I have read that algebraic closures are unique up to isomorphism fixing the base field, although there is in general no natural isomorphism between algebraic closures, so we must speak of 'an' algebraic closure rather than 'the' algebraic closure to account for this.

But if we fix some algebraically closed extension $L$ of $K$ , does this mean that taking algebraic closure of $K$ within $L$ is in a sense 'natural' in this context? For example, is the field of algebraic numbers the 'natural' algebraic closure of $\mathbb{Q}$ within $\mathbb{C}$? Below is my attempt at making sense of this:

Let $K$ be a field with a given algebraically closed extension $L$. We say that $\alpha\in L$ is algebraic over $K$ if it is a root of a non-zero polynomial over $K$. Define $A$ to be the set of all $\alpha\in L$ that are algebraic over $K$. Then $A$ is both a subfield of $L$ and an algebraic closure of $K$.

Let $B$ be a subfield of $L$ that is also an algebraic closure of $K$ and let $\alpha\in B$. Since $B$ is an algebraic extension of $K$, we have that $\alpha$ is a root of a non-zero polynomial $p$ over $K$, so that $\alpha\in A$. Hence $B\subseteq A$.

Since $B$ is algebraically closed, it has no proper algebraic extensions, so $B=A$. Call this algebraic closure $\overline{K}$.

If we then choose any algebraic closure $C$ of $K$, we then have a field isomorphism $\varphi:C\to\overline{K}$. So given an algebraically closed extension $L$, any algebraic closure of $K$ is isomorphic to a unique algebraic closure of $K$ within $L$? (but the isomorphism is not unique).

Is this correct?

  • 1
    $\begingroup$ Yes, this all looks correct. $\endgroup$ Commented Jun 26, 2021 at 12:29

1 Answer 1


If $L$ is an algebraically closed extension of $K$, then there is a unique intermediate field $F$, $K\subseteq F\subseteq L$ such that $F$ is algebraically closed and algebraic over $K$.

Note that an algebraic closure of $K$ must be algebraic over $K$, by definition. So you can consider the set $F$ of all elements in $K$ that are algebraic over $K$.

Since the sum and product of two algebraic elements is algebraic and the inverse of a nonzero algebraic element is algebraic, we have that $F$ is a subfield of $L$ containing $K$.

Now we can show that $F$ is algebraically closed (and so uniqueness will follow).

Take a polynomial of positive degree $f(x)\in F[x]$, say $$ f(x)=a_0+a_1x+\dots+a_nx^n $$ Then $f$ has coefficients in $K'=K(a_0,a_1,\dots,a_n)$ which is a finite extension of $K$. If $r$ is a root of $f$ in $L$ (which surely exists), then $r$ is algebraic over $K'$ and therefore $K'(r)$ is a finite extension of $K'$. Thus $K'(r)$ is a finite extension of $K$ and therefore $r$ is algebraic over $K$. Hence $r\in F$.


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