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I know that there are many proofs of this, but I am wondering if someone could help me understand what is "swept under the rug" in Dummit and Foote. I couldn't find anything about that on this website.

I am currently self-studying field theory for the first time from Dummit and Foote, and just read the proof demonstrating the existence of an algebraic closure for all fields. The procedure that it takes is proving via Zorn's lemma that for any field $F$, there exists an algebraically closed field $K$ containing $F$. Then the subfield $\overline{F} \subset K$ of elements of $K$ algebraic over $F$ is shown to be an algebraic closure for $F$. After this proof, the authors note that it is easy to see that $\overline{F}$ is unique up to isomorphism by Zorn's lemma and from results about splitting fields.

I have tried for the last hour to see why this is true, but think I am missing something. I assume the fact that splitting fields are unique up to isomorphism is used, but I can't fit it in anywhere. Could someone explain why this is the case? Thanhs!

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Let's first state clearly the question you have.

Let $F$ be a field. An "algebraic closure" of $F$ is a field extension $C/F$ such that

  • $C$ is algebraic over $F$;
  • $C$ is algebraically closed.

Claim:

If $C_1/F, C_2/F$ are two algebraic closures of $F$, then there exists an $F$-isomorphism $\iota: C_1 \simeq C_2$.

Remark: This is not a universal property, as we don't require (and it is usually impossible) that the isomorphism $\tau$ be unique.

Proof: We first use Zorn's lemma to show that there is an $F$-embedding (i.e. a homomorphism of fields which is identity on $F$) from $C_1$ to $C_2$.

Let $A$ be the set of pairs $(K, \iota)$, where:

  • $K/F$ is a subextension of $C_1/F$;
  • $\iota: K \rightarrow C_2$ is an $F$-embedding.

The set $A$ is non-empty, as it contains the element $(F, id)$ where $id$ stands for the canonical embedding of $F$ into $C_2$.

A partial order $\leq$ is defined on $A$ as follows. We say that $(K, \iota) \leq (K', \iota')$ if $K$ is a subextension of $K'$ and $\iota'|_K = \iota$.

Now if we have a chain $(K_i, \iota_i)_{i \in I}$ in $A$, we build an element $(K, \iota)$ with $K = \bigcup_{i\in I}K_i$ and $\iota: K \rightarrow C_2$ defined by $\iota(x) = \iota_i(x)$ for any $i\in I$ and any $x\in K_i$, which is well defined because of the chain condition.

This is then an upper bound of the chain.

Therefore, our set $A$ satisfies the condition in Zorn's lemma, and consequently $A$ contains at least one maximal element, say $(K, \iota)$.

Suppose that $K \neq C_1$. Then there exists $x \in C_1 \backslash K$. We let $K'$ be the field $K(x)$.

Since $C_1$ is algebraic over $F$ and hence algebraic over $K$, we may take the minimal polynomial $f$ of $x$ over $K$. Now $f$ has at least one root $y$ in $C_2$, and from results of rupture fields (see below *), we see that $\iota$ extends to a homomorphism $\iota': K' \rightarrow C_2$ such that $\iota'|_K = \iota$ and $\iota(x) = y$.

This means that we have $(K, \iota)$ is strictly smaller than $(K', \iota')$, which contradicts the maximality of $(K, \iota)$.

We have thus proved that $K = C_1$ and hence the existence of an $F$-embedding $\iota: C_1\simeq C_2$.


It only remains to see that $\iota$ is surjective.

For any $y\in C_2$, let $g\in F[X]$ be the minimal polynomial of $y$ over $F$ (or any polynomial in $F[X]$ such that $g(y) = 0$).

Since $C_1$ is algebraically closed, $g$ decomposes in $C_1$ as a product of linear factors: $g = \prod_{i = 1}^d(X - x_i)$.

Taking images under $\iota$, we get an identity $g = \prod_{i = 1}^d(X - \iota(x_i))$ in $C_2[X]$.

Putting $X = y$ gives $0 = g(y) = \prod_{i = 1}^d(y - \iota(x_i))$ as an identity in $C_2$.

But $C_2$ is a field, hence $y$ must be equal to one of the $\iota(x_i)$.

Thus we have shown that $\iota$ is surjective.


(*) In fact, we have $K' = K(x) \simeq K[X]/(f)$, so we may first define a ring homomorphism $K[X]\rightarrow C_2$ sending $X$ to $y$. Since $f(y) = 0$, the ring homomorphism passes to quotient and gives $\iota'$.
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  • $\begingroup$ thank you very much! That makes much more sense $\endgroup$ – user65557289 2 days ago

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