LEC ALG3 19

Algebraic closures

Embeddings

Definition 338.1(Definition).

Let $E/k$ and let $\sigma :k\to L$ be an embedding. $\sigma$ induces an isomorphism of $k$ with its image $\sigma k$, written as $k^{\sigma }$. An embedding $\tau$ of $E$ into $L$ is said to be over $\sigma$ if $\tau {|}_k=\sigma$. If $\sigma$ is an inclusion, we say $\tau$ is an embedding of $E$ over $k$.

Remark 338.2(Remark).

Note that if $\alpha \in E$ is a root of $f(x)\in k[x]$, then ${\alpha }^{\tau }$ is a root of $f^{\sigma }$.

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Example 338.3(Example).

There does not exist an embedding of $\mathbb{Q}[\sqrt{2}]$ into $\mathbb{Q}[\sqrt{3}]$ over $\mathbb{Q}$. Suppose $\tau$ were such an embedding, with $\tau (\sqrt{2})=\alpha$. $\alpha$ must satisfy the following for all $a+b\sqrt{2},c+d\sqrt{2}\in \mathbb{Q}[\sqrt{2}]$:

$$\begin{array}{rl}& ac+2bd+\alpha (ad+bc)=ac+{\alpha }^{2}bd+\alpha (ad+bc)\\ ⟹& 2={\alpha }^2,\end{array}$$

which is not possible since $\sqrt{2}\notin \mathbb{Q}[\sqrt{3}]$.

Lemma 338.4(Lemma).

Let $E$ be an algebraic extension of $k$, and let $\sigma :E\to E$ be an embedding of $E$ into itself over $k$. Then $\sigma$ is an automorphism.

Proof.

It suffices to prove that $\sigma$ is surjective. Let $\alpha ={\alpha }_1\in E$, let $p(x)$ be its irreducible polynomial over $k$. Let ${\alpha }_1,\dots ,{\alpha }_n$ be the roots of $p(x)$ which lie in $E$, and let $E^{\prime }=k({\alpha }_1,\dots ,{\alpha }_n)$ ¹. $E^{\prime }/k$ is finite by Prp 322.8. By Rmk 2, $\sigma ({\alpha }_1),\dots ,\sigma ({\alpha }_n)$ are roots of $p(x)$, so $\sigma (\{{\alpha }_1,\dots ,{\alpha }_n\})=\{{\alpha }_1,\dots ,{\alpha }_n\}$. Since $E^{\prime }$ is the smallest field containing $k$ and ${\alpha }_1,\dots ,{\alpha }_n$, it follows that $E^{\prime }\subseteq \sigma (E^{\prime })$. Since there exists² a $k$-basis of $k({\alpha }_1,\dots ,{\alpha }_n)$ of the form

$$\{{\alpha }_1^{i_1}\dots {\alpha }_n^{i_n}:0⩽i_1⩽d_1-1,\dots ,0⩽i_n⩽d_n-1\},$$

we have $\sigma (E^{\prime })\subseteq E^{\prime }$ and hence $\sigma (E^{\prime })=E^{\prime }$ ³. Since $\alpha \in E^{\prime }$, it follows that $\alpha$ is in the image of $\sigma$.□

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Question

In the proof of Lem 4, is it always true that $k({\alpha }_1,\dots ,{\alpha }_n)\ne k({\alpha }_s:s\in S)$ for all $S⊊[n]$? More concretely, can you give me an irreducible polynomial $p(x)$ over $\mathbb{Q}$ and a field extension $F/\mathbb{Q}$ such that $p(x)$ has at least three roots ${\alpha }_1,{\alpha }_2,{\alpha }_3$ in $F$, and $\mathbb{Q}({\alpha }_i)$ and $\mathbb{Q}({\alpha }_i,{\alpha }_j)$ are strictly contained in $\mathbb{Q}({\alpha }_1,{\alpha }_2,{\alpha }_3)$ for all $i,j$?

Extensions in which a given polynomial splits

Proposition 338.5(Proposition).

Let $k$ be a field and $f$ a polynomial in $k[x]$ of degree $⩾1$. Then there exists an extension $E$ of $k$ in which $f$ has a root.

Proof.

WLOG, we can assume $f(x)$ is irreducible. Let $K$ be the field $k[x]/(f(x))$. Consider the canonical homomorphism

$$\sigma :k[x]\to K.$$

Note that $\sigma {|}_k$ is an embedding of $k$ into $K$. Let $x^{\sigma }=x+(f(x))$ be the image of $x$ under $\sigma$. Let $f(x)={\sum }_{i=0}^n{a}_i{x}^i$, and let⁴

$$f^{\sigma }(X):=\sum _{i=0}^n{a}_i^{\sigma }{X}^n\in k^{\sigma }[X].$$

Then

$$\begin{array}{rl}{f}^{\sigma }(x^{\sigma })& =\sum _{i=0}^n{a}_i^{\sigma }(x^{\sigma }{)}^n\\ & =\sigma \left(\sum _{i=0}^n{a}_i{x}^n\right)\\ & =\sigma (f(x))\\ & =0.\end{array}$$

Thus, $x^{\sigma }$ is a root of $f^{\sigma }$. If you’re like me, you’d say we’re done: $k\cong k^{\sigma }$, and we just found an extension $K/k^{\sigma }$ in which $f^{\sigma }$ has a root. To be extra rigorous, you can define a set $E\supset k$ such that $|E|=|K|$, extend $\sigma$ to a bijection $E\to K$, and pull back the field structure of $K/k^{\sigma }$ to $E/k$ to get a “true” field extension of $k$; see Lang (2002) 5.2.3.

Note that $K/k^{\sigma }$ is a finite extension, since $\{1,x,x^2,\dots ,x^{\text{deg}f-1}\}$ is a $k^{\sigma }$-basis for $K$.□

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Corollary 338.6(Corollary).

Let $k$ be a field and let $f_1,\dots ,f_n$ be polynomials in $k[x]$ of degrees $⩾1$. Then there exists an extension $E/k$ in which each $f_i$ has a root.

Proof.

Using Prp 5, let $E_1$ be an extension in which $f_1$ has a root. We may view $f_2$ as a polynomial over $E_1$. Let $E_2$ be an extension of $E_1$ in which $f_2$ has a root. Proceed inductively.□

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Algebraically closed fields and algebraic closures

Definition 338.7(Definition).

1. A field $k$ is said to be algebraically closed if every polynomial in $k[x]$ of degree $⩾1$ has a root in $k$.
2. An algebraic closure of a field $k$ is an algebraic extension of $k$ which is algebraically closed.

Proposition 338.8(Proposition).

Every algebraic extension of an algebraically closed field is trivial.

Proof.

Suppose $k$ is algebraically closed and $E/k$ is an algebraic extension. Let $\alpha \in E$ and $p(x)\in k[x]$ be its irreducible polynomial. Since $k$ is algebraically closed, $p(x)$ splits into linear factors in $k[x]$, so all of its roots lie in $k$. It follows that $\alpha \in k$, and $E=k$.□

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Theorem 338.9(Existence of algebraically closed extensions, Lang (2002) 5.2.5).

Let $k$ be a field. Then there exists an algebraically closed extension $E/k$.

Proof.

We first construct an extension $E_1$ of $k$ in which every polynomial in $k[x]$ of degree $⩾1$ has a root. To each polynomial $f(x)\in k[x]$ of degree $⩾1$, associate a symbol $X_f$ and let $S$ be the set of all such symbols. Form the polynomial ring $k[S]$. We claim that the ideal generated by all the polynomials $f(X_f)$ in $k[S]$ is not the unit ideal. It it is. then there is a finite combination of elements in our ideal which is equal to $1$:

$$g_1{f}_1(X_{f_1})+\cdots +g_n{f}_n(X_{f_n})=1$$

with $g_i\in k[S]$. We will write $X_i$ instead of $X_{f_i}$. The polynomials $g_i$ will involve only a finite number of variables, say $X_1,\dots ,X_N$ (with $N⩾n$). Our relation then reads

$$\sum _{i=1}^n{g}_i(X_1,\dots ,X_N)f_i(X_i)=1.$$

By Cor 6, there exists a finite extension $F/k$ in which each polynomial $f_1(x),\dots ,f_n(x)\in k[x]$ has a root, say ${\alpha }_i$ is a root of $f_i$ in $F$. Let ${\alpha }_i=0$ for $i>n$. Take the image of both sides under the evaluation map $X_i\mapsto {\alpha }_i$. We get $0=1$, a contradiction.

By Thm 111.1 there exists a maximal ideal $\mathfrak{m}$ containing the ideal generated by all polynomials $f(X_f)$ in $k[S]$. $k[S]/\mathfrak{m}$ is a field, and we have a canonical map

$$\sigma :k[S]\to k[S]/\mathfrak{m}.$$

As in the proof of Prp 5, $\sigma {|}_k$ is an embedding of $k$ into $k[S]/\mathfrak{m}$, which is an extension of $k^{\sigma }$. For any polynomial $f(x)\in k[x]$ of degree $⩾1$, $X_f^{\sigma }\in k[S]/\mathfrak{m}$ is a root of $f^{\sigma }(X)\in (k[S]/\mathfrak{m})[X]$ ⁴:

$$\begin{array}{rl}{f}^{\sigma }(X_f^{\sigma })& =\sum _{i=0}^k{a}_i^{\sigma }(X_f^{\sigma }{)}^i=\sigma (f(X_f))=0.\end{array}$$

Using the same type of set-theoretic argument⁵ as in Prp 5, we conclude there exists an extension $E_1$ of $k$ in which every polynomial $f(x)\in k[x]$ of degree $⩾1$ has a root.

Inductively, we can form a sequence of fields

$$\dots /E_n/\dots /E_3/E_2/E_1/k$$

such that every polynomial in $E_n[x]$ of degree $⩾1$ has a root in $E_{n+1}$. Now, define

$$E=\bigcup _{n=1}^{\infty }{E}_i.$$

Then $E$ is naturally a field, for if $x,y\in E$ then there exists some $n$ such that $x,y\in E_n$, and we can take their product or sum in $E_n$. Every polynomial in $E[X]$ has its coefficients in some subfield $E_n$, hence a root in $E_{n+1}$, hence a root in $E$, as desired.□

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Theorem 338.10(Existence of algebraic closures, Lang (2002) 5.2.6).

Let $k$ be a field. There exists an algebraic extension $k^{\text{a}}$ of $k$ which is algebraically closed, called the algebraic closure of $k$.

Proof.

Let $E/k$ be the algebraically closed extension supplied by Thm 9. Using Prp 334.1, let $k^{\text{a}}$ be the subextension formed by the elements of $E$ that are algebraic over $k$. Then $k^{\text{a}}$ is algebraic over $k$. If $\alpha \in E$ and $\alpha$ is algebraic over $k^{\text{a}}$, then by Prp 119.19 $\alpha$ is algebraic over $k$. If $f$ is a polynomial of degree $⩾1$ in $k^{\text{a}}[x]$, then $f$ has a root $\alpha$ in $E$, and $\alpha$ is algebraic over $k^{\text{a}}$. Hence $\alpha$ is in $k^{\text{a}}$ and $k^{\text{a}}$ is algebraically closed.□

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Proposition 338.11(Proposition).

If $k$ is a field which is not finite, then any algebraic extension of $k$ has the same cardinality as $k$.

Proof.

Let $\aleph$ be the cardinality of $k$. It follows that $k[x]$ has cardinality $\aleph$. If $E/k$ is an algebraic extension, Each $\alpha \in E$ has an associated $p(x)\in k[x]$. Since any $p(x)\in k[x]$ can accommodate only finitely many roots, it follows that $E$ has cardinality $\aleph$.□

Uniqueness of algebraic closures

Lemma 338.12(Lemma).

Let $k$ be a field and $\sigma :k\to L$ an embedding of $k$ into $L$⁶. Let $E=k(\alpha )$ be an algebraic extension of $k$ generated by one element. Let $p(x):=\text{Irr}(\alpha ,k)={\sum }_{i=0}^n{a}_i{x}^i\in k[x]$. Let $p^{\sigma }(y):={\sum }_{i=0}^n{a}_i^{\sigma }{y}^i\in k^{\sigma }[y]$. Then, there exists a bijection

$$\{\psi :k(\alpha )\to L|\psi \text{ is an embedding extending }\sigma \}⟷\{\beta \in L:p^{\sigma }(\beta )=0\}.$$

Proof.

Let $\beta$ be a root of $p^{\sigma }$ in $L$. By Prp 119.13, $\{1,\alpha ,\dots ,{\alpha }^{n-1}\}$ is a $k$-basis for $k(\alpha )$. Thus, every element of $k(\alpha )$ can be uniquely expressed in the form ${\sum }_{i=0}^{n-1}{c}_i{\alpha }^i$, $c_i\in k$. Define an extension $\psi :k(\alpha )\to L$ of $\sigma$ by ${\sum }_{i=0}^{n-1}{c}_i{\alpha }^i\mapsto {\sum }_{i=0}^{n-1}{c}_i^{\sigma }{\beta }^i$. We do not need to prove that $\psi$ is well-defined owing to the unique representation afforded by the basis. Let $a,b\in k(\alpha )$.

$$\begin{array}{rl}\psi (ab)& =\psi \left(\left(\sum _{i=0}^{n-1}{a}_i{\alpha }^i\right)\left(\sum _{i=0}^{n-1}{b}_i{\alpha }^i\right)\right)\\ & =\psi \left(\sum _{i=0}^{n-1}{c}_i{\alpha }^i+p(\alpha )g(\alpha )\right)\\ & =\psi \left(\sum _{i=0}^{n-1}{c}_i{\alpha }^i\right)\\ & =\sum _{i=0}^{n-1}{c}_i^{\sigma }{\beta }^i\\ & \\ \psi (a)\psi (b)& =\left(\sum _{i=0}^{n-1}{a}_i^{\sigma }{\beta }^i\right)\left(\sum _{i=0}^{n-1}{b}_i^{\sigma }{\beta }^i\right)\\ & =\sum _{i=0}^{n-1}{c}_i^{\sigma }{\beta }^i+p^{\sigma }(\beta )g^{\sigma }(\beta )\\ & =\sum _{i=0}^{n-1}{c}_i^{\sigma }{\beta }^i\end{array}$$

$\psi$ clearly satisfies $\psi (a+b)=\psi (a)+\psi (b)$. Thus, $\psi$ is a homomorphism. It is also clear from the above computation that $p^{\sigma }(\beta )$ ($=p^{\sigma }({\alpha }^{\psi })$) must be zero for $\psi$ to be a homomorphism, forcing $\psi$ to map $\alpha$ to a root of $p^{\sigma }$.□

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Next, we inspect extensions of $\sigma$ to arbitrary algebraic extensions of $k$.

Theorem 338.13(Theorem).

Let $E/k$ be an algebraic extension, and $\sigma :k\hookrightarrow L$ an embedding of $k$ in an algebraically closed field $L$. Then there exists an embedding of $E$ into $L$ over $\sigma$. If $E$ is algebraically closed and $L$ is algebraic over $k^{\sigma }$, then any such embedding over $\sigma$ is an isomorphism of $E$ onto $L$.

Proof.

Let $S$ be the set of all pairs $(F,\tau )$ where $F$ is a subfield of $E$ containing $k$, and $\tau$ is an extension of $\sigma$ to an embedding of $F$ into $L$. We write a partial order $⩽$ by $(F,\tau )⩽(F^{\prime },{\tau }^{\prime })$ if $F\subseteq F^{\prime }$ and ${\tau }^{\prime }{|}_F=\tau$. Note that $S$ is not empty since it contains $(k,\sigma )$. If $\{(F_i,{\tau }_i)\}$ is a totally ordered subset, we let $F=\bigcup F_i$ and define $\tau$ on $F$ to be equal to ${\tau }_i$ on each $F_i$. Then $(F,\tau )$ is an upper bound for the totally ordered subset. By Zorn’s lemma, let $(K,\lambda )$ be a maximal element in $S$. Then $\lambda$ is an extension of $\sigma$.

We contend that $K=E$. Otherwise, there exists $\alpha \in E$, $\alpha \notin K$. If $p$ is the irreducible polynomial of $\alpha$ over $K$, $p^{\lambda }$ must have a root in $L$ since $L$ is algebraically closed. It follows from Lem 12 that there exists an extension $\psi :K(\alpha )\to L$ extending $\lambda$, contradicting the maximality of $(K,\lambda )$. This proves that there exists an extension $\lambda$ of $\sigma$ to $E$.

If $E$ is algebraically closed, and $L$ is algebraic over $k^{\sigma }$, then $E^{\sigma }$ is algebraically closed and $L$ is algebraic over $E^{\sigma }$, hence $L=E^{\sigma }$ by Prp 8.□

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An immediate corollary:

Corollary 338.14(Uniqueness of algebraic closures).

Let $k$ be a field and $E/k$, $E^{\prime }/k$ be algebraic extensions. Assume that $E,E^{\prime }$ are algebraically closed. Then there exists an isomorphism $\tau :E\to E^{\prime }$ over $k$.

Footnotes

1. It may very well be that $n\ne \text{deg}p(x)$. In other words, $p(x)$ being irreducible in $k[x]$ and $E/k$ being an extension containing a root of $p(x)$ does not imply that $E$ is the splitting field of $p(x)$. ↩

2. See Prp 119.13 and the proof of Prp 322.2. ↩

3. Knowing $\sigma (E^{\prime })\subseteq E^{\prime }$, an alternative way to glean $\sigma (E^{\prime })=E^{\prime }$ is to observe that since $\sigma$ is an injective $k$-module homomorphism, $\sigma (E^{\prime })$ is a subspace of $E^{\prime }$ having the same dimension $[E^{\prime }:k]$. ↩

4. Note that $f^{\sigma }$ is NOT the image of $f$ under $\sigma$ (which would be $0$)! ↩ ↩²

5. If we do not do this, we technically cannot take the union $\bigcup E_i$ later on in the proof; you can, however take the colimit instead. See Buzzard (n.d.) ↩

6. Lang (2002, p. 233) assumes $L$ is algebraically closed. I do not see where this this hypothesis is used, and hence have removed it. ↩

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References

Buzzard, K. (n.d.). Existence of Algebraic Closure of a Field. Retrieved November 11, 2025, from https://www.ma.imperial.ac.uk/~buzzard/maths/teaching/15Aut/M3P11/algclosure.pdf

Lang, S. (2002). Algebra (Vol. 211). Springer New York. https://doi.org/10.1007/978-1-4613-0041-0