Risultati esame Algebra 4

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Algebra 4 – June 12th – Tensor algebras

Tensor algebras

Definition of the tensor algebra T(M) of an R-module M. Definition of the symmetric (and antisymmetric) algebra S(M) (and $\Lambda(M)$ ).

If M is a cyclic module, then T(M)=S(M).

Let R=k[x,y] and I=(x,y). Show that $\Lambda^2 R=0$ and $\Lambda^2 I\not=0$ . Show that $\Lambda^2$ is not an exact functor. (Hint: The only non-trivial thing is the map $f:I\times I \to R/I$ defined by $f(ax+by,cx+dy)=ad-bc$ .)

References: Dummit and Foote “Abstract Algebra”

Algebra 4 – June 5th – Noetherian rings and formal power series

Noetherian rings and formal power series

Definition of the formal power series ring R[[x]] for any ring R and basic properties of its elements and ideals. Proof that if R is a local ring, R[[x]] is a local ring and, and if R is a field, R[[x]] is a euclidean ring.

If R is a Noetherian ring, then R[[x]] is a Noetherian ring.

The formal power series ring R[[x]] is the inverse limit of the rings $R[x]/(x^i)$ .

References: The Stacks Project

Algebra 4 – May 29th – Projective and injective modules

Projective and injective modules

Examples, counterexamples and relations between free, projective, flat and torsion free modules.

Projective modules over local rings are free.

In the ring $R=\mathbb{Z}[\sqrt{-5}]$ the ideals $I=(3, 1+\sqrt{-5})$ and $J=(3, 1-\sqrt{-5})$ are projective non-free modules.
In fact, $R/I\cong R/J\cong \mathbb{Z}_3$ therefore the ideals are maximal and we have a short exact sequence $0\to I\cap J\to I\oplus J\to R\to 0$ . But $I\cap J=IJ=(3)$ , and R is a projective module, so the sequence splits and I and J are summand of a free module, hence projective.

In the ring $R=k[x,y]$ where k is a field, the ideal $I=(x,y)$ is torsion free but not flat. Consider the short exact sequence $0\to I\to R \to R/I\to 0$ . Then tensor with I to get $I\otimes I\to I\to I/I^2$ . The element $x\otimes y + y\otimes(-x)$ is in the ker but it is not zero. For that we use the Lemma 6.4 in Eisenbud “Commutative algebra with a view toward algebraic geometry”, which states that $a_{11}m_1+ a_{12}m_2=x$ , $a_{21}m_1+ a_{22}m_2=y$
and $a_{11}y+ a_{21}(-x)=0$ and $a_{12}y+ a_{22}(-x)=0$ . Since k[x,y] is a UFD, we have $a_{11}y=a_{21}x$ and $(b_{11}x)y=(b_{21}y)x$ so $b_{11}=b_{21}$ and similarly $b_{12}=b_{22}$ . Plugging into the first ones $b_{11}m_1x+ b_{12}m_2x=x$ which yields $b_{11}m_1+ b_{12}m_2=1$ which means that 1 is a linear combination of elements in I which is impossible.

References: Rotman “An introduction to homological algebra”, http://en.wikipedia.org/wiki/File:Module_properties_in_commutative_algebra.svg, http://blog.jpolak.org/?p=363, http://stacks.math.columbia.edu/tag/058Z, Ex 3.25 from http://math.uga.edu/~pete/MATH8020C3.pdf

Algebra 4 – May 22th – More localization

More localization

The localization is an exact functor.

We have $S^{-1}A\otimes_A M\cong S^{-1}M$ as A-modules and $S^{-1}A$ -modules.

Local properties: an A-module M is zero if and only if its localizations at prime ideals are, if and only if the localizations at the maximal ideals are.

In $S^{-1}A$ all ideals are extension of ideals of A. An ideal I of A extends to the whole ring if and only if it meets S. Prime ideals of the localization are in 1-1 correspondence to prime ideals of A which do not meet S. The nilradical of the localization is the localization of the nilradical.

References: Atiyah-Macdonald chap 3.