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quadratic extension (Definition)

Let $ k$ be a field and $ K$ be its algebraic closure. Suppose that $ k\neq K$. A quadratic extension $ E$ over $ k$ is a field $ k< E\leq K$ such that $ E=k(\alpha)$ for some $ \alpha\in K-k$, where $ \alpha^2\in k$.

If $ a=\alpha^2$, we often write $ E=k(\sqrt{a})$. Every element of $ E$ can be written as $ r+s\sqrt{a}$, for some $ r,s\in k$. This representation is unique and we see that $ \lbrace 1,\sqrt{a}\rbrace$ is a basis for the vector space $ E$ over $ k$. In fact, we have the following

Proposition. If the characteristic of $ k$ is not $ 2$, then $ E$ is a quadratic extension over $ k$ iff $ \operatorname{dim}(E)=2$ (as a vector space) over $ k$.

Proof. One direction is clear from the above discussion. So suppose $ \operatorname{dim}(E)=2$ over $ k$ and $ \lbrace 1,\beta \rbrace$ is a basis for $ E$ over $ k$. Then $ \beta^2=r+s\beta$ for some $ r,s\in k$. Set $ \alpha=\beta-\frac{s}{2}$. Then clearly $ \alpha\in E-k$ and $ \lbrace 1,\alpha\rbrace$ is also a basis for $ E$ over $ k$. Furthermore, $ \alpha^2= r+\frac{s^2}{4}\in k$. Thus, $ k(\alpha)$ is quadratic extension over $ k$ and $ [k(\alpha):k]=2$. But $ k(\alpha)$ is a subfield of $ E$. Then $ 2=[E:k]= [E:k(\alpha)][k(\alpha):k]=2[E:k(\alpha)]$ implies that $ [E:k(\alpha)]=1$ and $ E=k(\alpha)$. $ \qedsymbol$

In the proposition above, the assumption that $ \operatorname{Char}(k)\ne 2$ can not be dropped. If fact, quadratic extensions of $ \mathbb{Z}_2$ do not exist, for if $ \alpha^2\in \mathbb{Z}_2$, then $ \alpha\in \mathbb{Z}_2$.

For the rest of the discussion, we assume that $ \operatorname{Char}(k)\ne 2$.

Pick any element $ \beta=r+s\sqrt{a}$ in $ E-k$. Then $ s\neq 0$ and $ (\beta - r)^2=s^2a\in k$. So $ \beta$ is a root of the irreducible polynomial $ m(x)=x^2-2rx+(r^2-s^2a)$ in $ k[x]$. If we define $ \overline{\beta}$ to be $ r-s\sqrt{a}$, then $ \overline{\beta}$ is the other root of $ m(x)$, clearly also in $ E-k$. This implies that the minimal polynomial of every element in $ E$ has degree at most 2, and splits into linear factors in $ E[x]$.

Since $ \operatorname{Char}(k)\ne 2$, $ \beta\neq\overline{\beta}$ are two distinct roots of $ m(x)$. This shows that $ k(\sqrt{a})$ is separable over $ k$.

Now, let $ f(x)$ be any irreducible polynomial over $ k$ which has a root $ \beta$ in $ E$. Then the minimal polynomial $ m(x)$ of $ \beta$ in $ k[x]$ must divide $ f$. But because $ f$ is irreducible, $ m=f$. This shows that $ k(\sqrt{a})$ is normal over $ k$. Since $ k(\sqrt{a})$ is both separable and normal over $ k$, it is a Galois extension over $ k$.

Let $ \phi$ be an automorphism of $ E=k(\sqrt{a})$ fixing $ k$. Then $ \phi(\sqrt{a})$ is easily seen to be a root of the minimal polynomial of $ \sqrt{a}$. As a result, either $ \phi=1$ on $ E$ or $ \phi$ is the involution that maps each $ \beta$ to $ \overline{\beta}$. We have just proved

Theorem. Suppose $ \operatorname{Char}(k)\neq 2$. Any quadratic extension of $ k$ is Galois over $ k$, whose Galois group is isomorphic to $ \mathbb{Z}/2\mathbb{Z}$.

Remark. A quadratic extension (of a field) is also known in the literature as a $ 2$-extension, a special case of a p-extension, when $ p=2$.



"quadratic extension" is owned by CWoo. [ full author list (2) ]
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See Also: $p$-extension

Other names:  $2$-extension
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Cross-references: p-extension, isomorphic, Galois group, maps, involution, automorphism, Galois extension, normal, irreducible, divide, separable, factors, degree, minimal polynomial, irreducible polynomial, root, implies, subfield, clear, iff, characteristic, proposition, vector space, basis, representation, algebraic closure, field
There are 15 references to this entry.

This is version 17 of quadratic extension, born on 2006-02-26, modified 2008-04-04.
Object id is 7656, canonical name is QuadraticExtension.
Accessed 2956 times total.

Classification:
AMS MSC: 12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)
 12F10 (Field theory and polynomials :: Field extensions :: Separable extensions, Galois theory)
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