# On the meaning of being algebraically closed

The definition of algebraic number is that $\alpha$ is an algebraic number if there is a nonzero polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $p(\alpha)=0$. By algebraic closure, every nonconstant polynomial with algebraic coefficients has algebraic roots; then, there will be also a nonconstant polynomial with rational coefficients that has those roots. I feel uncomfortable with the idea that the root of a polynomial with algebraic coefficients is again algebraic; why are we sure that for every polynomial in $\mathbb{\bar{Q}}[x]$ we could find a polynomial in $\mathbb{Q}[x]$ that has the same roots?

I apologize if I'm asking something really trivial or my question comes from a big misunderstanding of basic concepts.

• +1; indeed there is something nontrivial to be proved here. Oct 9, 2011 at 22:22
• I read the proof of this in Herstein's upper-division undergraduate algebra textbook, and then felt I knew it. Then later I realized there ought to be an algorithm: given the polynomials with rational coefficients whose solutions are the algebraic numbers that are coefficients in yet another equation, one ought to be able to find a polynomial with rational coefficients whose solutions are those of that equation with algebraic coefficients. It turns out that from the proof in Herstein, one can actually construct the algorithm. Oct 10, 2011 at 0:23

Let $p(x) = a_0+a_1x+\cdots +a_{n-1}x^{n-1} + x^n$ be a polynomial with coefficients in $\overline{\mathbb{Q}}$. For each $i$, $0\leq i\leq n-1$, let $a_i=b_{i1}, b_{i2},\ldots,b_{im_i}$ be the $m_i$ conjugates of $a_i$ (that is, the "other" roots of the monic irreducible polynomial with coefficients in $\mathbb{Q}$ that has $a_i$ as a root).

Now let $F = \mathbb{Q}[b_{11},\ldots,b_{n-1,m_{n-1}}]$. This field is Galois over $\mathbb{Q}$. Let $G=\mathrm{Gal}(F/\mathbb{Q})$. Now consider $$q(x) = \prod_{\sigma\in G} \left( \sigma(a_0) + \sigma(a_1)x+\cdots + \sigma(a_{n-1})x^{n-1} + x^n\right).$$ This is a polynomial with coefficients in $F$, and any root of $p(x)$ is also a root of $q(x)$ (since one of the elements of $G$ is the identity, so one of the factors of $q(x)$ is $p(x)$).

The key observation is that if you apply any element $\tau\in G$ to $q(x)$, you get back $q(x)$ again: \begin{align*} \tau(q(x)) &= \tau\left(\prod_{\sigma\in G} \left( \sigma(a_0) + \sigma(a_1)x+\cdots + \sigma(a_{n-1})x^{n-1} + x^n\right)\right)\\ &= \prod_{\sigma\in G} \left( \tau\sigma(a_0) +\tau\sigma(a_1)x+\cdots + \tau\sigma(a_{n-1})x^{n-1} + x^n\right)\\ &= \prod_{\sigma'\in G} \left( \sigma'(a_0) + \sigma'(a_1)x+\cdots + \sigma'(a_{n-1})x^{n-1} + x^n\right)\\ &= q(x). \end{align*}

That means that the coefficients of $q(x)$ must lie in the fixed field of $G$. But since $F$ is Galois over $\mathbb{Q}$, the fixed field is $\mathbb{Q}$. That is: $q(x)$ is actually a polynomial in $\mathbb{Q}[x]$.

Thus, every root of $p(x)$ is the root of a polynomial with coefficients in $\mathbb{Q}$.

For an example of how this works, suppose you have $p(x) = x^3 - (2\sqrt{3}+\sqrt{5})x + 3$. The conjugate of $\sqrt{3}$ is $-\sqrt{3}$; the conjugate of $\sqrt{5}$ is $-\sqrt{5}$. The field $\mathbb{Q}[\sqrt{3},\sqrt{5}]$ already contains all the conjugates, and the Galois group over $\mathbb{Q}$ has four elements: the one that maps $\sqrt{3}$ to itself and $\sqrt{5}$ to $-\sqrt{5}$; the one the maps $\sqrt{3}$ to $-\sqrt{3}$ and $\sqrt{5}$ to itself; the one that maps $\sqrt{3}$ to $-\sqrt{3}$ and $\sqrt{5}$ to $-\sqrt{5}$; and the identity. So $q(x)$ would be the product of $x^3 - (2\sqrt{3}+\sqrt{5})x + 3$, $x^3 - (-2\sqrt{3}+\sqrt{5})x+3$, $x^3 - (2\sqrt{3}-\sqrt{5})x + 3$, and $x^3 - (-2\sqrt{3}-\sqrt{5})x + 3$. If you multiply them out, you get \begin{align*} \Bigl( &x^3 - (2\sqrt{3}+\sqrt{5})x + 3\Bigr)\Bigl( x^3 + (2\sqrt{3}+\sqrt{5})x+3\Bigr)\\ &\times \Bigl(x^3 - (2\sqrt{3}-\sqrt{5})x + 3\Bigr)\Bigl( x^3 + (2\sqrt{3}-\sqrt{5})x + 3\Bigr)\\ &= \Bigl( (x^3+3)^2 - (2\sqrt{3}+\sqrt{5})^2x^2\Bigr)\Bigl((x^3+3)^2 - (2\sqrt{3}-\sqrt{5})^2x^2\Bigr)\\ &=\Bigl( (x^3+3)^2 - 17x^2 - 2\sqrt{15}x^2\Bigr)\Bigl( (x^3+3)^2 - 17x^2 + 2\sqrt{15}x^2\Bigr)\\ &= \Bigl( (x^3+3)^2 - 17x^2\Bigr)^2 - 60x^4, \end{align*} which has coefficients in $\mathbb{Q}$.

• Thank to you now I think I'm going to attend a Galois' Theory course :) Oct 9, 2011 at 23:05
• @EmanueleNatale: In that respect, Henning's argument is cleaner, if you already know that $a$ is algebraic over $\mathbb{Q}$ if and only if $Q[a]$ is a finite dimensional vector space over $\mathbb{Q}$. Oct 9, 2011 at 23:08
• @ArturoMagidin: Great answer! The usual proof (via the dimension of extensions) never seemed to have quite the right flavor for me (at least alongside Galois Theory). On a side note, what definition for Galois extension do you use? I recall that proving the equivalence of various definitions (and resulting properties) used the transitivity of [the property of being] algebraic extensions in a few spots (I don't think this is really a problem though). Oct 10, 2011 at 21:58
• @ArturoMagidin: I think in the first line we need p to have coefficients algebraic over \mathbb{Q} instead of in \mathbb{Q} Oct 10, 2011 at 22:06
• @RileyE: I use "normal and separable", or equivalently "splitting field of a set of separable polynomials". Separability is not an issue over $\mathbb{Q}$. Oct 10, 2011 at 22:22

Define $\mathbb A$ as the set of all complex roots of rational polynomials; then we want to prove that $\mathbb A$ is algebraically closed.

Let $f=X^n+a_{n-1}X^{n-1}+\cdots+a_0$ be a polynomial with coefficients in $\mathbb A$. I assume that we already know that $\mathbb A$ is a field, so taking the leading coefficient to be $1$ does not lose generality. Let $S$ be $\mathbb Q[a_0,\ldots,a_{n-1}]$, the smallest ring extension of $\mathbb Q$ that contains all of the coefficients of $f$. $S$ is a finite-dimensional vector space over $\mathbb Q$, because it is spanned by all products of powers of the $a_i$'s up to the degree of the rational polynomial each $a_i$ is a root in.

Now let $\beta\in\mathbb C$ be a root of $f$. Then $S[\beta]$, the smallest ring extension of $S$ that contains $\beta$, is a finite-dimensional vector space over $S$ (actually a finitely generated module, except that it turns out that $S$ is in fact a field), because it is spanned by powers of $\beta$ from $1$ up to $\beta^{n-1}$. Thus, in particular $S[\beta]$ is also a finite-dimensional vector space over $\mathbb Q$.

Now take sufficiently many powers of $\beta$, enough that there are more of them than the dimension of $S[\beta]$ over $\mathbb Q$. They must then be linearly dependent. But a nontrivial rational linear relation between powers of $\beta$ is exactly a rational polynomial that has $\beta$ as a root. So $\beta\in\mathbb A$.

HINT $\$ It boils down to the transitivity of algebraic extensions, which boils down to

$\rm\qquad\quad\ K\ =\ \mathbb Q(\alpha)\: =\ \mathbb Q\langle1,\alpha,\:\ldots\:,\alpha^{m-1}\rangle,\quad\ \ F\ =\ K(\beta)\: =\ K\langle1,\beta,\:\ldots\:,\beta^{\:n-1}\rangle$

$\rm\quad \Rightarrow\ \ F\ =\ \mathbb Q(\alpha)(\beta)\: =\ \mathbb{Q} \big\langle\{1,\alpha,\:\ldots\:,\alpha^{m-1} \}\cdot \{1,\beta,\:\ldots\:,\beta^{\:n-1}\}\big\rangle\: =\ \mathbb Q\big\langle\alpha^i\:\beta^j\big\rangle_{i\:<\:m,\ j\:<\:n}$

which follows simply be employing the minimal polynomials of $\rm\:\alpha,\:\beta\:$ as rewrite rules $\rm\:\alpha^m \to\ f(\alpha),\ \ \beta^{\:n}\to\ g(\beta)\:$ to reduce all powers of $\rm\:\alpha,\:\beta\:$ to powers $\rm\:< m,\:n\:,\:$ respectively.

Therefore, since $\rm\:[F:\mathbb Q]\:$ is finite, we deduce that $\rm\:\beta\:$ is algebraic over $\rm\:\mathbb Q\:.$ More explicitly, since, by above, $\rm\:dim\ F/\mathbb Q\ =\ m\:n\:,\:$ the $\rm\ m\:n+1\$ elements $\rm\:1,\:\beta,\:\ldots\:,\beta^{\:m\:n}$ are linearly dependent over $\mathbb Q\:,\:$ yielding a polynomial $\rm\:0\ne h(x)\in \mathbb Q[x]\:$ such that $\rm\:h(\beta) = 0\:.\:$ Therefore $\rm\:\beta\:$ is algebraic over $\rm\:\mathbb Q\:.$