with no zero divisiors,
and if the
with no zero divisors, ,
The permutations and combinations of taken at a time are respectively.
Show that the image of under the Segre embedding is actually irreducible.
Is function composition commutative?
If and are -algebras with ring morphisms and , and is an -algebra with morphism , then , where is .The map is not defined in the text, but my guess is it's .I don't understand why the diagram is commutative though. That would imply for all . Is that true, or is v something else?Added: On second thought, does this follow since and where ⋅ is the -module structure on ?
Let be rings, integral over ; let be prime ideals of such that and say. Then .
Question 1. Why ?
My attempt: Since , we have . But m is maximal in , which is not necessarily maximal in . I can't get by this.
Question 2. When we use that notation , which means the localization of at the prime ideal of . But in this corollary, doesn't necessarily be a prime ideal of . Why can he write ? Should we write rigorously?
Proof of commutative property in Boolean algebra
The theorem is stated in the context of commutative Banach (unitary) algebras, but the proof seems to show that it is valid for any commutative algebra defined as a linear space where a commutative, associative and distributive (with respect to the addition) multiplication is defined such that .
In any case, whether it concerns only commutative Banach unitary algebras or commutative algebras as defined above, I think we must intend contained as properly contained. Am I right?
Let be a Hilbert space and a commutative norm-closed unital -subalgebra of . Let be the weak operator closure of .
Question: For given a projection , is the following true?
It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than . Also, if the above is true, what happens if is non-commutative?
Just a simple question. What does Eisenbud mean by where a ring. An example on this is in the section 17 discussing the homology of the koszul complex. I assume it's something along the lines of for some .
Let be a commutative -algebra, where is a field of characteristic zero.
Could one please give an example of such which is also:
(i) Not affine (= infinitely generated as a -algebra).
(ii) Not an integral domain (= has zero divisors).My first thought was , the polynomial ring over k in infinitely many variables, but unfortunately, it satisfies condition (i) only. It is not difficult to see that it is an integral domain: If for some , then there exists such that , so if we think of in , we get that or , and we are done.
Let be a commutative Banach algebra. Let and be characters of .
I am having some difficulty seeing why the following statement is true:
If , then since , we have that .
Need to find a functor Set Set such that Alg(T) is concretely isomorphic to the category of commutative binary algebras.
The first idea is that the functor is likely to map object to the because then we have to get a binary algebra, i.e., the operation , which have to be commutative.
So the question (if these thoughts are right) is: how to map to to get later a commutative binary algebra?
I would like to know, under what condition on the group (abelian, compact or localement compact ...), the algebra is commutative?
Let be a -algebra.
(i) Let be a state on a -algebra . Suppose that for all unitary elements u∈A. Show that φ is a pure state. [Hint: ]
(ii) Let be a multiplicative functional on a -algebra . Show that is a pure state on .
(iii) Show that the pure and multiplicative states coincide for commutative .
I managed to work out the first two problems but I have no idea about the last one. How to see from being an extreme element in the state space of a commutative that the extreme element is multiplicative?
I am at a very initial stage of commutative algebra. I want to know whether the power of a prime ideal in a commutative ring is prime ideal or not?