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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.

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    $\begingroup$ Actually, how do you define $\overline K$? Nevertheless, you cannot conclude that $L$ is a subfield of $\overline K$ $\endgroup$ – Hagen von Eitzen Dec 30 '17 at 10:52
  • $\begingroup$ 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 $L\cong \overline{K}$? $\endgroup$ – ersin Dec 30 '17 at 14:49
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    $\begingroup$ @ersin, the field generated where? $\endgroup$ – Mariano Suárez-Álvarez Dec 30 '17 at 17:24
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You are too generous with equality and you start by assuming a not yet specifically defined $\overline K$.

I'd suggest you start with

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.

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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.

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