The argument quoted below comes from my textbook1. In the argument, $L$ is a finite extension of a field $K$.

Let $\{z_1, z_2, \dots, z_n\}$ be a basis for $L$ over $K$. Each $z_i$ is algebraic over $K$ 2, with minimum polynomial $m_i$ (say). Let $m = m_1m_2\dots m_n$ and let $N$ be a splitting field for $m$ over $L$. Then $N$ is also a splitting field for $m$ over $K$, since $L$ is generated over $K$ by some of the roots of $m$ in $N$.

I am puzzled by the last sentence ("Then $N$ is also..."). To prove that $N$ is a splitting field over $K$, one needs to show that

  1. $m$ splits completely over $N$; and
  2. $m$ does not split completely over any field $E$ such that $K \subset E \subset N$.

Assertion (1) follows from the choice of $N$.

I don't see how the observation "since $L$ is generated over $K$ by some of the roots of $m$ in $N$" proves assertion (2), or in any other way helps to prove that $N$ is a splitting field of $m$ over $K$. I would appreciate any additional detail that may help flesh out the argument.

1 Fields and Galois theory, by John M. Howie, p. 107.

2 This follows from theorem 3.12, on p. 59 of the book, and the assumption that $L$ is finite. This theorem asserts simply that "every finite extension is algebraic."

  • 1
    $\begingroup$ One usual definition of splitting field of a polynomial $f$ over a field $L$ is a field extension $K$ over which $f$ splits completely, and which is generated over $L$ by the roots of the polynomial, essentially. $\endgroup$ – Pedro Tamaroff Mar 6 '18 at 0:10
  • $\begingroup$ @PedroTamaroff: thanks; it's still fuzzy in my mind, but now at least I see the shape of the argument. $\endgroup$ – kjo Mar 6 '18 at 0:16

Let $r=\deg m$, and denote by $u_1, \dots u_r$ the roots of $m$ in the splitting field $M$ of $m$ over $L$. This means that $N=L(u_1, \dots u_r)$.

Now, by hypothesis, $L=K(z_1,\dots, z_n)=$, so $$M=K(u_{i_1},\dots , u_{i_n})(u_1, \dots u_r),$$ which of course is the same as $K(u_1, \dots u_r)$. As $m\in K[X]$, this proves $M$ is a splitting field of $m$ over $K$ as well.

  • $\begingroup$ Thank you, but it burned my eyes that the same variable ($N$) was being used for the name of a field extension and the degree of a polynomial. I went ahead and edited your post to eliminate this naming collision. I hope you don't mind. Also, the answer would be more helpful if it said something about the meaning of the indices $i_1, i_2, \dots, i_n$ in the equality $N = K(u_{i_1},\dots,u_{i_n})(u_1,\dots, u_r)$. $\endgroup$ – kjo Mar 6 '18 at 0:36
  • $\begingroup$ That's no problem. Indeed I didn't notice the collision (getting late here…). Thanks! $\endgroup$ – Bernard Mar 6 '18 at 0:44

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.