Past exam of the number theory course of the University of Oxford / 2010 / A / 3

Let the monic minimal polynomial of $\alpha$ over $L$ have coefficients $a_0,\dots ,a_{r-1}$, and put $F=K(a_0,\dots ,a_{r-1})$. Each coefficient is algebraic over $K$, so adjoining these finitely many coefficients successively gives $[F:K]<\infty$. The same polynomial makes $\alpha$ algebraic over $F$, giving $[F(\alpha ):F]\le r$. The tower law for field extensions yields $[F(\alpha ):K]<\infty$. Its subfield $K(\alpha )$ therefore also has finite degree over $K$. This holds for every $\alpha \in M$, proving $M/K$ is algebraic.

An algebraically closed field is a field in which every nonconstant polynomial has a root, equivalently every such polynomial splits. An algebraic closure of $K$ is an algebraic extension of $K$ which is algebraically closed.

For the displayed subset $\overline{K}\subseteq \mathbb{C}$, if $\alpha ,\beta$ are algebraic over $K$, then $K(\alpha ,\beta )/K$ is finite. Sums, differences, products and nonzero inverses of its elements are algebraic over $K$, so $\overline{K}$ is a subfield. If a nonconstant polynomial has coefficients in $\overline{K}$, those finitely many coefficients generate a finite extension $F/K$. It has a root $z\in \mathbb{C}$ by algebraic closure of $\mathbb{C}$, and $z$ is algebraic over $F$. Part (i) makes it algebraic over $K$, so $z\in \overline{K}$. Hence $\overline{K}$ is algebraically closed and is an algebraic closure of $K$.

Every $z\in \overline{K}$ lies in the finite subextension $K(z)$, and every element of a finite subextension is algebraic over $K$. Therefore $\overline{K}$ is exactly the union of all finite subextensions inside $\mathbb{C}$. The two examples follow.

Take $K=\mathbb{R}$. Every complex number is algebraic over $\mathbb{R}$, since $\mathbb{C}=\mathbb{R}(i)$, so its algebraic closure inside $\mathbb{C}$ is $\mathbb{C}$. This is a proper containment and

Take $K=\mathbb{Q}$. For every positive integer $n$, the real number $\sqrt[n]{2}$ lies in $\overline{\mathbb{Q}}$ and has degree $n$, by Eisenstein's criterion at $2$. If $\overline{\mathbb{Q}}/\mathbb{Q}$ had finite degree, all its subextension degrees would be bounded by that degree, contradicting this unbounded family. Therefore $[\overline{\mathbb{Q}}:\mathbb{Q}]$ is infinite.

Identify $\mathbb{C}$ with the Euclidean plane and start with the points $0,1$. A complex number is a constructible number if a finite sequence of drawing lines through known points and circles with known centres and known distances as radii, and selecting their intersections, produces that point.

Line-line intersections solve linear equations in the current coordinate field. Line-circle and circle-circle intersections require at most one quadratic equation after eliminating a coordinate. Hence all coordinates obtained lie in a finite tower of real quadratic extensions of $\mathbb{Q}$. If $\alpha =x+iy$ is constructible, choose a common such real field $F$ containing $x,y$. Adjoining $i$ if needed still gives an extension $F(i)/\mathbb{Q}$ of power-of-two degree. By the tower law for field extensions,A regular heptagon centred at $0$ with one vertex $1$ would have $\xi =e^{2\pi i/7}$ as another constructible vertex. But its cyclotomic degree is $\phi (7)=6$, which is not a power of two. Thus the regular heptagon is not constructible.