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Two similar questions were asked here and here; but I do not find the answers satisfactory.

In Ian Stewart's Galois Theory:

Theorem 5.3. The set of rational expressions $K(t)$ is a simple transcendental extension of the subfield K of C.

Proof. Clearly $K(t):K$ is a simple extension, generated by $t$. If $p$ is a polynomial over $K$ such that $p(t)=0$ then $p=0$ by definition of $K(t)$, so the extension is transcendental.

My question: why is $K(t):K$ a field extension? In the text, the definition of field extension is "a monomorphism $\iota: K \to L$, where $K$ and $L$ are subfields of $\mathbb{C}$". Now $t$ is a variable, hence $K(t)$ fails to even be a subset of $\mathbb{C}$ (just like $\{f(x)| f:\mathbb{R}\to\mathbb{R}\}$ is not a subset of $\mathbb{R}$), let along a subfield. What am I missing?

Please help if you can-many thanks in advance!

Dick Grayson
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    Huh, does he really define it that way (requiring subfields of C)? You might check if the publisher has errata. The 1994 edition defines a field extension to simply be a monomorphism of fields. – leslie townes Feb 04 '22 at 05:36
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    In the early editions, Stewart does everything in full generality first; in the later editions, he does everything in the complex numbers first, then goes back and does arbitrary fields. It's a field extension because there's a monomorphism from $K$ to $K(t)$. – Gerry Myerson Feb 04 '22 at 06:11
  • I see. Thank you both very much! – Dick Grayson Feb 04 '22 at 16:30

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