Field generated by elements

Hi,

Suppose that the ring

I can see that we could replace with for example and the ring we generate would be the same, However substituting for for example would not necessarily yield the same ring. Is there any general way of stating which combinations or operations we can perform on and and leave the ring they generate unchanged?

Thanks!

EDIT: replaced 'Field' with 'Ring'

Re: Field generated by elements

Quote:

Originally Posted by

**Ant** Hi,

Suppose that the field

I can see that we could replace

with

for example and the field we generate would be the same, However substituting

for

for example would not necessarily yield the same fiel. Is there any general way of stating which combinations or operations we can perform on

and

and leave the field they generate unchanged?

Thanks!

Polynomial rings are not fields...

Re: Field generated by elements

Quote:

Originally Posted by

**Gusbob** Polynomial rings are not fields...

Sorry, of course they're not.

Everywhere I said "field" I meant "Ring". I'll edit now, thanks

Re: Field generated by elements

Since and are variables, surely you can just replace and by two linearly independent variables which may or may not depend on and , say . Then gives a ring isomorphism between and .

EDIT: I realised the restriction of linearly independent is not enough (see and ). But if you specify that and are linearly independent for all choices of , it should be enough.

Re: Field generated by elements

Re: Field generated by elements

Quote:

Originally Posted by

**Ant** I can see that we could replace

with

for example and the ring we generate would be the same, However substituting

for

for example would not necessarily yield the same ring. Is there any general way of stating which combinations or operations we can perform on

and

and leave the ring they generate unchanged?

How familiar are you with Galois theory?

Re: Field generated by elements

Quite familiar, that's the course in which this question has come up.

After finding a splitting field for a polynomial with 4 roots, say, it's easy to eliminate two of the generators (roots) because they'll be complex conjugates. But if you wanted to simplify further (not necessarily to eliminate a root but just they're 'nicer' and so it's easier to find the min poly) I'm wondering what are the operations you can do on the roots which will leave the ring generated unchanged.

thanks

Re: Field generated by elements

Quote:

Originally Posted by

**Ant** Quite familiar, that's the course in which this question has come up.

After finding a splitting field for a polynomial with 4 roots, say, it's easy to eliminate two of the generators (roots) because they'll be complex conjugates. But if you wanted to simplify further (not necessarily to eliminate a root but just they're 'nicer' and so it's easier to find the min poly) I'm wondering what are the operations you can do on the roots which will leave the ring generated unchanged.

thanks

How much of the 'sameness' do you want (this is why we need to be careful with terminology)? For example as vector spaces, but not as fields.

For matter of field extensions, we'll need to thread very carefully. I'll illustrate the one-element case first, the second one generalises (or you can add the elements one at a time). Firstly, given isomorphic extensions it is NOT true in general that and have the same minimal polynomial. For example, are isomorphic extensions of the rationals, but definitely do not have the same minimal polynomial. In fact, even doing as you have suggested initially: taking the negative of , may change the minimal polynomial of the generator ( has minimal polynomial , but has minimal polynomial ).

It is true, however that there is an element in such that AND have the same minimal polynomial. In general, if you want to change your adjoined element to compute the minimal polynomial more easily, you can only change it to another root of the minimal polynomial, which kind of renders the entire exercise futile.

I can elaborate more on what kind of operations preserve the field, but not the minimal polynomial of the adjoined element if you wish, but I don't think that is what you asked for. (Think where )

Re: Field generated by elements

Thanks for the above, that's very helpful and it's made me think of a few things I've not considered before.

Generally, my task is given a polynomial 1) find it's splitting field 2) compute it's degree 3) List all distinct embeddings 4) find a primitive element (if one exists).

My usual technique is as follows:

Let be a field with characteristic 0.

1) Find the roots of the polynomial and thus the splitting field.

2) eliminate or simplify the way we write this splitting field. So for example if we find that with being the distinct roots of in some extension . We may find that we can eliminate some roots, or find a 'nicer' expression for . So yes, I'm definitley looking for preserve the field!

3) Using the Tower Law we spit into a series of simple extensions, for each compute the minimal polynomial and hence the degree of the extension. Take the product, this will give us the total number of distinct -embeddings . For some extension .

4) compute the embeddings step by step going up through the simple extensions.

does that make sense?