I thank the institute's teachers for teaching me knowledge and understanding of mathematics. I would also like to thank my family and friends for helping me in every possible way and encouraging me during this program. The main purpose of this project is to learn a branch of mathematics that studies commutative rings with unity.
Introduction
Let S be a subset of a ring A and I an ideal of A disjoint of S. Then the set of ideals in A containing I and disjoint of S contains a maximal element and if S is multiplicative, then every such maximum element prime. The set of ideals of Abevat I separate from S is not empty, because it containsI. Now we define by previous statement. T∈CT where C is a chain in F and T0 ∈F otherwise some element of S lies in T0 and therefore in T for some T which is a contradiction with the definition of F then by Zorn's lemma F has maximum element.
5 Now suppose that S is the multiplicative subset of A and let M be the maximal element in F Let bb0∈M and if b∈M/ then M⊂M+ (b).
Radical
An element c of A is in the Jacobson radical of A if and only if 1-acis is a unit for all a∈A. Let P1,P2,..,Pr,r≥1 be ideals in A so that Pi are prime ideals for i≥3. If an ideal I does not contain any of Pi, then I is not contained in the union of Pi. Then I mean but not in one of Pi. If z is in one of Pi for a given i≤r−1 then zr∈Pi contradicts tozr∈Pr. Now suppose z is inPr. Thenz1..zr−1 is in Pr.
If r is 2, we are done. If r≥3, then since Pr is a prime ideal, there is some zi,i≤r−1 in Pr, which is a contradiction, so our assumption zi∈Pi is false for all. So we're done.
Contraction and extension ideals
Nakayama's lemma Let A be a ring and I be an ideal in A. Let M be an A-module and assume that I is contained in all maximal ideals of A and that M is therefore finitely generated. And let M be a finitely generated module over A, the action of A onM/mM factor by K and elements a1, a2, .., and of M generate it as an A-module if the elements a1+mM, .., an+mM spanM /mM as a vector space above K. Ifa1, ., an generates M, then their images generate the vector space M/mM. Conversely, suppose a1+mM, .., an+mM spanM/mM and let N be a submodule of M, then the composite map is N →M →M/mM op and sumM =N +mM and then by lemmaM = N. Let A be a noetherian local ring with maximum ideal m.
Since Ais noetherian somis is finitely generated, then apply the previous proposition for M = m and we are done. a) The height ht(p) of a prime ideal p in A is the greatest length d of a chain of different prime ideals p=pd⊇..⊇p0. Let A be an integral domain, then dim(A) = 0 iff (0) is the maximum ideal of A iffA is field. Observation if A is an integral domain and S is any multiplicatively closed subset not containing 0, then f is injective.
Ice =I for all ideals of S−1A and Pec =P , if P is a prime ideal of A and disjoint from S. S−1u is well defined since u is a homomorphism of the module A. Now we observe S− 1 conservation of addition and multiplication. Let M be an A-module, then the canonical map M →Y. Read x∈M to zero at all Mm, then we will show xis zero.
Let A be a subring of B. An element b of B is said to be an integral over A. If it is a root of a nonzero monic polynomial with coefficients in A, it means that it satisfies the equaton. Such an equation is called an integral equation of overA. 2) A[b] is finitely generated as an A-module. Forr= 1 we are done with the previous statement. Assume inductively that B0 = A[b1, .., br−1] is finitely generated as an A-module . Since br is an integral overA, it is also an integral overB0.
27 B0[c] is finitely generated as an A-module by using B is integral over A, and then c is integralA.
Going Up Going Down Thereom
So the result holds for integral expansion, but may not be true for general rings. Let A⊆B be rings and I am an ideal of A and C is the ingraal closure, the integral closure of A in B. Then the set of all elements in B that are integral over I is the radical of IC =Ie.
Consider the ring M = A[b1, .., bn] , and it is finitely generated A module andxnM ⊆IM and consider the mapφxn;M →M by,. Suppose p1 ⊆p2 are two prime ideals of A and q2 is the prime ideal of B such that q2∩A=p2, then there exists a prime ideal of q1 contained in q2 such that q1∩A=p1. Also s ∈ B ⇒ s is integral over A and is now multiplied by equation y−n, which gives the equation sn+ (a1/y)sn−1+..+an/yn= 0, and since the above equatin is minimal, then this is also minimal equation for s.
Noether Normalization Thereom
Common to these examples is that the product in each case is a bilinear map. If V1 and V2 are two vector spaces over a field F, the tensor product is a bilinear map: V1×V2→V1⊗V2, where V1⊗V2 is a vector space over F. The tricky part is that to define this map, we need first construct this vector space V1⊗V2 We give two definitions.
The first is an axiomatic definition, in which we specify the properties that V1 ⊗V2 and the bilinear map must have. In a sense, this is all we need to work with tensor products in a practical way. The condition (∗) does not actually need to be checked for every possible base pair β1, β2, it is sufficient to check it for every single base pair.
One way is to think of the space V⊗W abstractly and to use the axioms to manipulate the objects. The second way is actually to identify the space V1 ⊗V2 and the map V1 ×V2 → V1 ⊗V2 with some known object. There are many examples that it is possible to make such an identification naturally.
Note that when we do this, it is crucial that we specify not only the vector space we identify as V ⊗V2 , but also the product (bilinear map) we use to make the identification.
Constructive definition of tensor product
Constructive definition of tensor product element inAand (which is irrelevant in the definition of F ree(A)). To help keep it straight in situations where there is a danger of confusion, we will write, and when we mean the operation in F ree(A). It is called 0 because it happens to be the zeroth element in N, but it is completely irrelevant in the construction of the free vector space.
For some some m ∈ N, in other words the elements of the vector space of polynomials in a single variable. If P is the space of all linear combinations of symbols (v, w), then R is the space of all those linear combinations that can be simplified to the zero vector by bilinearity. Thus, P/R is the set of all terms where two terms are equal if one can be simplified to the other using bilinearity.
Universal mapping property of tensor product
To prove that the condition (∗) holds, we use the following important lemma from the theory of quotient spaces. Suppose β is a basis for V and γ is a basis for W, then we must show that β⊗γ is a basis of V ⊗W.
Tensor product on modules
Properties of Tensor products
45 NOTE to show that Q⊗ZA= 0, we do not need A to be finite, but instead that every element of A has a finite order and thus Q⊗ZQ/Z= 0. For an ideal I i R and M is an R-module that is unique R-module isomorphism(R/I)⊗M ∼=M/IM. Then φ is well-defined clear , so by universal mapping property we get a linear map f : (R/I)⊗M →M/IM such that the diagram commutes.
Then φ is well defined, and then by the universal mapping property we get a linear mapf :R/I⊗R/J →R/(I+J) such that the diagram commutes. If and are additive functions, you can check f(g(t)) = t for all tensors by checking it only on elementary tensors, but it would be wrong to think that you can check the injectivity of a linear map f :M⊗ N →P have only been proven by looking at elementary tensors. Suppose R is a domain with fraction field K and V is a vector space over K, then there is an R-module isomorphism K⊗V ∼=V.
To show that f is one one, we first show that every tensor in K⊗V is elementary with 1 in the first component. Notice how we moved x∈K across, even though x doesn't have to be in R, we used the scaling of K to V to create b and 1/b on the right-hand side of ⊗and yield b.
Questions
Since this bilinear map actually sends (m, n) to m⊗n, we obtain m⊗n= 0. 3) The tensor productM⊗N is 0 if and only if every bilinear map from M×N (to all modules) is equal is identical to 0. Let Q be a primary ideal of A. We know that the radical of Q is the intersection of all prime ideals containing Q. Now it is sufficient to show that r(Q) is prime, and this is clear because Q is prime. Now our next claim is that Q is not a power of a prime ideal. First suppose that Q=Pn for a prime ideal P and also note that r(Q) = P and P2 (Pn (P which is impossible, so Q is not a power of prime ideal.
Claim: P is prime ideal of B but P2 is not prime ideal. Idea is B/P is integral domain implies P is prime. Claim: Every zero divisor of A/Q is zeropotent. Let if possible x =a+Q∈ A/Q is a zero divisor but not zero potent. Then x7→x¯6= 0∈A/M, which is not a zero divisor, implies x is not a zero divisor contradiction so x is nullpotent so Q is prime. If M is any maximal ideal of A , then r(Mn) =M implies Mn is primary. Let S be the set of ideals that are not finite intersections of irreducible ideals.
Since the intersection of two finite intersections of irreducible ideals, I is the intersection of irreducible ideals, i.e. In /∈S. Let G be an abelian group, then G is said to be topologically abelian if both the maps G×G→G and G→G defined by (x, y)7→. Let A be a Noetherian ring, I be an ideal of A, M a finitely generated A-module together with an I-filteringM =M0 ⊆M1.
By assumption ⊕Mn is a finitely generated A-module, i.e. an A-Noetherian module, so the chainPK must stop, i.e., 3N such that PN =PN+1=. Let A be a Noetherian ring, I an ideal of A, and M a finitely generated A-module with stable M of I-filling.
