*Algebra*

## Section 15.10. The Fundamental Theorem of Algebra

Exercise 15.10.1 Let \(\overline{\Q}\) denote the algebraic numbers. Suppose \(f \in \overline{\Q}[x]\). Then \(f\) has a root \(\alpha \in \C\). The extension \(\overline{\Q}(\alpha)/\overline{\Q}\) is finite, therefore algebraic, and the extension \(\overline{\Q}/\Q\) is also algebraic by definition of \(\overline{\Q}\). We showed in Exercise 15.3.10 that algebraic extensions are transitive, so \(\overline{\Q}(\alpha)/\Q\) is algebraic. Therefore there exists \(g \in \Q[x]\) such that \(g(\alpha) = 0\). But \(g\) splits completely over \(\C\) and its roots are algebraic numbers, so \(\alpha\) is one of these, and is itself algebraic.

Exercise 15.10.2 Fix \(p\). Let \(F_i\) denote the finite field of order \(p^{i!}\). By Theorem 15.7.3(e), each \(F_i\) contains a copy of \(F_{i-1}\). We can therefore form the ascending union \(F = \bigcup_{i=1}^\infty F_i\), which itself has the structure of a field. If \(f \in \mathbb{F}_p[x]\) is irreducible of degree \(d\), then \(f\) appears as a factor of \(x^{p^d} - x\), and therefore has a root in \(\mathbb{F}_{p^d} \subseteq F_d \subseteq F\).

We need to show that \(F\) is algebraically closed. The argument is very similar to that used in the previous problem. Since each \(F_i\) is a finite extension of \(\mathbb{F}_p\), it is algebraic, and since each element of \(F\) is an element of some \(F_i\), \(F\) is an algebraic extension of \(\mathbb{F}_p\). Let \(g \in F[x]\). Then \(g\) has a root \(\alpha \in K\) where \(K\) is some extension of \(F\). Since both \(F(\alpha)/F\) and \(F/\mathbb{F}_p\) are algebraic extensions, the extension \(F(\alpha)/\mathbb{F}_p\) is also algebraic. This implies that there exists \(h \in \mathbb{F}_p[x]\) such that \(h(\alpha) = 0\). But \(h\) splits completely over \(F\), therefore \(h\) splits completely over \(K\) and all of its roots lie in \(F\). This implies that \(\alpha \in F\). So \(F\) is algebraically closed.

Exercise 15.10.3 When \(r\) corresponds to the magnitude of some root of \(f\), the curve \(f(C_r)\) will pass through the origin, so its winding number will be undefined. It is when \(f(C_r)\) passes through such singular curves that the winding number can change discontinuously from just above \(r\) to just below \(r\).

For example, using Mathematica, we
can plot the curves \(f(C_r)\) where \(r = 1.2, 1.0, 0.8\) respectively and
\(f(z) = z^2 + 2z + 1\):

As \(r\) decreases from above 1 to below 1, its winding number around the
origin changes discontinuously from 2 to 1 as \(r\) passes through 1, where
the curve \(f(C_r)\) passes through the origin. Moreover, a similar issue
occurs with the total curvature, which may still be defined for curves that
pass through the origin, but not for curves that exhibit a cusp, as the middle
curve does above.