# Field generated by elements

• Apr 1st 2013, 04:23 AM
Ant
Field generated by elements
Hi,

Suppose that the ring $F := \mathbb{Q}[X,Y]$

I can see that we could replace $X,Y$ with $-X,-Y$ for example and the ring we generate would be the same, However substituting $X$ for $X^2$ for example would not necessarily yield the same ring. Is there any general way of stating which combinations or operations we can perform on $X$ and $Y$ and leave the ring they generate unchanged?

Thanks!

EDIT: replaced 'Field' with 'Ring'
• Apr 1st 2013, 04:50 AM
Gusbob
Re: Field generated by elements
Quote:

Originally Posted by Ant
Hi,

Suppose that the field $F := \mathbb{Q}[X,Y]$

I can see that we could replace $X,Y$ with $-X,-Y$ for example and the field we generate would be the same, However substituting $X$ for $X^2$ for example would not necessarily yield the same fiel. Is there any general way of stating which combinations or operations we can perform on $X$ and $Y$ and leave the field they generate unchanged?

Thanks!

Polynomial rings are not fields...
• Apr 1st 2013, 04:57 AM
Ant
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
• Apr 1st 2013, 05:06 AM
Gusbob
Re: Field generated by elements
Since $X$ and $Y$ are variables, surely you can just replace $X$ and $Y$ by two linearly independent variables which may or may not depend on $X$ and $Y$, say $f(X,Y),g(X,Y)$. Then $X\mapsto f(X,Y),Y\mapsto g(X,Y),1\mapsto 1$ gives a ring isomorphism between $\mathbb{Q}[X,Y]$ and $\mathbb{Q}[f(X,Y),g(X,Y)]$.

EDIT: I realised the restriction of linearly independent is not enough (see $X$ and $X^2$). But if you specify that $f(X,Y)^i$ and $g(X,Y)^j$ are linearly independent for all choices of $i,j\in \mathbb{N}$, it should be enough.
• Apr 1st 2013, 05:36 AM
Ant
Re: Field generated by elements
Quote:

Originally Posted by Gusbob
Since $X$ and $Y$ are variables, surely you can just replace $X$ and $Y$ by two linearly independent variables which may or may not depend on $X$ and $Y$, say $f(X,Y),g(X,Y)$. Then $X\mapsto f(X,Y),Y\mapsto g(X,Y),1\mapsto 1$ gives a ring isomorphism between $\mathbb{Q}[X,Y]$ and $\mathbb{Q}[f(X,Y),g(X,Y)]$.

EDIT: I realised the restriction of linearly independent is not enough (see $X$ and $X^2$). But if you specify that $f(X,Y)^i$ and $g(X,Y)^j$ are linearly independent for all choices of $i,j\in \mathbb{N}$, it should be enough.

Sorry I should really have been more clear, $X,Y$ are not variables but constants. They are element of some field different from $\mathbb{Q}$, so $\mathbb{Q}[X,Y]$ is a field extension of $\mathbb{Q}$.

So actually, I did mean field! I just confused my self by using X and Y to represent elements and not indeterminates. Sorry. Let me try again!

Let $\alpha,\beta \in \mathbb{C}$. Suppose that the field $F := \mathbb{Q}[\alpha,\beta]$

I can see that we could replace $\alpha,\beta$ with $- \alpha, - \beta$ for example and the ring we generate would be the same, However substituting $\alpha$ for $\alpha{^2}$ for example would not necessarily yield the same ring. Is there any general way of stating which combinations or operations we can perform on $\alpha$ and $\beta$ and leave the ring they generate unchanged?
• Apr 1st 2013, 07:03 PM
Gusbob
Re: Field generated by elements
Quote:

Originally Posted by Ant
I can see that we could replace $\alpha,\beta$ with $- \alpha, - \beta$ for example and the ring we generate would be the same, However substituting $\alpha$ for $\alpha{^2}$ for example would not necessarily yield the same ring. Is there any general way of stating which combinations or operations we can perform on $\alpha$ and $\beta$ and leave the ring they generate unchanged?

How familiar are you with Galois theory?
• Apr 2nd 2013, 04:03 AM
Ant
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
• Apr 2nd 2013, 05:17 AM
Gusbob
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 $\mathbb{Q}[\sqrt3]\cong\mathbb{Q}[\sqrt2]$ as $\mathbb{Q}$ 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 $\mathbb{Q}[\alpha],\mathbb{Q}[\beta]$ it is NOT true in general that $\alpha$ and $\beta$ have the same minimal polynomial. For example, $\mathbb{Q}[\sqrt2]\cong\mathbb{Q}[7000\sqrt2 + 5000000]$ 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 $\alpha$, may change the minimal polynomial of the generator ( $1+i$ has minimal polynomial $x^2-2x+2$, but $-(1+i)$ has minimal polynomial $x^2+2x+2$).

It is true, however that there is an element $\gamma$ in $\mathbb{Q}[\beta]$ such that $\mathbb{Q}[\alpha]\cong\mathbb{Q}[\gamma]$ AND $\alpha,\gamma$ 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 $m\alpha + n$ where $m,n\in \mathbb{Q}$)
• Apr 4th 2013, 06:19 AM
Ant
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 $K$ 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 $\Sigma_{f} = K[\alpha_{1},\alpha_{2},...,\alpha_{n}]$ with $\alpha_{i}$ being the distinct roots of $f$ in some extension $L/K$. We may find that we can eliminate some roots, or find a 'nicer' expression for $\Sigma_{f}$. 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 $K$-embeddings $K \to C$. For some extension $C/K$.

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

does that make sense?