Why is the "root field" premise necessary in this theorem about field extensions?

Question

In Charles Pinter's "A Book of Abstract Algebra," chapter 31, Theorem 8 (p316-17), he states:

Suppose $I \subseteq E \subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $E$, then $K$ is also the root field of some polynomial over $I$.

His short proof is then:

Suppose $K$ is a root field of some polynomial over $E$, and let $K = I(a)$. If $p(x)$ is the minimum polynomial of $a$ over $I$, its coefficients are certainly in $E$ and it has a root $a$ in $K$, so by Theorem 7, all its roots are in $K$. Therefore, $K$ is the root field of $p(x)$ over $I$.

I think "certainly in $E$" is a typo that should read "certainly in $I$," but here's my question:

Why is it required as a premise that $K$ be a root field of some polynomial over $E$?

Since $K$ is a finite extension of $E$ which is a finite extension of $I$, $K$ must also be a finite extension of $I$ (by Ch29 Th2). Therefore it must also be a simple extension (by Ch31 Th2). So we can write $K=I(a)$ for some $a \in K$, and be certain a minimum polynomial of $a$ over $I$ exists (since finite and simple extensions are defined to be in terms of algebraic elements). And from here, Pinter's argument (starting at sentence 2) still seems to hold. Am I missing something?

Answer 1

After reviewing this more, I believe the actual theorem from Pinter is wrong! Counterexample: Let $I=\mathbb{Q}$ and $E=K=\mathbb{Q}(\sqrt[4]{2})$.

I found the 2nd edition of the same book and discovered the theorem is different:

Suppose $I\subseteq E\subseteq K$ where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.

This makes much more sense now.

Answer 2

This answer refers to the theorem as written in the 2nd. edition of Pinter's book:

"Root field" seems to be the same of what many of us call "splitting field": $\;K\;$ is a root field of some polynomial $\;f(x)\in I[x]\;$ means $\;K=I(a_1,...,a_n)\;,\;a_1,...,a_n\;$ all the roots of $\;f(x)\;$ (in some algebraic closure of $\;I\;$) and any subfield $\;I\le L\lneq K\;$ fulfills that there exists at least one root $\;a_n\;$ such that $\;a_n\notin L\;$.

Observe then that $\;f(x)\in I[x]\subset E[x]\;$ . Either $\;E=K\;$ and there's nothing to prove, or else $\;E\lneq K\;$ and thus there is some root of $\;f(x)\;$ whic is not in $\;E\;\implies$ $\;K\;$ is the splitting field (root filed) of $\;f(x)\;$ as polynomial in $\;E[x]\;$ as $\;K\;$ contains all the roots of $\;f\;$ ...