PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: High Entry average rating: Very high
primitive element theorem (Theorem)
Theorem 1   Let $ F$ and $ K$ be arbitrary fields, and let $ K$ be an extension of $ F$ of finite degree. Then there exists an element $ \alpha\in K$ such that $ K=F(\alpha)$ if and only if there are finitely many fields $ L$ with $ F\subseteq L\subseteq K$.

Note that this implies that every finite separable extension is not only finitely generated, it is generated by a single element.

Let $ X$ be an indeterminate. Then $ \mathbb{Q}(X,i)$ is not generated over $ \mathbb{Q}$ by a single element (and there are infinitely many intermediate fields $ \mathbb{Q}(X,i)/L/\mathbb{Q}$). To see this, suppose it is generated by an element $ \alpha$. Then clearly $ \alpha$ must be transcendental, or it would generate an extension of finite degree. But if $ \alpha$ is transcendental, we know it is isomorphic to $ \mathbb{Q}(X)$, and this field is not isomorphic to $ \mathbb{Q}(X,i)$: for example, the polynomial $ Y^2+1$ has no roots in the first but it has two roots in the second. It is also clear that it is not sufficient for every element of $ K$ to be algebraic over $ F$: we know that the algebraic closure of $ \mathbb{Q}$ has infinite degree over $ \mathbb{Q}$, but if $ \alpha$ is algebraic over $ \mathbb{Q}$ then $ [\mathbb{Q}(\alpha):\mathbb{Q}]$ will be finite.

This theorem has the corollary:

Corollary 1   Let $ F$ be a field, and let $ [F(\beta,\gamma):F]$ be finite and separable. Then there exists $ \alpha \in F(\beta,\gamma)$ such that $ F(\beta,\gamma)=F(\alpha)$. In fact, we can always take $ \alpha$ to be an $ F$-linear combination of $ \beta$ and $ \gamma$.

To see this (in the case of characteristic 0), we need only show that there are finitely many intermediate fields. But any intermediate field is contained in the splitting field of the minimal polynomials of $ \beta$ and $ \gamma$, which is Galois with finite Galois group. The explicit form of $ \alpha$ comes from the proof of the theorem.

For more detail on this theorem and its proof see, for example, Field and Galois Theory, by Patrick Morandi (Springer Graduate Texts in Mathematics 167, 1996).



"primitive element theorem" is owned by alozano. [ full author list (3) | owner history (3) ]
(view preamble)

View style:

See Also: simple field extension, primitive element of biquadratic field

Keywords:  number theory

Attachments:
proof of primitive element theorem (Proof) by alozano
Log in to rate this entry.
(view current ratings)

Cross-references: Galois theory, proof, Galois group, minimal polynomials, splitting field, contained, characteristic, separable, algebraic closure, algebraic, sufficient, clear, roots, polynomial, isomorphic, generate, transcendental, indeterminate, generated by, finitely generated, separable extension, implies, degree, extension, fields
There are 3 references to this entry.

This is version 13 of primitive element theorem, born on 2001-10-15, modified 2007-06-13.
Object id is 214, canonical name is PrimitiveElementTheorem.
Accessed 5311 times total.

Classification:
AMS MSC12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)

Pending Errata and Addenda
None.
[ View all 6 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | prove | add result | add corollary | add example | add (any)