A \(k\dash\)algebra \(A\) is said to be rigid if it has no infinitesimal deformations, or equivalently, by (Ex. 9.8), if \(T^1(A) = 0\).
Let \(A = k[x,y,z,w] / (x,y) \intersect (z,w)\), and show that \(A\) is rigid. This corresponds to two planes in \(\AA^4\) which meet at a point.
Solution
Put \[ S=k[x,y,z,w], \qquad J=(x,y)\cap(z,w), \qquad A=S/J. \]
1. The ideal \(J\) is\[ J=(xz,xw,yz,yw). \]
Both \((x,y)\) and \((z,w)\) are monomial ideals, and their intersection is generated by the least common multiples of one generator from each ideal. Hence \[ (x,y)\cap(z,w) =(xz,xw,yz,yw) =(x,y)(z,w). \]
2. If \(e_1,e_2,e_3,e_4\) map respectively to\[ xz,\quad xw,\quad yz,\quad yw, \]then the syzygy module of these four generators is generated by\[ \begin{aligned} s_1&=(w,-z,0,0),& s_2&=(y,0,-x,0),\\ s_3&=(0,y,0,-x),& s_4&=(0,0,w,-z). \end{aligned} \]
Each displayed vector is visibly a syzygy. Conversely, suppose \[ (a_1,a_2,a_3,a_4)\in S^4 \] satisfies \[ xz a_1+xw a_2+yz a_3+yw a_4=0. \] Then \[ x(za_1+wa_2)+y(za_3+wa_4)=0. \] Since \(x\) and \(y\) are relatively prime, there is \(h\in S\) such that \[ za_1+wa_2=yh, \qquad za_3+wa_4=-xh. \] The first equality shows \(yh\in(z,w)\). Since \((z,w)\) is prime and \(y\notin(z,w)\), we have \(h\in(z,w)\), so write \[ h=zu+wv. \] Then \[ z(a_1-yu)+w(a_2-yv)=0, \] so relative primeness of \(z,w\) gives some \(r\in S\) with \[ a_1-yu=wr, \qquad a_2-yv=-zr. \] Similarly, \[ z(a_3+xu)+w(a_4+xv)=0 \] gives some \(t\in S\) with \[ a_3+xu=wt, \qquad a_4+xv=-zt. \] Therefore \[ (a_1,a_2,a_3,a_4) =r s_1+u s_2+v s_3+t s_4, \] as required.
3. An element of \(\Hom_A(J/J^2,A)\) is therefore exactly a quadruple\[ (a,b,c,d)\in A^4 \]satisfying\[ \boxed{ wa=zb, \qquad ya=xc, \qquad yb=xd, \qquad wc=zd. } \]
Step 2 gives a presentation \[ S^4\longrightarrow S^4\longrightarrow J\longrightarrow0, \] where the second free module maps its basis to \(xz,xw,yz,yw\) and the first map has columns \(s_1,s_2,s_3,s_4\). Tensoring with \(A=S/J\) gives the right-exact presentation \[ A^4\longrightarrow A^4\longrightarrow J/J^2\longrightarrow0. \] Applying \(\Hom_A(-,A)\) identifies a homomorphism \(J/J^2\to A\) with the four values \[ a=\phi(xz),\quad b=\phi(xw),\quad c=\phi(yz),\quad d=\phi(yw) \] that annihilate the four syzygies. Those four conditions are precisely the displayed equations.
4. Let\[ B=A/(z,w)\cong k[x,y], \qquad C=A/(x,y)\cong k[z,w]. \]The natural map\[ A\longrightarrow B\oplus C \]is injective.
An element of \(A=S/J\) maps to zero in both quotients exactly when a lift in \(S\) lies in \[ (z,w)\cap(x,y)=J. \] Hence its class in \(A\) is zero.
5. Every quadruple \((a,b,c,d)\) satisfying the equations in step 3 has the form\[ \begin{aligned} a&=zp+xr,& b&=wp+xs,\\ c&=zq+yr,& d&=wq+ys \end{aligned} \]for some \(p,q,r,s\in A\).
Write \(a_B,a_C\) for the images of \(a\) in \(B,C\), and similarly for \(b,c,d\).
Projecting \(wa=zb\) to \(C=k[z,w]\) gives \[ w a_C=z b_C. \] Since \(z\) and \(w\) are relatively prime in the UFD \(C\), there is \(p_C\in C\) such that \[ a_C=zp_C, \qquad b_C=wp_C. \] Likewise, projecting \(wc=zd\) to \(C\) gives some \(q_C\in C\) with \[ c_C=zq_C, \qquad d_C=wq_C. \]
Projecting \(ya=xc\) to \(B=k[x,y]\) gives some \(r_B\in B\) with \[ a_B=xr_B, \qquad c_B=yr_B, \] and projecting \(yb=xd\) to \(B\) gives some \(s_B\in B\) with \[ b_B=xs_B, \qquad d_B=ys_B. \]
Choose lifts \[ p,q,r,s\in A \] of \(p_C,q_C,r_B,s_B\), respectively. The elements \(a\) and \(zp+xr\) have the same images in both \(B\) and \(C\). By the injectivity in step 4, \[ a=zp+xr. \] The same argument gives the other three displayed identities.
6. The map\[ \Hom_A(\Omega_{S/k}\tensor_SA,A) \longrightarrow \Hom_A(J/J^2,A) \]is surjective.
3.Choose \(p,q,r,s\) as in step 5 and define a \(k\)-derivation\[ \delta:S\longrightarrow A \]by\[ \delta(x)=p, \qquad \delta(y)=q, \qquad \delta(z)=r, \qquad \delta(w)=s. \]Under the universal property of Kähler differentials, \(\delta\) is an elementof\[ \Hom_A(\Omega_{S/k}\tensor_SA,A). \]Its restriction to \(J/J^2\) sends the four generators to\[ \begin{aligned} \delta(xz)&=zp+xr=a,\\ \delta(xw)&=wp+xs=b,\\ \delta(yz)&=zq+yr=c,\\ \delta(yw)&=wq+ys=d. \end{aligned} \]Thus it maps to \(\phi\), proving surjectivity.
7. The algebra \(A\) is rigid.
By Exercise III.9.8, \[ T^1(A) =\operatorname{coker}\!\left( \Hom_A(\Omega_{S/k}\tensor_SA,A) \longrightarrow \Hom_A(J/J^2,A) \right). \] Step 6 shows that this map is surjective, hence \[ \boxed{T^1(A)=0}. \] Exercise III.9.8 identifies \(T^1(A)\) with the infinitesimal deformation space of \(A\), so \(A\) has no nontrivial infinitesimal deformations and is rigid.
8. Q.E.D.
Steps <1>1–<1>6 compute \(T^1(A)\), and step 7 shows that it vanishes and hence that \(A\) is rigid.