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This is exercise I.2.14 in Hartshorne's Algebraic Geometry:

Define $\psi : \mathbb{P}^{n}\times \mathbb{P}^{m}\longrightarrow \mathbb{P}^{N}$ where $N=rs+r+s$ by $(a_0,\dots,a_r)\times (b_0,\dots,b_s)=(\dots,a_ib_j,\dots)$ Show that $Im\psi$ is a subvariety of $\mathbb{P}^N$.

Hartshone gave a hint as follows:

Let the homogeneous coordinate of $\mathbb{P}^N$ be $z_{ij}, i=0,..r; j=0,...,s$ and let $\mathfrak{a}$ be the kernel of the homomorphism $k[{z_{ij}}]\rightarrow k[x_0,\dots,x_r,y_0,\dots,y_s]$ which sends $z_{ij}$ to $x_iy_j$. Then show that $Im\psi=Z(\mathfrak{a})$

My idea is to prove that $I(Im\psi)\subseteq I(Z(\mathfrak{a}))$ and $I(Z(\mathfrak{a}))\subseteq I(Im\psi)$ but this lead to a calculation of $\mathfrak{a}$.

So, how can I use the hint of Hartshone to solve it? Please give me some hint.

Thank for reading my question!

KReiser
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Arsenaler
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    Would it not be easier to show that $Im,\psi\subseteq Z(\frak a)$ and $Z(\frak a)\subseteq Im,\psi$? I mean, surely you have spotted that $z_{ij}z_{k\ell}-z_{i\ell}z_{kj}$ are in $\frak{a}$ for all $i,j,k,\ell$. – Jyrki Lahtonen Feb 15 '13 at 09:43

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Recall the definition of an algebraic subset in projective space, it is the zero locus of some set of homgeneous polynomials. In this case, these polynomials are $z_{ab} z_{cd} - z_{ad} z_{cb}$ ("exchange the first coordinates") and $z_{ab} z_{cd} - z_{cb} z_{ad}$ ("exchange the second coordinates") for $0 \leq a,c \leq r$ and $0 \leq b,d \leq s$. Since we have $x_a y_b x_c x_d = x_a y_d x_c x_b$ and $x_a y_b x_c x_d = x_c y_b x_a x_d$, the image of $\psi$ lies in the zero locus of the polynomials. Conversely, assume that $p=[p_{ij}] \in \mathbb{P}^N$ lies in the zero locus, i.e. we have $p_{ab} p_{cd} = p_{ad} p_{cb}$ and $p_{ab} p_{cd} = p_{cb} p_{ad}$. Some coordinate is $\neq 0$, say w.l.o.g. $p_{00}=1$. Now try to deduce $p_{ab} = p_{a0} p_{0b}$ and $p_{0b}=p_{b0}$, and infer that $p$ lies in the image.

One can also check that $\psi$ is an isomorphism onto its image; in particular the image is irreducible, i.e. a variety.

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    Is the polynomial $z_{ab}z_{cd} - z_{ad}z_{bc}$ really different from the polynomial $z_{ab}z_{cd} - z_{bc}z_{ad}$? Naively, I would expect $z_{ad}z_{bc}=z_{bc}z_{ad}$, but maybe I'm not allowed to assume commutativity here? – Thomas Klimpel Apr 11 '14 at 14:57
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    No, you are correct. One only needs one of the both sets of relations. – Martin Brandenburg Apr 16 '14 at 04:57