DU 1. Denote by A^xJ the ring of formal power series with coefficients in A, i.e. its elements are formal expressions
f{x) = ao + a\x + ci2X2 + • • • = anxn.
1. Let J C R be an ideal and define the limit of the diagram
----> R/J2     R/J2 R/J
to be the completion of R at the ideal J and denote it Rj. Show that A\x^ = ^4[^]mo, the completion of the polynomial ring at the maximal ideal mo = (x). Consider all the rings in the diagram equipped with the discrete topology and endow A^xJ with the limit topology. Give an explicit criterion for a sequence (or a net or a filter) of formal power series fk(x) = ^aknxn to converge to f{x) = ^anxn in terms of the coefficients a&n and an.
2. Show that for a field k the ring k[x]] is a UFD. As one of the ingredients, prove more generally that f(x) G A^xJ is invertible iff ao £ Ax is invertible; in the nontrivial direction, reduce to the case ao = 1, write f(x) = l+g(x) so that g(x) = a\x+a2X2 + - ■ ■, and show that the following makes sense and defines the inverse:
f(x)-1 = (1 + g(x))-1 = 1 - g(x) + g(x)2
(using the previous point, one can make sense of the infinite sum on the right hand side as the limit of the sequence of partial sums; this indeed converges).
3. The goal is to prove that k[x]] is a UFD. As for the polynomials, one observes that for a tuple of variables x = (x',t) , one gets k[x]] = k[x']]|[t]]. However, it is generally not true that A being a UFD implies A^xJ being a UFD, so one has to argue differently. We say that /(x) G k[x]] is a not-necessarily-monic Weierstrass polynomial w.r.t. t if it lies in
and as such is a polynomial of degree n with leading coefficient invertible in k[x']] and all other coefficients non-invertible, i.e. lying in mo [[£]]. Classically, a Weierstrass polynomial is additionally assumed to be monic. The Weierstrass preparation theorem says that every formal power series /(x) G k[x']]|[£]] \ mo|[£]] is associated to a (unique monic) Weierstrass polynomial (the condition simply means that / is non-zero along the f-axis, i.e. that as an element of k[x]] it contains some monomial tn with a nonzero coefficient).1 Since we can assume that k[x']][i] is a UFD by induction, this can be
1Here is an idea of the proof: We want /(s) = a(is)(tn + fe(s)) with fe(s) G mo[t] of degree n — 1. Now rewriting this as tn = a(s;)_1/(s;) — fe(s) we want to divide tn by /(s) "with a remainder". Now one can easily divide by tn, for one can write canonically any formal power series as ao(~x.)tn + &o(s) with &o(s) G k[s'][t] of degree n — 1. Expressing /(s) in this way as /(s) = u(is.)tn + v(is.), now with v(is.) G mo[t] - this determines n, we see that /(s) is roughly u(x.)tn and we start by dividing by u(is.)tn instead:
ao{^.)tn + &o(s) = ao(s;)w(s;)_1 w(s;)tn + &o(s) = ao(s;)w(s;)_1/(s;) + (—ao(s;)w(s;)_1u(s;) +&o(s)).
Continuing in this way we see that the quotient is (ao(s) + ffli(s;) + ■ ■ ■ and the remainder is &o(s) +
&i(x) + ■ ■ ■ provided that these converge. But one can see easily, using G mo[i], that if afe(s) G mo[i] then both afe+i(s),fefe+i(s) G m£+1[[t].
ultimately used to show that k[x]] is a UFD: Assume that M C N are two (cancellative) monoids satisfying:
• The mapping M/ass —> N/ass is bijective. In detail, every element of ./V is associated to some element of M and every two elements of M that are associated in ./V are associated also in M.
Show that if M has a unique factorization property then so does N. Apply this to M consisting of the (not-necessarily-monic) Weierstrass polynomials (these easily inherit the unique factorization property from k[x']][t] for they are closed both under multiplication and factorization) and ./V = k[x']]|[t]] \ moM- Finally, show that for any non-zero element /(x) G k[x]] one can set up the coordinates so that /(x) G ./V (look at the lowest non-zero homogeneous degree and apply the theorem from algebraic geometry about polynomials).
DU 2. Here we work over M. (or better over C). Let C = V(f) C A2 be a curve that is smooth at xo G C in the sense that d/(x) 7^ 0. Then one can parametrize C near xo locally as x = (£(£), a(£)), passing through xo for t = to, and any two parametrizations differ by a diffeomorphism of the parameter space near to, so the following definition makes sense: The multiplicity of the intersection point xo of C = V(f) and D = V(g) is defined as the order of zero at to °f the function <?(£(£), &(t)) (here the function g is defined uniquely up to a constant if it is required to be polynomial - since it is the generator of 1(D) - and up to a multiple by a nonzero smooth function otherwise). It is not obvious why this definition is symmetric in C and D; you may try to give it a thought.
1. Take some concrete example like / = x2 + y2 — 1 and g = (x — \/2)2 + (y — V%)2 — 1 and use the parametrization using sin and cos or any other to compute the multiplicity.
2. Now assume that the coordinate system has been set up so that / has a non-zero coefficient at yr, the x-coordinates of all intersection points are distinct and also distinct from all the points of C for which fx = 0 (these form the set V(f,f'x) which must be finite unless (x — a) \ f - in other words, we want all the linear factors of / depend on y). Then near each x-coordinate of an intersection point, one can write the branches of V(f) as y = ai(x), i.e. the local parametrizations as above are (x,a.i(x)). Plug the above into the formula
Res(/, g; y) = asr • g(x, at(x)) • • • g(x, ar(x))
and prove that the number of intersections is bounded by rs even when counted with multiplicity in the sense of this exercise. In the concrete example above, plugging in the non-polynomial parametrization into the formula must give a polynomial expression (the resultant is a polynomial after all), try it.
DU 3. Let /: X —> Y be a regular map and denote by </? = /*: k[Y] —> k[X] the induced map on coordinate k-algebras. We showed that V(tp*I) = /_1(l/(/)).
1. Prove symmetrically that
I(f(S)) = <p*I(S)
2
(this should be quite obvious) and deduce that f(S) = V(p*(I(S)). Now specialize to s = v{j) and use Hilbert Nullstellensatz to prove that
WUV) = v{^j).
(I don't know if Hilbert Nullstellensatz is really necessary - you may try to figure it out.)
2. Now assume in addition that / is dominant, i.e. that p is injective. We showed (using Nakayama lemma) that if p is finite then / is surjective and closed. Prove that fibres of / are all finite (this does not use the previous point, and it should have been done in the tutorial).
3. You may know the following theorem: Let A be a finitely generated Ik-algebra, acted upon by a finite group G of Ik-algebra automorphisms, from the right and denoted as ag. If n = \G\ is not divisible by char Ik then the subalgebra
AG = {a G A | a9 = a}
of invariants (or fixed points) is also finitely generated. In addition, A is a finite AG-algebra.2
4. Assume now that an affine variety X has a (left) action of a finite group G by polynomial maps. This induces a (right) action of G on k[X] and we may apply the previous point to conclude that there exists an affine variety Xq such that k[X(j] = k[X]G. Show that the canonical map X —> Xq is surjective with finite fibres and identify them as orbits of the G-action. Conclude that Xq = X/G as sets.
2Start by observing that any a G A is a root of a monic polynomial pa = IlgeG^ — ffl9) wnose coefficients are the elementary symmetric polynomials 04(a) = ai{a9 \ g G G}, up to a sign, and as such lie in AG. Denote the generators of A as Uj and denote by B the subalgebra generated by tji(uj). Then pUj (uj) = 0 implies that u™ G B{1, Uj,..., m"-1}. For this reason,
A = SJm™1 ... u"r | «i,..., ar < n} = B{ua},
so A is already finite over B. Explicitly, for any a £ A, we have an expression a = ^ baua with ba G B. Apply the symmetrization S: A —> AG, S(a) = ^ ^2geG a9', assuming that a G AG, we get
a = S(a) = J2b<^S(ua)
and thus A° = B{S(ua)} = h[ai(Uj), S(aa)}.
3
DÚ 1. Dokažte následující izomorfismy:
• Ala-1} ~ A[t]/(at—t)y
• (A/I)[t] = A[t]/J a popište ideál J,
• A/{I I J) = (A/I)/Jr a popište ideál J' ve stylu "je to v zásadě J, jenom... ". DU 2. Pomocí Grobncrovy báze vyřešte soustavu polynomiálních rovnic
x2 + y + z = 1 x + y2 + z = 1 x + y + z2 = 1
DU 3. Nechť k je algebraicky uzavřené těleso. Studujte vztah mezi nenulovými kvadratickými polynomy / C k[íEi,..., xn] a příslušnými afinními varietami V{f) C An; konkrétně se zabývejte tím, nakolik je zobrazení / i > V{f) injektivní. Dále proveďte analogickou studii pro kubické polynomy.
DU 4. Dokažte následující tvrzení:
• Afinní varieta X je ireducibilní, právě když pro libovolné afinní variety X±, X2 platí
X C X\ U X2 ==>■ (X C X\ VI C X2).
• Ideál J je prvoideál, právě když pro libovolné ideály J±, J2 platí
J D Ji J2 =^(JDJiVJ3 J2).
• Pomocí předchozích dvou tvrzení dokažte, že X je ireducibilní, právě když I{X) je prvoideál (není k tomu potřeba Hilbertova věta o nulách, ale klidně ji použijte).
DÚ 5. Označme I = (g \ g €ž I) ideál generovaný homogenizacemi g = xQdegs'<?(f^, • • •, f^)-
(Platí Faf(/) = l/pr(/), to ukážeme na příští přednášce; toto tvrzení je důvodem, proč se tímto ideálem zabýváme.) Uvažujme následující uspořádání monomů
xa>gIxP \a\ > |/3| V(|a| = |/3| f\xa > x?).
Dokažte, že v případě, že / = (51,... ,gr) je Gróbnerova báze vzhledem k >gr, je také / = (gi,..., gr) Gróbnerovou bází vzhledem k podobném uspořádání >gr, jen s xq navíc a menším než zbylé proměnné, tj. x\ > • • • > xn > xq.
DU 6. Ukažte, že zobrazení /: P2— -tF2, (x$ : x\ : x2) 1—> (x\x2 : x2xq : xqX\) je biracionální ekvivalence a najděte otevřené podmnožiny P2, na nichž je / izomorfismus. (Nápověda: napíšete-li si zobrazení afinně, inverze by měla být jasná.)
DU 7. Dokažte, že obraz regulárního zobrazení A1 —> An je uzavřený (nápověda: použijte, že P1 ->• Pn je re gulární a zkoumejte obraz nevlastního bodu).
DU 8. Řekneme, že fc-rovina K a /-rovina L se protínají transverzálně v Pn, jestliže jejich průnik je {k + 1 — n)-rovina. Ukažte, že obecná dvojice (K, L) G G(k, n) x G(l, n) se protíná transverzálně.
DU 9. Nechť /: X—> Y je surjcktivní uzavřené zobrazení mezi Nocthcrovskými topologickými prostory takové, že pro každou dvojici uzavřených podmnožin A ^ B C X, kde B je ireducibilní, je f (A) Q f (B). Dokažte, že / je otevřené.
4