Esame gennaio 2020, Commutative Algebra, Kloosterman

REMKE KLOOSTERMAN

If both there properties hold then dim A A = dim A + dim A .

Suppose that R is a ring and M, N are R-module such that M N .

Then there may exist R-modules P such that M P is not a submodule of N P , i.e., the tensor product of an injective morphism may have a non-trivial kernel, however if P is flat over R then this does not happen.

Recall that A and A are K-algebras and hence are K-modules. If we take the tensor product of two K-algebras is just the tensor product as K-modules together with the algebra structure. Every K-module is flat, hence the fact that K[x , . . . , x ] is a submodule of A implies that K[x , . . . , x ] A is a submodule of A A . From the fact K[y , . . . , y ] is a submodule of A it follows that K[x , . . . , x ] K[y , . . . , y ] is a submodule of K[x , . . . , x ] A and therefore of A A . This yields the first point.

The second

The point follows from the following observation. If R is an ₀⊗integral extension of R and S an R-algebra then R S is an integral ring₀⊗ ⊗ ⊗extension of R S: The ring R S is generated by elements a b, with₀∈ ∈a R , b S hence it suffices to show that each such element is integral₀⊗over R S. Since R is integral over R it follows that there exists a monicn iP ∈polynomial f (t) = c t , (with c R, c = 1) such that f (a) = 0 Leti i ni=0n n−i iP ⊗g(t) = (c b )t . Thenii=0n nX Xn−i i i n n⊗ ⊗ ⊗ ⊗ ⊗g(a b) = (c b )(a rb) = (c r b ) = g(r) b = 0i ii=0 i=0⊗ ⊗Hence a b is a zero of a monic polynomial with coefficients from S R.⊗Since A is integral over K[y , . . . , y ] we find that A A is integral2 1 r 1 2⊗ ⊗over A K[y , . . . , y ] and similarly that A K[y , . . . , y ] is integral over1 1 r 1 1 s⊗K[x , . . . , x ] K[y , . . . , y ].1 r 1 r(3) Let R be a

local ring with the maximal ideal m.

(a) Formulate Nakayama’s lemma.

Solution: Let R be a local ring with maximal ideal m. Let M be a finitely generated R-module and let m₁, . . . , m_n be generators for M. Let k = M/mM. Then the m₁, . . . , m_n lift to a set of generators for M. (There are many alternative formulations; one is that if M = MmM holds, then M = 0.)

(b) Show that if R is Noetherian, then M = 0.

Solution: Since R is Noetherian, we have that I = m₁, . . . , m_n is finitely generated. Moreover, one can show that I = m₁, . . . , m_n = m², . . . , m_n², . . .

Assuming this, we may apply Nakayama’s lemma and obtain I = 0.

It remains to show that M = I.

Since R is Noetherian, we have that I has a primary decomposition I = Q₁ ∩ . . . ∩ Q_t.

Suppose now that M = I. Then there is a b ∈ I such that b ∉ Q_i for every i.

Since b ∈ I, it follows that b^m ∈ mI for every i.

Since b ∉ Q_i, it follows that b^m ∉ Q for every i.

Therefore, b^m ∉ P.

Since Q is primary, P is a prime ideal.

and b it follows that m i i