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In my book they show that if $K \subset L$ is a finite normal extension, then $L$ is the splitting field for some polynomial $f(X) \in K[X]$.

They do so as follows:

Suppose $a_1, ... ,a_n$ is a basis for $L$ as vector space over $K$, hence $L = K(a_1,\ldots,a_n)$. Now let $f_i$ be the minimal polynomial of $a_i$. Since $a_i$ is a root of $f_i$ and since $f_i$ is irreducible, $f_i$ splits completely over $L$, hence $f = f_1\cdots f_n$ also splits completely over $L$. Thus $L$ is the splitting field of $f(X)$.

Now my question. My definition in my book says that $L$ is a splitting field of $f(X)$ over $K$, if

  • $f(x) = a(X-\lambda_1)^{m_1}\ldots(X-\lambda_q)^{m_q}$ where $a \in K^*, m_i \in \mathbb{N}$
  • $L = K(\lambda_1,\ldots,\lambda_q)$

Now in the proof when $f(X)$ splits into linear factors in $L[x]$ it could have more roots than just $a_1,\ldots,a_n$, hence according to the definition the splitting field would equal to $K(a_1,\ldots,a_n,\lambda_1,\ldots,\lambda_p)$, where $\lambda_1,\ldots,\lambda_p$ are the remaining roots of $f$. Now, I wonder whether my reasoning is correct:

$$L = K(a_1,\ldots,a_n) \subseteq K(a_1,\ldots,a_n,\lambda_1,\ldots,\lambda_p)\subseteq L$$

hence $L$ is the splitting field.

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@DylanMoreland, fixed was a typo. –  sxd Feb 16 '12 at 0:15
    
You're overlooking that knowledge of the dimension of $L$ as a $K$-vector space tells you something about the degree of $f$. –  Hurkyl Feb 16 '12 at 0:24
    
@Hurkyl, oh yes indeed the dimension of $L$ as $K$-vector space is exactly the degree of $f$. Thanks! –  sxd Feb 16 '12 at 0:32
    
Almost -- the degree of $K(a_1)$ over $K$ (which divides the degree of $L$ over $K$) is the degree of $f_1$. –  Hurkyl Feb 16 '12 at 1:01

2 Answers 2

up vote 2 down vote accepted

The proof shows precisely that $K(a_1,\ldots,a_n)=K(a_1,\ldots,a_n,\lambda_1,\ldots,\lambda_p)$, which follows from the assumption that $L/K$ is normal. If you read the proof carefully, that's exactly what it says: since $L/K$ is normal, and one root of $f_i$ is in $L$, they are all in $L$.

Maybe, it's easier to parse if you assume that $L=K(\alpha_1)$. Then, if $f$ is the minimal polynomial of $\alpha_1$ and if $\alpha_2,\ldots,\alpha_r$ are the remaining roots of $f$, then $L$ being normal implies that $K(\alpha_1) = K(\alpha_1,\ldots,\alpha_r)$.

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Do you mean my addition at the end or directly from their proof? –  sxd Feb 16 '12 at 0:14
    
Directly from the proof, or rather from the definition of normality: $L/K$ being normal means by definition that if the field is generated by a root of an irreducible polynomial, then all the other roots are in that field, too. –  Alex B. Feb 16 '12 at 0:18
    
So my argument makes no sense then? –  sxd Feb 16 '12 at 0:19
    
@Dimitri I don't know. What is your argument, beyond using the definition of normality? –  Alex B. Feb 16 '12 at 0:20
    
Never mind, I know what you mean now. Thanks for the help! –  sxd Feb 16 '12 at 0:21

By definition of splitting field of $f(x)$, we say $N$, we have that $L$ is containing to $N$. On the onther hand, but by the way that $f(x)$ is writing down, is obvious that $a_1,a_2,...,a_n$ are in $L$ and $k$ too. So by definition of $L$, $N$ is containing to $L$. That is $N=L$

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