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.

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

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 $K(\alpha_1) = K(\alpha_1,\ldots,\alpha_r)$.


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 $N=L$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.