4
$\begingroup$

Consider following two statements from same book of Cohn: Basic Algebra.

Proposition 7.3.2: If $k$ is any field and $\mathcal{F}$ is a set of polynomials over $k$, then any two minimal splitting fields for $\mathcal{F}$ over $k$ are isomorphic.

Theorem11.8.3: If $k$ is any field then $k$ has algebraic closure and any two algebraic closures are isomorphic.

My question is, if we take $\mathcal{F}$ to be set of all the monic irreducible polynomials over $k$, then by 7.3.2, doesn't it imply that any two algebraic closures are isomorphic?

Further, just after Theorem 7.3.2, the author gives also the existence of minimal splitting field for $\mathcal{F}$ (it is not visible in link). Then doesn't it imply that algebraic closure of any field exists and is unique up to isomorphism?

In other words, I was thinking that Theorem 7.3.2 and existence of minimal splitting field actually gives proof of theorem 11.8.3 above. Why it was differently stated and proved I didn't understand. Are these two theorems really different?

$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

An algebraic closure must be a splitting field of all polynomials over itself (coefficients from the algebraic closure). So existence and uniqueness of a splitting field of all polynomials over a field $K$ does not trivially imply existence and uniqueness of an algebraic closure of $K$. However, it is not hard to prove that an algebraic extension of an algebraic extension is also an algebraic extension, which indeed bridges the gap, and hence the minimal splitting field of all polynomials over $K$ is indeed the same as the algebraic closure of $K$ (and there is no need to restrict to monic irreducible polynomials).

$\endgroup$
9
  • $\begingroup$ now I got clarification (by both the answers). Thank you very much user21820. $\endgroup$
    – p Groups
    Jun 25, 2016 at 4:16
  • 1
    $\begingroup$ @pGroups: As I said in my comment on the other answer, it has subtle flaws. It implicitly suggests (by comparison with algebraic closure) that constructing the splitting field of a set of polynomials does not need AC, which is false in general. It is true if the set is well-orderable, since we can then iterate through and extend the current field to split each of them according to how they factorize over the current field. This does not need AC if the base field is countable, but needs AC in general. $\endgroup$
    – user21820
    Jun 25, 2016 at 4:22
  • 1
    $\begingroup$ @pGroups: Also, there is a simple proof via Zorn's lemma. Consider the set of all fields $M$ that extend $K$ and have elements from the set $S = K \cup \{ \langle f,k \rangle : \text{$f$ is a polynomial over $K$ and $k \in \mathbb{N}$} \}$ such that $f(\langle f,k \rangle) = 0$ in $M$ for any polynomial $f$ over $K$. This set exists and is non-empty since $K$ is in it. The partial order on this set ordered by inclusion satisfies the property that every non-empty chain has an upper bound. Thus it has a maximal element $C$. [continued] $\endgroup$
    – user21820
    Jun 25, 2016 at 4:33
  • 1
    $\begingroup$ [continued] Every polynomial $f$ over $K$ has at most $deg(f)$ roots in $C$, so it must split completely over $C$ otherwise we could extend $C$ by adding a suitable element from $S$, namely some $\langle f, k \rangle$ that isn't already in $C$. Hence $C$ is a splitting field of all polynomials over $K$, and by the last remark in my answer $C$ is algebraically closed. Done. $\endgroup$
    – user21820
    Jun 25, 2016 at 4:35
  • $\begingroup$ sorry, still I have a little doubt, posted as a question in comment to other answer. (I don't know whether I should post it as a different question or not.) For example, can we look any difference the following two extensions of rational field $Q$? Let $\mathcal{F}$ be set of all the irreducible polynomials over $Q$. Then what is difference between algebraic closure of $Q$ and minimal splitting field of $\mathcal{F}$ over $Q$? (We can look them inside $C$, complex field, and give some element in one of these two extensions which is not in other.) $\endgroup$
    – p Groups
    Jun 25, 2016 at 9:39
3
$\begingroup$

When you take the splitting field of all irreducible polynomials in a given field $K$ then all those polynomials split in the bigger field, but there is no guarantee that all polynomials in the bigger field split since there are many more new polynomials now. The difficulty with proving an algebraic closure exists is precisely this: it is easy to add the roots of 0any given polynomial, but then you must consider many new polynomials. This is why a simple Zorn's lemma does not work, and one must use the axiom of choice more carefully. In contrast, constructing the splitting field of a polynomial is easy, and does not require choice.

$\endgroup$
8
  • 1
    $\begingroup$ He said "splitting field of a set of polynomials", which does require AC. Also, a simple Zorn's lemma does work if you do it carefully. $\endgroup$
    – user21820
    Jun 25, 2016 at 3:51
  • 1
    $\begingroup$ @user21820 I believe what you comment here is precisely what I wrote in my answer. We even both use the word 'carefully'. Your emphasizing certain words comes over as a bit aggressive, which is strange as we seem to agree. $\endgroup$
    – Ittay Weiss
    Jun 25, 2016 at 4:48
  • $\begingroup$ @pGroups you are most welcome. $\endgroup$
    – Ittay Weiss
    Jun 25, 2016 at 4:49
  • $\begingroup$ I did not intend any agression, so sorry if it came across like that. But I honestly don't think we agree (yet). Your first sentence is correct. Your second sentence is not quite right, since the difficulty is not in the new polynomials (that doesn't have anything to do with AC), but in the fact that you must not do the polynomials one by one since you must only add roots that are not already there in the current field. Your third sentence seems to imply that Zorn's lemma does not work and one needs to go back to using AC directly (or via transfinite induction). $\endgroup$
    – user21820
    Jun 25, 2016 at 5:13
  • 1
    $\begingroup$ @IttayWeiss: Yes I think we would agree on the mathematics, if we write down everything precisely, and it is perhaps in the choice of words that we differ. (Anyway I made a typographical error in my comment and said "must not do ... one by one"; the "not" should not have been there.) $\endgroup$
    – user21820
    Jun 25, 2016 at 9:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .