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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."

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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