Let \(k\) be algebraically closed, let \(n\ge0\), and put \(S\coloneqq k[x_0,\ldots,x_n]\) and \(S_+=(x_0,\ldots,x_n)\).
Show that \(Y \mapsto I(Y)\) and \(\mfa \mapsto Z(\mfa)\) give a one-to-one inclusion-reversing correspondence between algebraic sets in \(\PP^n\) and homogeneous radical ideals of \(S\) not equal to \(S_+\). Because \(S_+\) does not occur in this correspondence, it is called the irrelevant maximal ideal of \(S\).
An algebraic set \(Y \subseteq \PP^n\) is irreducible if and only if \(I(Y)\) is a prime ideal.
Show that \(\PP^n\) is itself irreducible.
Solution
The ideal \(I(Y)\) is generated by the homogeneous polynomials vanishing on \(Y\), and \(Z(J)\) is the projective zero set of a homogeneous ideal \(J\). Irreducible spaces are nonempty, and prime ideals are proper.
1. The maps in (a) are inverse inclusion-reversing bijections, with \(\varnothing\) corresponding to \(S\).
For every projective algebraic set \(Y\), the identity \(Z(I(Y))=Y\) follows from P-AGH23ZIPROPS, part (e), since \(Y\) is closed. If \(Y\ne\varnothing\), part (d) of that card, applied to \(I(Y)\), gives \[ I(Y)=I(Z(I(Y)))=\sqrt{I(Y)}. \] Thus \(I(Y)\) is homogeneous and radical. It cannot equal \(S_+\), whose projective zero set is empty. For \(Y=\varnothing\), every homogeneous polynomial vanishes vacuously, including \(1\), so \(I(Y)=S\); this is also a homogeneous radical ideal distinct from \(S_+\).
Conversely, let \(J\) be homogeneous and radical with \(J\ne S_+\). When \(Z(J)\) is nonempty, the same homogeneous Nullstellensatz calculation gives \(I(Z(J))=\sqrt J=J\). When \(Z(J)\) is empty, each coordinate \(x_i\) vanishes on it, so P-AGH21HOMNSS gives a power of \(x_i\) in \(J\). Radicality implies \(x_i\in J\) for every \(i\), hence \(S_+\subseteq J\). Since \(S/S_+=k\) is a field, \(S_+\) is maximal. The assumption \(J\ne S_+\) therefore forces \(J=S\), and again \(I(Z(J))=I(\varnothing)=S=J\).
Both maps reverse inclusions by P-AGH23ZIPROPS, parts (a) and (b). Their two inverse identities prove (a), including the unit ideal rather than restricting the correspondence to proper ideals.
2. If \(Y\) is irreducible, then \(I(Y)\) is prime.
Nonemptiness gives \(1\notin I(Y)\). For homogeneous \(f,g\in S\) with \(fg\in I(Y)\), the equality of zero sets \[ Z(fg)=Z(f)\cup Z(g) \] holds because the values lie in a field. Thus \(Y\) is the union of its closed subsets \(Y\cap Z(f)\) and \(Y\cap Z(g)\). Irreducibility forces one of them to be all of \(Y\), so \(f\in I(Y)\) or \(g\in I(Y)\).
This homogeneous test implies ordinary primality. In the nonnegatively graded ring \(R=S/I(Y)\) it says that a product of two nonzero homogeneous elements is nonzero. For arbitrary nonzero \(a,b\in R\), let \(a_d,b_e\) be their highest nonzero homogeneous components. The component of degree \(d+e\) of \(ab\) is \(a_db_e\), which is nonzero by that test. Hence \(ab\ne0\), so \(R\) is a domain and \(I(Y)\) is prime.
3. If \(I(Y)\) is prime, then \(Y\) is irreducible.
The empty set has ideal \(S\), which is not prime, so \(Y\) is nonempty. Suppose \(Y=Y_1\cup Y_2\) for closed proper subsets \(Y_1,Y_2\) of \(Y\). Since \(Y\) is projectively closed, both \(Y_i\) are projective algebraic sets. Step 1 makes each strict inclusion \(Y_i\subsetneq Y\) correspond to a strict inclusion \(I(Y)\subsetneq I(Y_i)\). Choose homogeneous \[ f\in I(Y_1)\setminus I(Y),\qquad g\in I(Y_2)\setminus I(Y). \] Such homogeneous choices exist because all these ideals are homogeneous: an element outside the smaller ideal has a homogeneous component outside it. The product \(fg\) vanishes on \(Y_1\cup Y_2=Y\), so \(fg\in I(Y)\). This contradicts primality, since neither factor belongs to \(I(Y)\). Thus no such decomposition exists, proving this implication and completing (b).
4. Projective space \(\PP_k^n\) is irreducible.
It is nonempty, and it equals \(Z((0))\). By P-AGH23ZIPROPS, part (d), \[ I(\PP_k^n)=I(Z((0)))=\sqrt{(0)}=(0), \] since the polynomial ring \(S\) is a domain. The zero ideal is therefore prime, and step 3 proves (c), also when \(n=0\).
5. Q.E.D.
Step 1 proves (a), steps <1>2–<1>3 prove (b), and step 4 proves (c).