Aluffi - Algebra, Chapter 0



AMS GSM 104

Algebra

Errata available online:

"This text presents an introduction to algebra suitable for upper-level undergraduate or beginning graduate courses. While there is a very extensive oﬀering of textbooks at this level, in my experience teaching this material I have invariably felt the need for a self-contained text that would start ‘from zero’ (in the sense of not assuming that the reader has had substantial previous exposure to the subject), but impart from the very beginning a rather modern, categorically-minded viewpoint, and aim at reaching a good level of depth. Many textbooks in algebra satisfy brilliantly some, but not all of these requirements. This book is my attempt at providing a working alternative."

=Chapter I : Preliminaries: Set theory and categories =

=Chapter II : Groups, ﬁrst encounter=

=Chapter III : Rings and modules=

=Chapter IV : Groups, second encounter=

=Chapter V : Irreducibility and factorization in integral domains=

=Chapter VI : Linear algebra=

=Chapter VII : Fields=

=Chapter VIII : Linear algebra, reprise=

=Chapter IX : Homological algebra=

Section IX.1 (Un)necessary categorical preliminaries

 * After Example 1.11: The coproduct $$A \sqcup B$$ is endowed with morphisms $$i_A: A \to A \sqcup B$$, $$i_B: B \to A \sqcup B$$ which are easily checked to be monomorphisms (do this!) : We have $$\pi_A \circ i_A = 1_A$$ by definition, or in other words $$\pi_A$$ is a left-inverse of $$i_A$$. Since monomorphisms are defined by cancellation, this is actually even a bit stronger than being a monomorphism (roughly the equivalent of "unit implies non-zerodivisor" in a commutative ring.)


 * Proposition 1.12, $$\pi_B$$ is the cokernel of $$i_A$$ : The proof given here is bizarrely stated. I had to work things out on my own, after which I eventually realized that my method matched the books. Don't, as suggested, stare at the diagram; the diagram is weird and unhelpful. Here's what it says: We already know that $$\pi_B \circ i_A = 0$$, so what remains to show is that for any object C and any map $$\gamma : A \sqcup B \to C$$ with $$\gamma \circ i_A = 0$$, there exists a unique map $$\delta : B \to C$$ such that $$\delta \circ \pi_B = \gamma$$. In particular, we claim that $$\delta = \gamma \circ i_B$$. To see this, consider the (unique!) map $$g : = (0 \sqcup (\gamma \circ i_B)) : A \sqcup B \to C$$; that is, the unique map such that $$g \circ i_A = 0$$ and $$g \circ i_B = \gamma \circ i_B$$. By definition, $$\gamma$$ satisfies both equations, so $$g = \gamma$$. On the other hand, we can also check that $$(\gamma \circ i_B \circ \pi_B)$$ satisfies both equations, so $$g = \gamma \circ i_B \circ \pi_B$$. Therefore $$\gamma = \gamma \circ i_B \circ \pi_B = \delta \circ \pi_B$$, which is what we wanted to show.


 * Exercise 1.14: The word 'sheaf' should be replaced by 'presheaf' here.