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I'm reading J.J. Rotman's Advanced Mondern Algebra.

Here's the sketch of the proof in this book that every field k has an algebraic closure.

Making a variable $t_f$ for each non-constant polynomial $f(x) \in k[x]$, all these variables together, call it set $S$, adjoin $S$ with $k$, we get ring $k[S]$.

In $k[S]$, consider the ideal generated by all $f(t_f)$. Using Zorn's lemma to show there's a maximal ideal M containing it. so we get field $K = k[S]/M$.

The book then shows that every $f(x) \in k[x]$ splits in K by saying "$t_f + M$ is a root of $f(x)$, so by induction on degree , $f(x)$ splits over $K$"

What I can't understand is how the induction works? This construction only guarantees one root, the other factor is a polynomial in K instead of k, on which induction hypothesis can't apply.

Martin Sleziak
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Chen
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    Your sketch does not seem to be an accurate representation of the proof in Rotman's book. He does not claim that $K$ is algebraically closed. Instead, he defines $k = k_0$, $K = k_1$, and then builds a chain of fields, where $k_{i+1}$ is constructed to extend $k_i$ just like $K$ was constructed to extend $k$. Then he argues that the union $E = \bigcup_{n\in \mathbb{N}} k_n$ is algebraically closed. See the paragraph at the top of p. 329. I'm looking at the second edition of the book. – Alex Kruckman Jan 13 '20 at 14:00
  • Thank you. I'm reading first edition, it's on page 355, without mentioning this fields tower. but knowing this now I think I can work out the proof. – Chen Jan 13 '20 at 14:05
  • Ah, I see. In light of that, I've turned my comment into an answer, with a bit more information. – Alex Kruckman Jan 13 '20 at 14:27

2 Answers2

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This seems to be an error in the first edition of Rotman's book, which was corrected in the second edition.

In the second edition (see the paragraph at the top of p. 329), Rotman does not claim that $K$ is algebraically closed. Instead, he defines $k = k_0$, $K = k_1$, and then builds a chain of fields, where $k_{i+1}$ is constructed to extend $k_i$ just like $K$ was constructed to extend $k$. Then he argues that the union $E = \bigcup_{n\in \mathbb{N}} k_n$ is algebraically closed. This is a very common argument for the existence of the algebraic closure.

It turns out (due to a theorem of Gilmer) that $K = k_1$ actually is already algebraically closed, so the above chain construction is not necessary. But proving this less trivial than the wrong proof in Rotman's first edition. See here for references.

Alex Kruckman
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The induction is irrelevant: $K$ is already algebraically closed. This is a theorem of Robert Gilmer. See here or here. For a similar construction where it is less technical to see you get an algebraic closure after one step, see here.

KCd
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