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The question I am posting here is not about "constructing" the algebraic closure, but it is about the failure of the natural way one tries to prove it. The proof is not so simple that an undergraduate will do it independently at first time.

Let the base field be $F$.

Way 1: Start with an irreducible polynomial $f_1$ over $F$, and construct an extension $E_1$ which contains root of $f_1$. If $E_1$ is algebraiclly closed, we are done, otherwise, if $f_2$ in $F[x]$ is irreducible over $E_1$, obtain algebraic extension $E_2$ of $E_1$ where $f_2$ has root, and continue this process. Then is it not true that $\cup_i E_i$ is an algebraic closure of $F$?

Way 2: (Exercise from book Algebra-by L. Grove) Let $\mathfrak{L}$ be the set of all algebraic extensions of $F$. Define partial ordering on $\mathfrak{L}$ by subset-relation. Applyting Zorn's lemma, we find a maximal element in $\mathfrak{L}$ which turns out to be an algebraic closure of $F$.

However, in Way 2, the Grove says, that this argument is wrong. He says (which I don't understood): suppose , $|\mathfrak{L}|=\alpha$. For each subset of $\mathfrak{L}$, construct an algebraic closure of $F$, and arrive at a contradiction. Thus, the statement let $\mathfrak{L}$ be the set of all algebraic extensions of $F$ in way 2 doesn't make sense. (I didn't arrive at contradiction, can you help me?)

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    $\begingroup$ I don't understand what you mean in Way 2, but Way 1 will work as long as you remember to use all irreducible polynomials (there might be more than countably many of these, so it's not true that you can always arrange the $E_i$ to be indexed by the natural numbers). $\endgroup$ – Qiaochu Yuan Oct 14 '15 at 5:27
  • $\begingroup$ What is your question? $\endgroup$ – Eric Wofsey Oct 14 '15 at 5:37
  • $\begingroup$ @Yuan, Wofsey: sorry. I have done addition of some statements which would make Way 2 clear. $\endgroup$ – Groups Oct 14 '15 at 6:56
  • $\begingroup$ The issue being raised is that there is no set of all algebraic extensions. But that is not a huge problem. $\endgroup$ – Zhen Lin Oct 14 '15 at 7:08
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The issue with the first way is first that you have to check that if every polynomial over $F$ has a root in $E$, then every polynomial over $F$ splits in $E$. Then this procedure only works if your field is countable. If it's uncountable, you will need some transfinite recursion of some kind, so you need to well-order the set of irreducible polynomials over $F$ and you need the Axiom of Choice for that.

The issue with the 2nd way is that if $\mathfrak L$ is a set then so is the reunion of the underlying sets of the algebraic extensions. But this would be the set of all sets :

Let $x \notin F$ be a set. Then we have an obvious bijection $f : F \to F_x = (F \setminus \{0_F\}) \cup \{x\}$. We can give a field structure to $F_x$ so that $f$ is a field isomorphism. Thus $f: F \to F_x$ is an algebraic field extension of $F$. When you perform the reunion of all the underlying sets of all the algebraic extensions, $x$ will be in it. Obviously, $F$ will also be included, so you really get the set of all sets.
Resolving this problem is easy, you only need to restrict the possible underlying set to a fixed set of large enough cardinality. In fact you can use $F$ itself unless $F$ is finite, then you can use $\Bbb N$. Once you fix an underlying set, the possible field structures you can put on it form a set, and the possible algebraic extensions with it are again a set.

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