# Generation of isomorphic fields by separate algebraic elements

• Simfish
In summary: If any of the following things differ then the extensions will not be isomorphic.- degrees of the extensions are different (same as the degrees of the minimal polynomials of a and b).- Galois groups are different.- For extensions of Q, algebraic number theory gives other invariants such as class group, the group of units (a finite group), the Zeta function, etc, although this is getting quite advanced.
Simfish
Gold Member
So this sentence kind of confuses me:

"On the other hand, if $$\alpha$$ is a root of $$x^3 - x + 1$$, then $$\beta = \alpha^2$$ is a root of $$x^3 - 2x^2 + x - 1$$. The two fields Q(\alpha) and Q(\beta) are actually equal though if we were presented only with one of the two polynomials, it might take us some time to notice how they are related."

Okay but how _are_ those two fields actually equal?

Degree argument. Certainly $\beta = \alpha^2 \in \mathbb{Q}(\alpha)$, and
$$[\mathbb{Q}(\alpha) : \mathbb{Q}(\beta)] = \frac{[\mathbb{Q}(\alpha) : \mathbb{Q}]}{[\mathbb{Q}(\beta) : \mathbb{Q}]} = 1.$$

Also, using $\alpha\beta - \alpha+1=0$ gives $\alpha=1/(1-\beta)\in \mathbb{Q}(\beta)$ and $\beta=\alpha^2\in\mathbb{Q}(\alpha)$.

did you want a proof or an explanation why it can happen?

Philosophically, in general, for a finite degree field extension of Q, almost any element not in Q generates it, while almost all of these generators have different irreducible polynomials.

Yeah, so from Artin 2.9 (chap. 13) it says that if \alpha and \beta are algebraic elements of two extension fields of F, there is an isomorphism of fields F(\alpha) -> F(\beta) which sends \alpha -> \beta iff the irreducible polynomials for \alpha and \beta over F are equal. This seems to imply that different fields are uniquely determined by irreducible polynomials. But the two polynomials above have no common factors...

If alpha and beta have different minimal polynomials, there can still be an isomorphism F(alpha)->F(beta) where alpha doesn't go to beta. E.g. when alpha->1/(1-beta) as in the original example.

Okay I see. So do field isomorphisms occur iff the degrees of the fields are equal? (or are there additional conditions you need to satisfy?) (especially if you try to map alpha into a potentially infinite number of betas). where did the \alpha*\beta - \alpha + 1 example come from??

The degrees of the extensions $\mathbb{Q}\mapsto\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}\mapsto\mathbb{Q}(\sqrt{3})$ are the same, but they are not isomorphic (there is no solution to x^2=3 inside $\mathbb{Q}(\sqrt{2})$).
If their degrees are the same *and* one extension is contained inside the other, then you can conclude that they are the same extension.

I used $\beta=\alpha^2$ and $\alpha^3-\alpha+1=0$ above to get
$$0=\alpha\alpha^2-\alpha+1=\alpha\beta-\alpha+1$$
and then write alpha in terms of beta.

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Er...aren't those fields are isomorphic? Take a map that's identity on the rationals, and root 2 mapped to root 3, right?

I mean, strictly speaking I don't think that the degree being the same is enough to guarantee an isomorphism (if I had to guess, I'd say the automorphism groups of the extension over the base field would have to be isomorphic, but I don't know), but I don't think this is a counterexample to the claim.

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Mystic998 said:
Er...aren't those fields are isomorphic? Take a map that's identity on the rationals, and root 2 mapped to root 3, right?

No, they're not. If a^2=2 and b^2=3 then you can't map a to b by an isomorphism. If f(a)=b then f(a)^2=b^2=3 and f(a^2)=f(2)=2, so f(a)^2 != f(a^2).
A necessary condition for mapping a to b is that they have the same minimal polynomial.

Ah, of course. I must have been thinking of something else. I've got to stop posting when I haven't slept.

oh, ok, no problem :)

A necessary condition for mapping a to b is that they have the same minimal polynomial.

OH I see. So what you did is set up a relation relating alpha and beta in such a way that you used the relation to establish the minimal polynomial (in alpha) of the same form as beta. Is there an easy way to test when such a move IS NOT possible?

Simfish said:
Is there an easy way to test when such a move IS NOT possible?

If any of the following things differ then the extensions will not be isomorphic.
- degrees of the extensions are different (same as the degrees of the minimal polynomials of a and b).
- Galois groups are different.
- For extensions of Q, algebraic number theory gives other invariants such as class group, the group of units (a finite group), the Zeta function, etc, although this is getting quite advanced.

Also, there's the following:
If f and g are minimal polynomials of a and b, you can check to see if f(x)=0 has a solution in Q(b). If it doesn't, then the extensions are not isomorphic. If it does, then there is a homomorphism Q(a)->Q(b) and it is an isomorphism if the degrees of f and g are the same. I'm not sure if there's a simple way to check this though.

## 1. What is the meaning of "isomorphic fields"?

Isomorphic fields refer to fields that are structurally identical. In other words, they have the same algebraic properties and can be transformed into one another through a bijective mapping.

## 2. How are isomorphic fields generated by separate algebraic elements?

Isomorphic fields are generated by separate algebraic elements through the use of isomorphisms. Isomorphisms are bijective mappings that preserve the algebraic properties of the elements, allowing for the generation of isomorphic fields.

## 3. What are separate algebraic elements?

Separate algebraic elements are distinct elements that can be used to generate isomorphic fields. They can be any algebraic structure, such as numbers, polynomials, or matrices.

## 4. What are some applications of generating isomorphic fields by separate algebraic elements?

Isomorphic fields have various applications in mathematics and computer science, such as in the study of group theory, cryptography, and coding theory. They also have practical applications in fields such as data encryption and error correction.

## 5. Are there any limitations to the generation of isomorphic fields by separate algebraic elements?

While isomorphic fields can be generated by separate algebraic elements, there are limitations to this process. For instance, not all algebraic structures have isomorphic fields, and the generation of isomorphic fields may not always be efficient or practical.

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