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When proving that every field has an algebraic closure, you have to be careful. In this article https://proofwiki.org/wiki/Field_has_Algebraic_Closure, and as I have been told on this site, if we have a field F. The "collection of all algebraic extensions of F" is not a set.

Is there a simple way to explain why this is not a set, and we can not apply zorns lemma on it? Or do you need a lot of reading in deep set-theory and logic to understand this? I have seen the russel paradox, but that is basically how much I know about this.

What also is very confusing is that in real analysis we have that "the space of continuous functions on [0,1] is a vector space". So there is a set of continuous functions? This doesn't sound any more mysterious than "all algebraic extensions of a given field F", however one of them gives rise to a set, and one doesn't?

Martin Sleziak
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user119615
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3 Answers3

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Here's a reformulation of the argument that avoids the notion of cardinality. Suppose $F$ is not algebraically closed, let $K$ be some algebraic extension of it, and let $a$ be any element of $K$ that isn't in $F$. Then, for any entity $x\notin K$, we can produce another algebraic extension of $F$, isomorphic to $K$, by replacing the element $a$ in $K$ with $x$. So every $x\notin K$ is in an algebraic extension of $F$, and so is every $x\in K$ (because $K$ itself is an algebraic extension of $F$). If there were a set of all algebraic extensions of $F$, then the union of this set of fields would be a set (by the axiom of union) and yet would contain everything. That contradicts the theorem (of standard set theory) that no set can contain everything.

Andreas Blass
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  • I never saw this type of thing argued in this form, which is really neat. – quid Apr 30 '15 at 23:06
  • So what you have done here is that you can let the x vary over every set that exists. And then there is an axiom that says that since the elements of "the set of all algebraic extensions" is itself a set of sets, you can take the union of this(axiom of union you call it). And then suddenly you have a set that contain everything, and hence you get Russels paradox? – user119615 Apr 30 '15 at 23:09
  • And if you have the time, can you please answer this?: If we don't take the set of all algebraic extensions, but "the set of all algebraic extensions, up to isomorphisms", that is, if two algebraic extensions are isomorphic, only one of them is in the set/collection, is this then a set? – user119615 Apr 30 '15 at 23:17
  • @user119615 Your first comment is a good summary of the argument. For your second comment, the answer is yes, you can have a set that contains isomorphic copies of all algebraic extensions of $F$. Let $K$ be any algebraic closure of $F$, and form the set $S$ of all subfields of $K$. Every algebraic extension of $F$ can be embedded in $K$, so it has an isomorphic copy in $S$. – Andreas Blass Apr 30 '15 at 23:24
  • Thank you so much. The thing is that I wanted to use this set to simplify the argument of proving that every field has an algebraic closure. But you argue that this set exists because you have an algebraic closure of F, but if I am trying to prove that every field has an algebraic closure, I can not use this fact. Can we only use that the collection S is a set after we have proved that the set K exists, and hence not be used in the proof of "all fields have an algebraic closure"(by making chains in S, and then using zorns lemma)? – user119615 Apr 30 '15 at 23:32
  • @user119615 There are alternative ways to get a set $S$ containing isomorphic copies of all algebraic extensions of $F$, without presupposing that $F$ has an algebraic closure. As far as I can see, they depend on the notion of cardinality. First prove (by counting polynomials and roots thereof) that there is a cardinal $M$ such that all algebraic extensions of $F$ have cardinality $\leq M$. Then fix some set $X$ of cardinality $M$ and let $S$ be the set of all fields whose underlying set is a subset of $X$. – Andreas Blass Apr 30 '15 at 23:39
  • Thank you very much, what you are describing seems very much like what how this proof was constructed, and what I tried to escape from: http://math.stackexchange.com/questions/1257985/using-zorns-lemma-to-show-that-every-field-has-an-algebraic-closure/ , it seems you have to do something like this no matter what. My last question for you is if this stuff about cardinality and if a collection is a set, is that studied in very advanced courses, or do you have a reccomandation about an ok book I can read so I can learn about this about cardinality and when collections are sets? – user119615 Apr 30 '15 at 23:46
  • @user119615 There are several good books for the level of set theory that's involved here. The first two that come to mind are Enderton's "Elements of Set Theory" and Halmos's "Naive Set Theory". – Andreas Blass May 01 '15 at 00:06
  • I'll check them out, thanks. – user119615 May 01 '15 at 00:08
  • On first reading I thought you meant you could union the fields together, but actually you mean only to union the underlying sets together. Do you think you should make that clearer in your answer? – user21820 May 01 '15 at 13:53
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If an algebraic closure $K$ of $F$ has cardinality $\kappa$, then every set of cardinality $\kappa-|F|$ can be unioned with $F$ to give an algebraic closure, by just defining the addition and multiplication by transport via the bijection with $K$. Thus, the class of sets of cardinality $\kappa-|F|$ (which we'll assume is nonzero, i.e., $F$ is not closed already) injects into the class of algebraic closures of $F$. But the class of all sets of a given nonzero cardinality is not a set, and so the class of algebraic closures is not a set either.

Continuous functions $[0, 1]\to\mathbb R$ form a set because $[0, 1]\times\mathbb R$ is a set, and the set of continuous functions $[0, 1]\to\mathbb R$ form a subset of the power set of $[0, 1]\times\mathbb R$.

Nishant
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  • Thank you for your help, I don't know enough about cardinality to completely understand this, but I got a good feeling for why we have a set of functions. – user119615 May 01 '15 at 00:14
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Suppose you have a proper algebraic extension $K$ of $F$ (so $F$ is not algebraically closed).

If $X$ is any set disjoint from $F$ with $|X|=|K\setminus F|$, and $f_X\colon K\setminus F\to X$ is a bijection, then we can build a bijection $g_X\colon K\to F\cup X$ so that $$ g_X(a)=\begin{cases} a & \text{if $a\in F$}\\ f_X(a) & \text{if $a\in K\setminus F$} \end{cases} $$ and transport the field structure from $K$ to $F\cup X$ so that $g_X$ is a field isomorphism. Thus $F\cup X$ is an algebraic extension of $F$.

Since the class of sets equipotent to $K\setminus F$ and disjoint from $F$ is not a set, we're done.

The situation is completely different in the case of the continuous functions from $[0,1]$ to $\mathbb{R}$. But there is no “set” of all continuous functions to $\mathbb{R}$ with an arbitrary metric space as domain.

egreg
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  • Thank you for your help egreg, and also for your help in my other thread. I don't have the required knowledge to completely understand what you write, but I will come back and try to read your answers fully when I get that knowledge. – user119615 May 01 '15 at 00:13