# uniqueness of algebraic closure

I denote the field of algebraic elements over a field $K$ by $\overline{K}$. An algebraic closure of a field $K$ is an algebraic extension of $K$ that is algebraically closed. I want to prove algebraic closure of a field is unique up to isomorphism.

My attempt: If $L$ is an algebraic extension of $\overline{K}$ then it is algebraic extension of $K$. So by definition of $\overline{K}$, $L$ is a subfield $\overline{K}$, which implies that $L=\overline{K}$. This means $\overline{K}$ has no nontrivial algebraic extension. Hence $\overline{K}$ is an algebraic extension of $K$ that is algebraically closed, thus, $\overline{K}$ is an algebraic closure of $K$.

I conclude that any algebraic closure of a field $K$ is isomorphic to our specific field $\overline{K}$ which is the field of algebraic elements over $K$. So all closures are isomorphic.

Is there any mistake at my attempt? Any comment and suggestion will be appreciated.

• Actually, how do you define $\overline K$? Nevertheless, you cannot conclude that $L$ is a subfield of $\overline K$ Dec 30, 2017 at 10:52
• Ok, let us define $\overline{K}$ as the field generated by all roots of polynomials over $K$. Since these generators are algebraic over $K$, it follows that every element of $\overline{K}$ is algebraic over $K$. So $\overline{K}$ is an algebraic extension of $K$. If $\overline{K}$ has an algebraic extension $L$ then $L$ is an algebraic extension of $K$. So every element of $L$ would be a root of a polynomial over $K$. Then $L$ is a subfield of $\overline{K}$. Is this imply $L=\overline{K}$ or $L\cong \overline{K}$? Dec 30, 2017 at 14:49
• @ersin, the field generated where? Dec 30, 2017 at 17:24

You are too generous with equality and you start by assuming a not yet specifically defined $\overline K$.

Lemma. Let $K$ be a field, $L/K$ algebraic (but not necessarily algebraically closed), $M/K$ algebraically closed (but not necessarily algebraic). Then there exists at least one field homomorphism $f\colon L\to M$.

Proof sketch. Consider the set of all field homomorphisms $\phi\colon L'\to M$ where $K\subseteq L'\subseteq L$ and define a partial order on this set by saying $\phi\le \psi$ if $\psi$ is an extension of $\phi$ (i.e., the domain of $\phi$ is a subfield of the domain of $\psi$ and $\phi$ is the restriction of $\psi$ to that subfield). Verify that Zorn's lemma can be applied. Pick a maximal element $\phi_\max\colon L_\max\to M$ and show that in fact $L_\max=L$ (using that $L$ is algebraic and $M$ is algebraically closed).

Once you have that, apply the lemma to the case that $L$ and $M$ are both algebraic and algebraically closed and show that any $f$ the lemma gives you is in fact onto.

To elaborate on Hagen von Eitzen's point, you're missing the point that field extensions don't have to all be contained in one "universal extension". In particular, you commented:

let us define $\overline{K}$ as the field generated by all roots of polynomials over $K$.

This definition doesn't make sense! Assuming $K$ is not algebraically closed, given any object $x$, you can construct a field extension of $K$ which contains $x$ as an element which is a root of some polynomial over $K$ (just take your favorite nontrivial algebraic field extension of $K$, and then replace one of its elements with $x$). So taken literally, everything in the mathematical universe is a root of a polynomial over $K$.

Your definition would make sense if you fixed some appropriate field extension $M$ of $K$ (in particular, with the property that every polynomial with coefficients in $K$ splits over $M$), and you instead defined $\overline{K}$ as the subfield of $M$ generated by all roots of polynomials over $K$ in $M$. But then the rest of your argument falls apart. Given an algebraic extension $L$ of $\overline{K}$, you would only to able to conclude that $L=\overline{K}$ if you already knew that $L$ was a subfield of $M$, since $\overline{K}$ only contains the roots of polynomials over $K$ that are in $M$.

So instead, you really need to talk explicitly about homomorphisms between fields, not just fields being subfields of other fields. Using Hagen von Eitzen's lemma, you can embed $L$ in $M$ even if it is not originally a subfield of $M$, and then apply your argument to the image of the embedding.