By Andreas Gathmann

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For simplicity, let us restrict to short exact sequences. 11 (Recovering modules from an exact sequence). Consider an exact sequence ϕ ψ 0 −→ M −→ N −→ P −→ 0 of R-modules, and let us ask whether we can determine one of the three modules if we know the rest of the sequence. 10 (a) tells us that P ∼ = N/ ker ψ = N/ im ϕ. In the same way, M is uniquely determined if we know N and P together with the map ψ, since by the injectivity of ϕ we have M ∼ = im ϕ = ker ψ. The most interesting question is thus if we can recover the middle term N if we know M and P (but none of the maps).

Mn ∈ M}. Note that IM is a submodule of M, and M/IM is an R/I-module in the obvious way. 3 (d)) is defined to be N : N := {a ∈ R : aN ⊂ N } R. In particular, for N = 0 we obtain the so-called annihilator ann N := annR N := {a ∈ R : aN = 0} R of N. The same definition can also be applied to a single element m ∈ M instead of a submodule N: we then obtain the ideals N : m := {a ∈ R : am ∈ N } and ann m := {a ∈ R : am = 0} of R. 13. (a) If M, N, and N are submodules of the R-module R, i. e. 1.

10 (a). 24. 19 (a)). 20 (c))? It is actually easy to see that in this case an injective morphism ϕ : M → M need not be bijective: the map ϕ : Z → Z, m → 2m is a simple counterexample. 28 below. The main ingredient in the proof is the following generalization of the Cayley-Hamilton theorem from linear algebra. 25 (Cayley-Hamilton). Let M be a finitely generated R-module, I an ideal of R, and ϕ : M → M an R-module homomorphism with ϕ(M) ⊂ IM. Then there is a monic polynomial (i. e. its leading coefficient is 1) χ = xn + an−1 xn−1 + · · · + a0 ∈ R[x] with a0 , .