Math Is Fun Forum

1. Hopefully you know that the prime ideals in K[x] are generated by (f) where f is irreducible and monic.  Now when we move to the larger ring K[x,y]/(y^2-x^2-x^3), you really only have to consider multiplication by a single polynomial to get an ideal.  Now all that remains is to check if it's prime.

2. Recall that in a localized ring, an element a/s is zero if and only if there exists a d in R with da = 0.  So we assume there is a d in B with da = 0 for some a in A.  Prove that there is an element c in A with ca = 0 as well.  (Hint: Integral extension)

Ah, I see.  Your post was spot on, bobby.

soroban, though differentials are written like this because they behave like multiplication/division most of the time, one can not prove an identity by using arithmetic.

Identity, you sure you have it right?  I am getting that (assuming equality of mixed partials)

Look at the elements (1,0) and (1, 1).  Since the first coordinate in each of these elements are the same, (1,0)S(1,1).  The same goes for (1,0)S(1,5).  So we consider them to all be the same, which we call a "class".  Your task is to find a way to represent exactly one element from each class.  So (1,0) is a representative of the class {(1,0), (1, 1), (1, 5), ...}.

Solvable won't do it for you.  Think "Fundamental theorem of Galois theory".

The same way you prove anything is a module.  Make sure your multiplication is well-defined, and most of the other properties you don't have to check because P/IP is an R-module.

Any projective module is a direct summand of a free module.

Hint:

Also, the function N, called the norm, has the very nice property that N(ab) = N(a)N(b).  This is quite useful when talking about irreducibility.

For any subgroup H, the corresponding subfield is all the elements which are fixed by any permutation in H.  Hence if you take H = G you get the base field (assuming the extension is Galois), and if you take H = 1, you get the entire field extension.  These are just the trivial cases.

I'm not familiar with the notation N • p

These are not the same.

Yes.

1.  If you know the division algorithm works in F[x], then this proof becomes easy.  Prove it's a Euclidean domain, and this implies it is a principal ideal domain.

2b "show that Rs isomorfic with the subring of Q(R)" With a subring of Q(R)?  When you have a subset, there is always a very easy way to injectively define a map.

3 f(ab) = f(a)f(b)

This is what you call, "deciding whether you like the taste of broccoli"?  That is an exaggeration to the point where it does have anything to do with actually liking it.  The hypnosis will only work while in a hypnotic state, which may not even occur in the first place.  Typically a person has to be "suggestible" in order to induce a trance.

My point stands, one can not choose whether or not they like the taste of broccoli.  Over time it can change however.

While the word typically has a Christian connotation to it, it has changed to also mean anyone who focuses on proselytizing their beliefs.

Getting back to the point, why is someone who "preaches" atheism not an atheist?

Dawkins preaches in the same way political organizations, science educators, and D.A.R.E. employees do.