How do I show that $\overline{\mathbb{Q}} := \{\alpha \in \mathbb{C}\mid \alpha\text{ is algebraic over }\mathbb{Q} \}$ is algebraically closed? I am thinking about solving this using that $\mathbb{C}$ is algebraically closed but I don't know hot to proceed from here.


2 Answers 2


Let $f=a_0+a_1x+\dots+a_nx^n$ be a polynomial with coefficients in $\bar{\mathbb{Q}}$; then $f$ has coefficients in $K=\mathbb{Q}[a_0,\dots,a_n]$, which is finite dimensional over $\mathbb{Q}$. A splitting field for $f$ over $K$ is finite dimensional over $K$, hence over $\mathbb{Q}$. So the roots of $f$ are in a finite dimensional extension of $\mathbb{Q}$, which means they're algebraic over $\mathbb{Q}$.


  1. If $b$ is algebraic over $\mathbb{Q}$, then $\mathbb{Q}[b]=\mathbb{Q}(b)$ is finite dimensional over $\mathbb{Q}$

  2. By induction, if $b_1,\dots,b_k$ are algebraic over $\mathbb{Q}$, then also $\mathbb{Q}[b_1,\dots,b_k]$ is finite dimensional over $\mathbb{Q}$.

  3. The splitting field $K$ can be assumed to be a subfield of $\mathbb{C}$ because $\mathbb{C}$ is algebraically closed.

  4. The splitting field of a polynomial in $K[x]$ is finite dimensional, because it is generated by the roots of $K$, which are algebraic over $K$ and the same argument as in 1 and 2 can be used.

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    $\begingroup$ Note that for the definition of $\bar{\mathbb{Q}}$ given in the problem, you must also use the fact that $\mathbb{C}$ is algebraically closed in order to know that the roots you refer to exist in $\mathbb{C}$ at all. $\endgroup$ Oct 2, 2015 at 17:53
  • $\begingroup$ What you used was: Take the ring $K = \mathbb{Q}[a_0,\ldots,a_n]$. This is the ring where $f$ has its coefficients. Even more, it is a finite-dimensional vector space over $\mathbb{Q}$. Then, a splitting field for $f$ \over $K$ has to be finite over $K$, why? Where did you use that the coefficients belongs to $\bar{\mathbb{Q}}$? $\endgroup$ Oct 2, 2015 at 18:26
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    $\begingroup$ @LeonardoFranciscoCavenaghi I added some notes. $\endgroup$
    – egreg
    Oct 2, 2015 at 18:33
  • $\begingroup$ Very nice! I got it! Thanks a lot for the answer. $\endgroup$ Oct 2, 2015 at 18:37

It is a proposition more general: Let L be a extension of K.If L is algebraically closed then the set of elementes from L algebraic over K is also algebraically closed. In your case L is C and K is Q!


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