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Dimension and Integral Extensions

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Dimension and Integral Extensions (c)Karen E. Smith 2018 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Dimension and Integral Extensions Through out, R and S denote commutative rings with identity, and K denotes a field. Definition: The (Krull) dimension of R is the supremum of the lengths of chains in the poset Spec R. Explicitly, dim R := Supft j 9Pi 2 Spec R; s:t: P0 ( P1 ( ··· ( Ptg Lying Over Theorem: If R,! S is integral, then Spec S ! Spec R is surjective, and the fiber over any P 2 Spec R consists of incomparable primes in Spec S. Going Up Corollary: If R,! S is integral, then given any chain of primes in R, say P0 ⊂ P1 ⊂ · · · ⊂ Pt and Q0 contracting to P0, there exists a chain of primes Q0 ⊂ Q1 ⊂ · · · ⊂ Qt in S with the property that each Qi contracts to Pi under the induced map of Spectra. Dimension Corollary: If R,! S is integral, then dim R = dim S. (1) For each of the rings below, find the dimension using the definition/theorem: (a) The field C(t) of meromorphic functions on the Riemann sphere. (b) Z (c) The algebraic integers A, i.e., the ring of all elements in C which satisfy an equation of integral dependence over Z. (d) Q[x] (e) Q[x; y]=hy17 − x2y3 + 5x5i (f) The polynomial ring K[x1; : : : ; xn]. For this one, just guess, and prove your guess is a lower bound. To show it's an upper bound is quite hard; we'll do it soon enough. (g) The polynomial ring K[x1; : : : ; xn;::: ] (h) Q[x; y; z; w]=hw3 − w2x − yz; z4 − xyi (use (f)). (2) Explain why dim W −1R ≤ dim R and dim R=I ≤ dim R. (3) Explain why every principle ideal domain has dimension one. (4) Prove that the ring R × S has dimension equal to the larger of dim R and dim S. (5) Prove that the ring R and its reduced ring Rred have the same dimension. (6) Proofs of the Corollaries. Let R,! S be an integral extension of rings. Assume the Lying Over Theorem. (a) Suppose that Q0 2 Spec S contracts to P0 2 Spec R. Prove that the map R=P0 ,! S=Q0 is also an integral extension. (b) Given a chain of primes P0 ⊂ P1 in R, prove that there exists a chain of primes Q0 ⊂ Q1 in S such that Pi = Qi \ R for i = 0; 1. [Hint: Use (a)] (c) Show that the Going Up Corollary follows from the Lying Over Theorem. (d) Show also that the Dimension Corollary follows. 2 (7) Incomparable Fibers. Let R,! S be an integral extension. (a) Prove that if R and S are domains, then every u 2 S has a non-zero multiple in R. (b) Show that if Q0 ⊂ Q1 in Spec S both contract to P in Spec R, then Q0 = Q1. [Hint: Use 5a.] (c) Conclude that the fibers of the map Spec S ! Spec R consist of incomparable primes, and that the ring S ⊗R RP =P RP has Krull dimension zero. p p (8) Consider the extension Z ,! Z[ 2] and the corresponding contraction map Spec Z[ 2] ! Spec Z. (a) Compute the fiber over the points h0i; h2i; h3i; h7i of Spec Z. (b) For p > 2 a prime integer, prove that there are at most two points over hpi, with exactly one if and only if 2 is not a square mod p. (9) The Cayley-Hamilton Theorem, revisited. Let R,! S be an integral extension of rings. Let I be an ideal of R. Our goal in this problem is to prove: If u 2 S is contained IS, then there is an equation of integral dependence n n−1 u + r1u + ··· + rn = 0 i where each ri 2 I ⊂ R. (a) For u 2 IS, write u = i1s1 + ··· + itst where ij 2 I ⊂ R and sj 2 S. Explain why R,! 0 S = R[s1; : : : ; st] is module-finite. Now, for any set fa1; : : : ; ang of R-module generators for 0 0 S over R, explain why IS = Ia1 + ··· + Ian. (b) Show that the n × n matrix over R representing the R-linear map S0 ! S0 given by multi- plication by u can be assumed to have all entries in I. n n−1 i (c) Prove that u + r1u + ··· + rn = 0 where each ri 2 I ⊂ R. [Hint: Cayley-Hamilton!] (10) Easy Lemma. Let S be an arbitrary ring, and let U be a multiplicative set and J an ideal of S. Consider the two subsets, V(J) and fP j P \ U = ;g, of Spec S. Prove that their intersection is non-empty if and only if J \ U = ;. [Hint: Consider the ring SU −1=JSU −1:] (11) The Proof of Lying Over. Let R,! S be anp integral extension. (a) Let I ⊂ R be any ideal. Prove that IS \ R ⊂ I. In particular, if I is radical, IS \ R = I. [Hint: Use 8.] (b) Take P 2 Spec R. Show that, as subsets of S,(R n P ) \ PS = ;. [Hint: Use 9.] (c) Use (b) to show that for all P 2 Spec R, there exists Q 2 Spec S such that Q \ R = P . (d) Conclude that the Lying Over Theorem holds. (12) Let U be a multiplicative set of a ring R that contains some non-unit. Prove that R ! U −1R is not an integral extension. [Hint: Consider the map of Spec.] X X (13) Let F2 be the ring of F2-valued functions on an infinite set X. Prove that F2 is integeral over 2 F2, and compute its dimension. [Hint: Consider the polynomial x − x = 0.] (14) Consider the ring extension C[y] ,! C[x; y]=hy − x2i. Let π : Y ! X be the induced map of Spectra. (a) Is this ring homomorphism integral? (b) Explain why the closed points of Y all have the form hx − a; y − bi where (a; b) lies on the parabola y = x2. Compute the contraction of hx − a; y − bi to Spec C[y]. (c) Compute the fiber of π over each point of X. In particular, what cardinalities are possible? (d) Draw a picture representing map on the subspace of closed points. Which ones are special, and in what ways?.
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