*Algebra*

## Section 15.1. Examples of Fields

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 \(\C\).