Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 21 / 1

Use the usual algebraically closed field convention for classical algebraic sets, or equivalently interpret their points geometrically. The proof also gives the scheme-theoretic statement over any field; the difference from an arbitrary-field rational-point image is explained below.

Choose an affine open cover $X={\bigcup }_i{U}_i$. On each $U_i$, the subset $Y_i=Y\cap (U_i\times {\mathbb{P}}^n)$ is closed relative to $U_i\times {\mathbb{P}}^n$. If the affine-base assertion holds, its projected image is closed in $U_i$. ButThus the complement of $\pi (Y)$ is open on each member of the open cover and hence open in $X$. Therefore it suffices to prove closedness for affine $X$. No finite-cover assumption is needed for this localization argument.

Put $I=(f_1,\dots ,f_r)$ and $\mathfrak{m}=(y_0,\dots ,y_n)$. If ${\mathfrak{m}}^N\subseteq I$, a projective common zero would have some coordinate $y_j\ne 0$. For $N>0$, the polynomial $y_j^N\in I$ could not vanish there; for $N=0$, the inclusion puts $1$ in $I$, also impossible. Thus the projective zero set is empty.

Conversely, over an algebraically closed field, emptiness of the projective zero set means the affine common zero set is contained in the origin. Every $y_j$ vanishes there, so the Hilbert Nullstellensatz puts $y_j$ in $\sqrt{I}$. Choose integers $e_j\ge 1$ with $y_j^{e_j}\in I$, and setEvery degree-$N$ monomial has some exponent at least the corresponding $e_j$, by the pigeonhole principle, and is consequently in $I$. Those monomials generate ${\mathfrak{m}}^N$, provingThe argument includes the unit-ideal case. This proves the Projective Nullstellensatz criterion directly from its affine counterpart.

Over a general field, this same criterion concerns geometric zeros: apply the proof after extension to an algebraic closure. Inclusion of each degree-$N$ monomial descends back to $k$, since its class in $k[y]/I$ is zero if and only if it becomes zero after a field extension; tensoring a vector space with an extension field is faithful. If instead one counts only $k$-rational projective points, the assertion is false. For example $y_0^2+y_1^2$ has no real projective zero but has the complex zero $[1:i]$, so its ideal cannot contain any power of $\mathfrak{m}$.

For an affine base, use the homogeneous equations furnished by the question, with fixed degrees $d_i$ in the projective coordinates. At a point $x$, the fiber $Y_x$ is the projective zero set of their specializations; a specialization may be zero, but is still assigned its original degree. LetThen $x\in \pi (Y)$ exactly when $Y_x$ is nonempty. Part (b) says this is equivalent to $I_x$ containing no power of the irrelevant ideal of projective space. In logical form, noncontainment must hold for every $N$, soFor schemes over an arbitrary field, replace $k$ in this fiber calculation by the residue field $\kappa (x)$ and use geometric fiber points. A point belongs to the scheme-theoretic topological image exactly when its fiber is nonempty, which is equivalent to nonemptiness after algebraic closure of that residue field.

By part (a), work first with an affine base and its homogeneous equations. Put $S=k[y_0,\dots ,y_n]$, write $S_N$ for its finite-dimensional degree-$N$ vector space, and use $S_j=0$ for $j<0$. At each base point define the graded multiplication map for a projective fiberIts image is exactly $(I_x{)}_N$, since taking the degree-$N$ part of any expression in the homogeneous ideal discards all other degrees. Moreoverbecause degree-$N$ monomials generate ${\mathfrak{m}}^N$.

Choose monomial bases for the domain summands and codomain. Every entry of the resulting matrix is a coefficient of one of the $f_i$, hence a regular function of $x$. If $h_N=\dim S_N$, failure of surjectivity is the vanishing of all $h_N$-by-$h_N$ matrix minors. It is therefore a closed rank-defect locus $C_N$. If the matrix has fewer than $h_N$ columns, the failure is automatic and $C_N=X$, also closed. Constants and $N=0$ are covered by the same construction.

Part (c) gives $\pi (Y)={\bigcap }_{N\ge 0}{C}_N$, an intersection of closed subsets, hence closed. Apply part (a) to pass back to arbitrary $X$. Thus projection from $X\times {\mathbb{P}}^n$ sends every closed subset to a closed subset. This proves the closedness of projection from projective space without assuming the properness result being proved.

The field convention matters: a rational-point projection need not be Zariski closed. Over $\mathbb{R}$, the closed equation $y_0^2=t{y}_1^2$ in ${\mathbb{A}}^1\times {\mathbb{P}}^1$ has real-point image $[0,\infty )$. Indeed $y_1=0$ would force $y_0=0$, so a solution exists precisely when $t$ is a square in $\mathbb{R}$. This infinite proper subset of the affine line is not Zariski closed. The valid arbitrary-field theorem uses scheme or geometric images, for which the matrix argument remains valid at every residue field.