Brian Bi
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Exercise 15.1.1 Let $$r$$ be a nonzero element of $$R$$. The map $$\varphi : R \to R$$ given by $$\varphi(x) = rx$$ is an $$F$$-linear endomorphism. If $$\varphi$$ is surjective, then $$x$$ has an inverse in $$R$$. If $$\varphi$$ is not surjective, then since the dimension is finite, $$\ker \varphi$$ is nontrivial, implying $$R$$ is not an integral domain.
Exercise 15.1.2 Suppose $$x = \frac{-b+\delta}{2}$$. Then \begin{align*} x^2 + bx + c &= \frac{b^2 - 2b\delta + \delta^2}{4} + \frac{-b^2+b\delta}{2} + c \\ &= \frac{1}{4} (b^2 - 2b\delta + b^2 - 4c - 2b^2 + 2b\delta + 4c) \\ &= 0 \end{align*} Conversely, suppose $$x^2 + bx + c$$ splits over $$F$$. Complete the square to obtain $$(x + b/2)^2 = b^2/4 - c$$. There is some $$x \in F$$ that solves this equation. Let $$\delta = 2x + b$$. Then $$\delta^2 = 4(b^2/4 - c) = b^2 - 4c$$.
Exercise 15.1.3 The subfields of $$\C$$ that are contained with $$\R$$ are obviously not dense subsets of $$\C$$. We claim that the converse is also true. Let $$K$$ be a subfield of $$\C$$ and suppose $$K$$ contains the element $$z_0 \notin \R$$. Then $$\{1, z_0\}$$ is an $$\R$$-basis of $$\C$$. Now $$K$$ must contain the rational numbers as well as all rational multiples of $$z_0$$. Let $$z \in \C$$ be given and write $$z = a + bz_0$$ where $$a, b \in \R$$. If $$r, s \in \Q$$, then $$r + sz_0 \in K$$. By choosing $$r$$ and $$s$$ to be arbitrarily close to $$a$$ and $$b$$, respectively, we can make $$r + sz_0$$ arbitrarily close to $$z$$. Therefore $$K$$ is dense in $$\C$$.