Let L/K be an algebraic extension (i.e. L is an algebraic extension of K), such that
$L\subset \overline{K}$ (i.e. L is contained in an algebraic closure of K).
Then L/K is called a normal extension if any one of the following holds:
- Every K-embedding of L in ${\overline {K}}$ induces an automorphism of L.
- L is the splitting field of a family of polynomials in $K[X]$
- Every irreducible polynomial of $K[X]$ which has a root in L splits into linear factors in L.
Questions:
Why the definition assumes $L\subset \overline{K}$? Can we drop the assumption?
It is my understanding that, for field extension L/K, sometime it can be viewed as an
embedding of K in L, that is the injective homomorphism of K in L, and K is not necesarily a subset of L. K can be identified as the image of embedding in L.
I dont understand why the definitions require L to be a subset $\overline{K}$ and not identified as the image of embedding in $\overline{K}$. I know that the algebraic closure of field K is unique up to isomorphism.
It may be useful to see that, in this textbook, John B. Fraleigh, Victor J. Katz – A first course in abstract algebra, it always assumes all algebraic extensions and all elements algebraic over a field F are contained in one fixed algebraic closure $\overline{F}$ of F.
I would be very appreciative if someone could explain the concept behind, or provide any reference for me, thank you.
Best Answer
It is not necessary to assume $L \subset \overline{K}$, and I would not make such an assumption. Instead, I would replace the first of those three conditions with "Every $K$-embedding of $L$ into $\overline{K}$ has the same image". I think that explains better what the idea is: even if $L$ has multiple $K$-embeddings into $\overline{K}$, the embeddings all have the same image field.
Example. Consider $\mathbf Q(\alpha)/\mathbf Q$ where $\alpha^2 = 2$. There are two $\mathbf Q$-embeddings of $\mathbf Q(\alpha)$ into $\overline{\mathbf Q}$, and both have image $\mathbf Q(\sqrt{2})$, since $\mathbf Q(\sqrt{2}) = \mathbf Q(-\sqrt{2})$.