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Basic Algebra Prof. D. P.Patil, Department of Mathematics, Indian Institute of Science, Bangalore May-July 2003 Basic Algebra 6. Homomorphisms of rings Ferdinand Georg Frobenius† Joseph Henry Maclagen Wedderburn†† (1849-1917) (1882-1948) X 6.1. Let X be a set ane let P(X) be the power set ring of X. The map e : P(X) → (K2) , that assigns every subset A ⊆ X to its characteristic funktion eA (which maps x → 1ifx ∈ A and x → 0ifx/∈ A), is an isomorphism of rings of the ring P(X) onto the X–fold direct product of K2. A permutation σ of X induces the permutation σ of P(X) with σ (A) := σ (A) ={σ(x) : x ∈ A} for A ∈ P(X).Forσ ∈ S(X), σ ∈ Aut P(X), and the map σ → σ is an isomorphism of groups from the group S(X) onto Aut P(X). 6.2. Let ϕ : A → A be a homomorphism of rings. For a ring B denote B (ϕ) : Hom(B, A) → Hom(B, A ) and B (ϕ) : Hom(A ,B) → Hom(A, B) the cannonical maps (on the sets of ring homomorphisms) with τ → ϕτ resp. σ → σϕ. a). ϕ is injective if and only if B (ϕ) is injective for all rings B. (Hint : Construct the ring Z × (Kern ϕ) and consider the ring homomorphism Z × (Kern ϕ) → A defined by (n, a) → n · 1A + a and (n, a) → n · 1A.) b). If ϕ is surjective, then B (ϕ) is injective for all rings B. (Remark : The converse does not hold: For every ring B, B (Z → Q) is injective, but the inclusion Z → Q is not surjective.) 6.3. Let m, n ∈ N∗ and n be a divisor of m. Then there exists exactly one ring homomorphism → × × → × ∈ Z ϕ :Am An. Further, both ϕ as well as ϕ :Am An are surjective. (Hint : If a with gcd(a, n) = 1, then there exists a r ∈ N such that gcd(a + rn,m) = 1, for example, the product of those prime factors of m, which do not divide a or n.) 6.4. a). Let n ∈ N. Show that the projections pi : (ai ) → ai ,i= 1,...,n, are the only ring n homomorphisms Z → Z. (Hint : Consider er es = 0 for r = s, where er ,r = 1,...,n, is the standard Z–basis of Zn.) N b). Show that the projections pi : (aj )j∈N → ai ,i∈ N, are the only ring homomorphisms Z → Z. (Hint : Use the following theorem : N Theorem (Specker) The projections πm : Z → Z,m∈ N, with (an)n∈N → am form a basis of the Z–module of all linear forms on ZN. Corollary (R. Baer, E. Specker) Let I be an infinite set. Then the ablelian group ZI is not a free group. In particular, ZN is not free. N Proof Suppose that Z is free with the basis fi ,i∈ I, then I will be necessarily uncountable. But then ∗ ∈ there will be uncountably many coordinate functions fi ,i I and hence there will be uncountably many linear forms on ZN, but by the theorem there are only countably many linear forms on ZN. • Proof of the theorem : (Due to E. Specker). First we shall show that πm,m ∈ N, are linearly Z ∈ N Z(N) ⊆ ZN = independent over . Let en,n , be the standard basis of . Then πm(en) δmn and from = ∈ Z = = = ∈ N m∈N amπm 0,am , it follows that 0 m∈N amπm(en) m∈N amδmn an for all n . N Now it remain to show that every linear form on Z is a linear combination of the πm,m ∈ N. Let ZN → Z = ∈ N h : be a given linear form and let bn : h(en). Let cn,n , be a sequence of positive natural ∈ N ≥ + + n | | numbers such that cn+1 is a multiple of cn for all n and such that cn+1 n 1 r=0 cr br for all D. P. Patil July 18, 2003 ,3:27 p.m. 2 Basic Algebra ; May-July 2003 ; 6. Homomorphisms of rings ∈ N = ∈ N n . Such a sequence can be defined recursively. We consider c : h((cn)n∈N). For every m , there ∈ ZN = m + = m + exists a ym such that (cn) n=0 cnen cm+1ym. Applying h we get c n=0 cnbn cm+1h(ym). | − m |= | | ≥ ≥|| | − m |≤ and soc n=0 cnbn cm+1 h(ym) which is either 0 or cm+1.Ifm c , then c n=0 cnbn | |+ m | |≤ + m | | = m ≥| | c n=0 cnbn m n=0 cnbn <cm+1 by definition of cn. Therefore c n=0cnbn frr all m c . = = | | − |c| This means that cnbn 0 and so bn 0 for all n> c . Therefore the linear form h n=0 bnπn vanishes on all elements of the standard basis en,n∈ N. We shall now show that such a linear from must be zero. N N Therefore let g be a linear form on Z with g(en) = 0 for all n ∈ N. Let (cn) ∈ Z be given. There n + n = = − 2n = exists integers vn,wn such that vn2 wn3 cn. (for example cn cn(3 2).) Then g((cn)) n + n ∈ N ∈ ZN n = m−1 n + m g((vn2 )) g((wn3 )). For every m there exists a zm with (vn2 ) n=0 vn2 en 2 zm. n m n It follows that g((vn2 )) ∈ 2 Z for all m ∈ N, and so g((vn2 )) = 0. Analogously it follows that n g((wn3 )) = 0. This proves that g((cn)) = 0, as desired. • c). What are all the automorphisms of the ring Zn,n∈ N, resp. ZN? 6.5. (Radical of an ideal) Leta be an ideal in a commutative ring A. The inverse image of the nilradical of A/a in A under the√ canonical projection A → A/a is called the radical ofa.It is usually denoted by r(a) or by a. r(a) is the set of all a ∈ A, for which there is (dependent on a) n ∈ N such that an ∈ a. The residue ring A/r(a) is canonically isomorphic to the reduction of A/a. Further, r(r(a)) = r(a). 6.6. (Theorem of M. H. Stone) Let A be a Boolean ring. Let M denote the set of all maximal ideals in A. Then m∈M m = 0. For every m ∈ M there exists a unique homomorphism → m → M → ϕm : A K2 with the kernel . The map ϕ : A K2 defined by a (ϕm(a))m∈M is an injective ring homomorphism. If A has only finitely many elements, then ϕ is bijective. (Hint: Distinct maximal ideals are relatively coprime.) From the theorem of Stone deduce that: Every Boolean ring is isomorphic to a subring of a full power set ring; every finite Boolean ring is isomorphic to a full power set ring. (see also exercise 6.1) There exists a Boolean ring, which is not isomorphic to the full power set ring. (Hint: Full power set ring is never countablly infinite.) 6.7. Let a1,...,an and b1,...,bm be two-sided ideals in a ring A with ai + bj = A for all i, j. Then a1 ···an + b1 ···bm = A.Deduce that: If a and b are relatively coprime two-sided ideals in a ring, then an and bm are also relatively coprime ideals for arbitrary m, n ∈ N. 6.8. For pairwise relatively coprime two-sided ideals a ,...,a in a ring A we have : 1 n a1 ∩···∩ an = aσ(1) ···aσ (n) . (Hint : Induction) σ ∈S n n a = n a If A is commutative, then i=1 i i=1 i . 6.9. Let A be a ring. The set Idp(Z(A)) of all idempotent elements in the center Z(A) of A is finite if and only if A is isomorphic to a finite direct product of indecomposable rings. Moreover, in this case |Idp(Z(A))|=2s , where s is the number of indecomposable components in the product representation of A. 6.10. Let ϕ : A → B be a homomorphism of commutative rings. ϕ induces a map ϕ1 : Idp(A) → Idp(B) between the idempotent elements of A resp. B.Ifϕ is surjective and the kernel of ϕ is conatined in the nilradical nA of A, then ϕ1 is bijective. (Hint: Use exercise 1.4) Corollary: A commutative ring A is indecomposable (see T6.8) if and only if A/nA is indecomposable. (Hint: If A/nA is indecomposable, then so is A even in the non-commutative case (proof!). However, in the non-commutative case the residue ring A/nAcan be decomposable without A being so. For example the ring ab B := : a, b, c∈ K ⊆ M (K) 0 c 2 of 2 × 2 upper triangular matrices over a field K (vgl. Kapitel V) is indecomposable, but the residue ring ∼ B/nB = K × K is decomposable.) D. P. Patil July 18, 2003 ,3:27 p.m. Basic Algebra ; May-July 2003 ; 6. Homomorphisms of rings 3 Below one can see (simple) test-exercises. Test-Exercises T6.1. Let A and B be rings and let f, g be homomorphisms of A in B. The subset A of all elements a ∈ A with f(a)= g(a) is a subring of A. Moreover, if A is a division ring, then A is also a division ring. T6.2. Let K and L be fields of characteristic = 2 and ϕ be a group homomorphism of (K, +) in (L, +) with the following properties: (1) ϕ(1) = 1. (2) ϕ(a)ϕ(a−1) = 1 for all a ∈ K, a = 0.
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