# 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. – hmakholm left over Monica Oct 9 '11 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. – Michael Hardy Oct 10 '11 at 0:23

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 $\beta\in\mathbb A$.

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 :) – Immanuel Weihnachten Oct 9 '11 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 $\mathbb{Q}$. – Arturo Magidin Oct 9 '11 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). – Riley E Oct 10 '11 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} – Riley E Oct 10 '11 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 $\mathbb{Q}$. – Arturo Magidin Oct 10 '11 at 22:22

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 $\rm\:\mathbb Q\:.$