Fulton 2.2

Posted on November 15, 2017

Bonus problem 2.a. Show the map \widetilde{\varphi}:\mathcal{F}(W,k)\to\mathcal{F}(V,k) is a homomorphism.

Observe that for any f,g\in\mathcal{F}(W,k) and any x\in W,  \begin{aligned}[t] ((f+g)\circ\varphi)(x) &= f(\varphi(x))+g(\varphi(x))\\ &= (f\circ\varphi)(x)+(g\circ\varphi)(x)\\ &= ((f\circ\varphi)+(g\circ\varphi))(x). \end{aligned} Therefore \widetilde{\varphi}(f+g)=\widetilde{\varphi}(f)+\widetilde{\varphi}(g). Similarly, \widetilde{\varphi}(fg)=\widetilde{\varphi}(f)\widetilde{\varphi}(g).

Bonus problem 2.b. If T_1,\ldots,T_m\in k[X_1,\ldots,X_n] determine a polynomial map T:\mathbb{A}^n\to\mathbb{A}^m, the T_i are uniquely determined by T.

Suppose F\in k[X_1,\ldots,X_n] such that F(P)=T_i(P) for all P\in\mathbb{A}^n. Then (F-T_i)(P)=0 for all P\in\mathbb{A}^n. So F=T_i by Problem 1.4.

Bonus problem 2.c. Let V\subset\mathbb{A}^n,W\subset\mathbb{A}^m be a varieties. Let \varphi:V\to W be a polynomial map. Then \widetilde{\varphi}:\Gamma(W)\to\Gamma(V) is a homomorphism. Moreover, F+I(W)\mapsto F\circ\varphi+I(V).

Suppose \varphi is defined by the polynomials T_1,\ldots,T_m\in k[X_1,\ldots,X_n]. For any polynomial F\in k[X_1,\ldots,X_m], \widetilde{\varphi}(F)=F\circ\varphi=F(T_1,\ldots,T_m) is a polynomial in k[X_1,\ldots,X_n].

We will now show that \widetilde{\varphi} induces a well defined homomorphism \Gamma(W)\to\Gamma(V), that is, if F+I(W),G+I(W)\in\Gamma(W) then \widetilde{\varphi}(F)+I(V)=\widetilde{\varphi}(G)+I(V). Let H\in I(W). Observe that (H\circ\varphi)(a_1,\ldots,a_n)=0 for all (a_1,\ldots,a_n)\in V. Thus \widetilde{\varphi}(H)\in I(V). Thus, by Problem 2.a, the map is a well defined homomorphism.

Although a slight abuse of notation, we use \widetilde{\varphi} to refer to the homomorphism \Gamma(W)\to\Gamma(V) induced by \widetilde{\varphi}. This falls in line with the dual use of \Gamma(V) as both the ring k[X_1,\ldots,X_n]/I(V) and the set of all polynomial maps in \mathcal{F}(V,k).

Note. Recall that Fulton assumes all homomorphisms fix the field k. That is, the following statement is a bijection between polynomial maps and homomorphisms of coordinate rings that fix k.

Proposition 1. Let V\subset\mathbb{A}^n, W\subset\mathbb{A}^m be affine varieties. There is a one-to-one correspondence between the polynomial maps \varphi:V\to W and the homomorphisms \widetilde{\varphi}:\Gamma(W)\to\Gamma(V). Any such \varphi is the restriction of a polynomial map \mathbb{A}^n\to\mathbb{A}^m.

By Problem 2.c, each polynomial map V\to W induces a homomorphism \Gamma(W)\to\Gamma(V). Moreover, any polynomial map V\to W is the restriciton of a polynomial map \mathbb{A}^n\to\mathbb{A}^m. Thus it remains to prove the converse.

Let \alpha:\Gamma(W)\to\Gamma(V) be a homomorphism. For each i\in\{1,\ldots,m\} select T_i\in k[X_1,\ldots,X_n] such that T_i+I(V)=\alpha(X_i+I(W)). Let T:\mathbb{A}^n\to\mathbb{A}^m be the polynomial map defined by (T_1,\ldots,T_m). Let \widetilde{T}:k[X_1,\ldots,X_m]\to k[X_1,\ldots,X_n] be the homomorphism defined by F\mapsto F\circ T. Let \pi_V:k[X_1,\ldots,X_n]\to\Gamma(V) and \pi_W:k[X_1,\ldots,X_m]\to\Gamma(W) be the natural homomorphisms.  \xymatrix{     k[X_1,\ldots, X_m] \ar[r]^{\widetilde{T}}     \ar[d]^{\pi_W}         & k[X_1,\ldots,X_n] \ar[d]^{\pi_V}\\     \Gamma(W) \ar[r]^{\alpha} & \Gamma(V) } Observe that  \begin{aligned}[t] \pi_V(\widetilde{T}(F)) &= \pi_V(F(T_1,\ldots,T_m))\\ &= F(T_1,\ldots,T_m)+I(V)\\ &= F(\alpha(X_1+I(W)),\ldots,\alpha(X_m+I(W)))+I(V)\\ &= \alpha(F+I(W))\\ &= \alpha(\pi_W(F)). \end{aligned} Thus the diagram above commutes. In particular, \pi_V(\widetilde{T}(I(W)))= \alpha(\pi_W(I(W)))=0+I(V). Thus \widetilde{T}(I(W))\subset I(V).

Let (a_1,\ldots,a_n)\in\mathbb{A}^n. Suppose F\in I(W). Then F(T(a_1,\ldots,a_n))= \widetilde{T}(F)(a_1,\ldots,a_n)=0, since \widetilde{T}(F)\in I(V). Thus T(a_1,\ldots,a_n)\in W.

Observe that \widetilde{T|_V} is the same as the homomorphism resulting from applying Lemma 1.1 to \pi_W and \pi_V\circ\widetilde{T}. Thus T|_V:V\to W is a polynomial map such that \widetilde{T|_V}=\alpha.

Corollary. Let V,W be affine varieties. Then V is isomorphic to W if and only if \Gamma(V) is isomorphic to \Gamma(W).

Suppose V\cong W, that is, there exist polynomial maps \varphi:V\to W and \psi:W\to V such that \psi\circ\varphi=\text{id}_V and \varphi\circ\psi=\text{id}_W. Note that \text{id}_{\Gamma(V)}=\widetilde{\text{id}_V}=\widetilde{\psi\circ\varphi}. Thus by Problem 2.6, \text{id}_{\Gamma(V)}=\widetilde{\psi\circ\varphi} =\widetilde{\varphi}\circ\widetilde{\psi}. Similarly, \text{id}_{\Gamma(W)}= \widetilde{\psi}\circ\widetilde{\varphi}. Thus \widetilde{\varphi},\widetilde{\psi} are isomorphisms and \Gamma(V)\cong\Gamma(W).

Conversely, suppose \Gamma(V)\cong\Gamma(W), that is, there exist isomorphisms \widetilde{\varphi}:\Gamma(W)\to\Gamma(V) and \widetilde{\psi}: \Gamma(V)\to\Gamma(W) such that \widetilde{\varphi}\circ\widetilde{\psi}= \text{id}_{\Gamma(V)} and \widetilde{\psi}\circ\widetilde{\varphi}=\text{id}_{\Gamma(W)}. By Proposition 1, \widetilde{\varphi},\widetilde{\psi} are induced by some polynomial maps \varphi:V\to W and \psi:W\to V. Since \widetilde{\varphi}\circ \widetilde{\psi}=\text{id}_{\Gamma(V)}, F\circ(\psi\circ\varphi)+I(V)=F+I(V) for any F+I(V)\in\Gamma(V). In particular, this is true for \sum_{i=1}^nX_i+I(V). Thus (\psi\circ\varphi)(a_1,\ldots,a_n)=(a_1,\ldots,a_n)+F(a_1,\ldots,a_n) for some F\in I(V). Thus (\psi\circ\varphi)(a_1,\ldots,a_n)=(a_1,\ldots,a_n) for every (a_1,\ldots,a_n)\in V. So \psi\circ\varphi=\text{id}_{V}. A similar argument shows \varphi\circ\psi=\text{id}_{W}, and therefore V\cong W.

2.6.* Let \varphi:V\to W, \psi:W\to Z. Show that \widetilde{\psi\circ\varphi}=\widetilde{\varphi}\circ\widetilde{\psi}. Show that the composition of a polynomial map is a polynomial map.

Let \varphi:V\to W, \psi:W\to Z. Let f\in\mathcal{F}(Z,k). Observe that  \begin{aligned}[t] \widetilde{\psi\circ\varphi}(f) &= f\circ(\psi\circ\varphi)\\ &= (f\circ\psi)\circ\varphi\\ &= \widetilde{\varphi}(f\circ\psi)\\ &= \widetilde{\varphi}(\widetilde{\psi}(f))\\ &= (\widetilde{\varphi}\circ\widetilde{\psi})(f). \end{aligned} Therefore \widetilde{\psi\circ\varphi}=\widetilde{\varphi}\circ\widetilde{\psi}. It is simple to verify that composing polynomial maps results in a polynomial map.

2.7.* If \varphi:V\to W is a polynomial map, and X is an algebraic subset of W, show that \varphi^{-1}(X) is an algebraic subset of V. If \varphi^{-1}(X) is irreducible, and X is contained in the image of \varphi, show that X is irreducible. This gives a useful test for irreducibility.

Observe that if P\in\varphi^{-1}(X) then \varphi(P)\in X. Thus F(\varphi(P))=0 for all F\in I(X). Moreover, if F(\varphi(P))=0, then \varphi(P)\in X, and P\in\varphi^{-1}(X). So \varphi^{-1}(X)=V(\widetilde{\varphi}(I(X))).

Suppose that \varphi^{-1}(X) is irreducible and X\subset\varphi(V). Suppose that I_W(X) is not prime, that is, there exist f,g\in\Gamma(W)\setminus I_W(X) such that fg\in I_W(X). Since f,g\not\in I_W(X) there exist P,Q\in X such that f(P)\ne 0, g(Q)\ne 0. Since X\subset\varphi(V), there exist P&39;,Q&39;\in\varphi^{-1}(V) such that \varphi(P&39;)=P, \varphi(Q&39;)=Q. Therefore (f\circ\varphi)(P&39;)=f(P)\ne 0 and \widetilde{\varphi}(f)\not\in I_V(\varphi^{-1}(X)). Similarly, \widetilde{\varphi}(g)\not\in I_V(\varphi^{-1}(X)). But, \widetilde{\varphi}(f)\widetilde{\varphi}(g)=\widetilde{\varphi}(fg)\in I_V(\varphi^{-1}(X)). So I_V(\varphi^{-1}(X)) is not prime, a contradiction to the assumption that \varphi^{-1}(X) is irreducible.

2.8. (a) Show that \{t,t^2,t^3\}\subset\mathbb{A}^3(k) is an affine variety.

Let \varphi:\mathbb{A}^1(k)\to\mathbb{A}^3(k) be defined by t\mapsto (t,t^2,t^3). Observe \varphi is a polynomial map. Let V=V(X-Y^2, X-Z^3)=\{(t,t^2,t^3)\in\mathbb{A}^3(k)\}. Note that \varphi(\mathbb{A}^1(k))=V and \varphi^{-1}(V)=\mathbb{A}^1(k). Recall \mathbb{A}^1(k) is irreducible by Problem 1.29. Therefore V is irreducible by Problem 2.7.

(b) Show that V(XZ-Y^2, YZ-X^3, Z^2-X^2Y)\subset\mathbb{A}^3(\mathbb{C}) is a variety.

Let V=V(XZ-Y^2, YZ-X^3, Z^2-X^2Y). It is simple to compute that Y^3-X^4, Z^3-X^5, z^4-Y^5\in I(V). Suppose (a,b,c)\in V. Then b^3=a^4, that is, b=t^4 whenever t is a third root of a in \mathbb{C}. Similalry, c=s^5 where s is a third root of a. Since c^4=b^5 we have s^{20}=t^{20}. But t^3=a=s^3. Thus t(s^{20})=t(t^{20})=(t^{3})^{7}= (s^{3})^7=s(s^{20}). So t=s and (a,b,c)=(t^3,t^4,t^5). Moreover, for any t\in \mathbb{C}, observe that (t^3,t^4,t^5)\in V. Let \varphi:\mathbb{A}^1(\mathbb{C})\to \mathbb{A}^3(\mathbb{C}) be the polynomial map defined by t\mapsto(t^3,t^4,t^5). Then V=\varphi(\mathbb{A}^1(\mathbb{C})) and thus V is irreducible by Problem 2.7.

2.9.* Let \varphi:V\to W be a polynomial map of affine varieties, and let V&39;\subset V, W&39;\subset W be subvarieites. Suppose \varphi(V&39;)\subset W&39;.

(a) Show that \widetilde{\varphi}(I_W(W&39;))\subset I_V(V&39;).

Suppose f\in I_W(W&39;). Then f(P)=0 for all P\in\varphi(V&39;)\subset W&39;. Thus \widetilde{\varphi}(f)\in I_V(V&39;).

(b) Show that the restriction of \varphi gives a polynomial map V&39;\to W&39;.

The restriction of any polynomial map is a polynomial map by the proof of Proposition 1. Thus since \varphi(V&39;)\subset W&39;, \varphi|_{V&39;} is a polynomial map V&39;\to W&39;.

2.10.* Show that the projection map \mathbb{A}^n\to\mathbb{A}^r with n\ge r is a polynomial map.

The projection map is defined by the polynomials T_i=X_i.

2.11. Let f\in\Gamma(V), V\subset\mathbb{A}^n a variety. Define      G(f)=\{(a_1,\ldots,a_n,a_{n+1})\in\mathbb{A}^{n+1} \ | \ (a_1,\ldots,a_n)\in V         \text{ and } a_{n+1}=f(a_1,\ldots,a_n)\}, the graph of f. Show that G(f) is an affine variety, and the map      (a_1,\ldots,a_n)\mapsto (a_1,\ldots,a_n,f(a_1,\ldots,a_n)) defines an isomorphism of V with G(f).

Let \varphi:\mathbb{A}^n\to G(f) be the map defined above. It is immediate that \phi is a bijection with inverse \pi|_{G(f)}, where \pi:\mathbb{A}^{n+1}\to\mathbb{A}^n is the projection map. Since both are polynomial maps, \varphi is an isomorphism.

2.12. (a) \varphi:\mathbb{A}^1\to V=V(Y^2-X^3)\subset\mathbb{A}^2 be defined by \varphi(t)=(t^2,t^3). Show that although \varphi is a one-to-one, onto polynomial map, \varphi is not an isomorphism.

Observe if (a,b)\in V the a^3=b^2. Since k is algebraically closed, there exists t\in k such that t^2=a. So b^2=t^6 and b=\pm t^3. Thus either \varphi(t)=(a,b) or \varphi(-t)=(a,b). So \varphi is onto.

Suppose t,s\in\mathbb{A}^1 such that \varphi(t)=\varphi(s). Then t^2=s^2 and t^3=s^3. Thus t(t^2)=s(s^2)=s(t^2), and t=s. Therefore \varphi is one-to-one.

It remains to show that \varphi is not an isomorphism, that is, its inverse is not a polynomial map. Observe that \widetilde{\varphi}(f)=f(T^2,t^3) for every f\in\Gamma(V). Therefore \widetilde{\varphi}(\Gamma(V))\subset (T^2,T^3) \subsetneq k[T]=\Gamma(\mathbb{A}^1). Thus \widetilde{\varphi}:\Gamma(V)\to\Gamma(\mathbb{A}^1) is not an isomorphism. Hence \varphi is not an isomorphism by Proposition 1.

(b) Let \varphi:\mathbb{A}^1\to V=V(Y^2-X^2(X-1)) be defined by \varphi(t)=(t^2-1,t(t^2-1)). Show that \varphi is one-to-one and onto, except that \varphi(\pm 1)=(0,0).

Let (a,b)\in V. Then b^2=a^2(a+1). Let t\in k such that t^2=a+1. Then a=t^2-1, b=(t^2-1)t. Therefore \varphi(t)=(a,b). So \varphi is onto.

Suppose t,s\in k such that \varphi(t)=\varphi(s). Then (t^2-1)=(s^2-1). Moreover, (t^2-1)t=(s^2-1)s. If t^2-1\ne 0, then t=s. If t^2-1=0, then t=\pm 1, s=\pm 1. Therefore \varphi is one-to-one and onto, except that \varphi(\pm 1)=(0,0).

2.13. Let V=V(X^2-Y^3, Y^2-Z^3)\subset\mathbb{A}^3 as in Problem 1.40, \overline{\alpha}:\Gamma(V)\to k[T] induced by the homomorphism \alpha of that problem.

(a) What is the polynomial map f from \mathbb{A}^1 to V such that \widetilde{f} =\overline{\alpha}?

Let f:\mathbb{A}^1\to\mathbb{A}^3 be the polynomial map (T^9,T^6,T^3). Observe that f(\mathbb{A}^1)\subset V so f:\mathbb{A}^1\to V is a polyomal map such that \widetilde{f}=\overline{\alpha}.

(b) Show that f is one-to-one and onto, but not an isomorphism.

First note that f is clearly one-to-one. Suppose (a,b,c)\in V. Let t\in k such that t^4=c. Then b^2=c^3=t^{12}, so b=\pm t^6. Moreover, a^2=b^3=\pm t^{18}, so a=\pm t^{9}, or \pm t^{9}\sqrt{-1}. Therefore (a,b,c) is either (\pm t^9,t^6,t^4), or (\pm t^9\sqrt{-1},-t^6,t^4). Thus (a,b,c) is equal to either f(\pm t) or f(\pm t\sqrt{-1}). So f is onto.

It remains to show that f is not an isomorphism. By Proposition 1 it suffices to observe that \widetilde{f} is not an isomorphism, since \widetilde{f}(\Gamma(V))\subset (T^9, T^6, T^4)\subsetneq k[T]=\Gamma(\mathbb{A}^1).