{Y}{n}}\times\PP{Y}{m}}(1)=p_1^*\OO_{\PP{Y}{n}}(1)\otimes p_2^*\OO_{\PP{Y}{m}}(1)$. Consider the morphism $X\ra \PP{Y}{n}\times \PP{Y}{m}$ given by $(\iota_1\times \iota_2)$. We see that $(\iota_1\times \iota_2)^*p_1^*\OO(1)=\mathcal L$ and $(\iota_1\times \iota_2)^*p_2^*\OO(1)=\mathcal M$, so that
$$(\iota\circ(\iota_1\times\iota_2))^*\OO_{\PP{Y}{N}}(1)=\mathcal L\otimes\mathcal M.$$
It remains to show that $\iota\circ(\iota_1\times \iota_2)$ gives a closed immersion into $\PP{Y}{N}$ over $Y$, and this is immediate from the facts that a composition of closed immersions is an immersion, and that $\iota_1\times\iota_2$ is a closed immersion. One can show the latter fact from drawing out the fibered product diagram, and using that if a composition of two morphisms is an immersion, then the first map must also be one.
(b) Let $j:\PP{Z}{n}\times\PP{Z}{m}\ra \PP{Z}{N}$ be the Segre Embedding, with $N=nm+n+m$. Let $\pi\circ p_1:\PP{Y}{n}\times \PP{Z}{m}\ra\PP{Z}{n}$ be the first projection followed by the projection $\pi:\PP{Y}{n}=\PP{Z}{n}\times_ZY\ra\PP{Z}{n}$ and let $p_2: \PP{Y}{n}\times \PP{Z}{m}\ra\PP{Z}{m}$ be the second projection. We have an immersion $\iota_1\times(\iota_2\circ f):X\ra\PP{Y}{n}\times \PP{Z}{m}$, where $\iota_1:X\ra\PP{Y}{n}$ and $\iota_2:Y\ra \PP{Z}{m}$ are the immersions induced by $\mathcal L$ and $\mathcal M$, respectively. Consider
$$h= j\circ ((\pi\circ p_1)\times p_2)\circ (\iota_1\times(\iota_2\circ f)):X\longrightarrow \PP{Z}{N}.$$
This is a composition of three immersions, and thus an immersion. Diagram chasing, we obtain,
$$h^*\OO(1)=(\iota_1\times(\iota_2\circ f))^*(p_1^*\pi^*\OO_{\PP{Z}{n}}\otimes p_2^*\OO_{\PP{Z}{n}})=\iota_1^*\pi^*\OO_{\PP{Z}{n}}(1)\otimes f^*\iota_2^*\OO_{\PP{Y}{m}}(1)=\mathcal L\otimes f^*\mathcal M.$$
Note that $\pi^*\OO_{\PP{Z}{n}}(1)=\OO_{\PP{Y}{n}}(1)$ basically from definition.
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