Chapter 27

[a(x)+b(x)]^{\prime}=a^{\prime}(x)+b^{\prime}(x)

If \(_{c}\) is algebraic over \(F\) so are \(c+1\) and \(k c\) (where \(\left.k \in F\right)\)

Prove that each of the following numbers is algebraic over \(\mathbb{Q}\) : (a) \(i\) (b) \(\sqrt{2}\) (c) \(2+3 i\) (d) \(\sqrt{1+\sqrt[3]{2}}\) (e) \(\sqrt{i-\sqrt{2}}\) (f) \(\sqrt{2}+\sqrt{3}\) (g) \(\sqrt{2}+\sqrt[3]{4}\)

\([a(x) b(x)]^{\prime}=a^{\prime}(x) b(x)+a(x) b^{\prime}(x)\)

If \(c\) is transcendental over \(F\), so are \(c+1, k c\) (where \(k \in F)\), and \(c^{2}\).

Find the minimum polynomial of the following numbers over the indicated fields: $$ \begin{array}{ll} \sqrt{3}+i & \text { over } \mathbb{R} ; \text { over } \mathbf{Q}: \text { over } \mathbb{Q}(i) ; \text { over } Q(\sqrt{3}) \\ \sqrt{i+\sqrt{2}} & \text { over } \mathbb{R} ; \text { over } Q(i) ; \text { over } \mathbb{Q}(\sqrt{2}) ; \text { over } \mathbb{Q} \end{array} $$

If \(F\) has characteristic 0 and \(a^{\prime}(x)=0\), then \(a(x)\) is a constant polynomial. Why is is conclusion not necessarily true if \(F\) has characteristic \(p \neq 0 ?\)

If \(c\) is transcendental over \(F\), every element in \(F(c)\) but not in \(F\) is transcendental ver \(F\).

Let \(a\) be a root of \(p(x+c)\). Then \(F[x] /\langle p(x+c)\rangle \cong F(a)\) and $$ F[x] /\langle p(x)\rangle \cong F(a+c) $$ onclude that \(F[x] /\langle p(x+c)\rangle \cong F[x] /\langle p(x)\rangle .\)

For each of the following polynomials \(p(x)\), find a number \(a\) such that \(p(x)\) is the nimum polynomial of \(a\) over \(\mathbb{Q}\) : (a) \(x^{2}+2 x-7\) (b) \(x^{4}+2 x^{2}-1\) (c) \(x^{4}-10 x^{2}+1\)

Find the derivative of the following polynomials in \(\mathbb{Z}_{5}[x]\) : $$ x^{6}+2 x^{3}+x+1 \quad x^{5}+3 x^{2}+1 \quad x^{15}+3 x^{10}+4 x^{5}+1 $$

Name a field \((\neq \mathbb{R}\) or \(C)\) which contains a root of \(x^{5}+2 x^{3}+4 x^{2}+6\)

Find a monic irreducible polynomial \(p(x)\) such that \(Q[x] /\langle p(x)\rangle\) is isomorphic to: (a) \(Q(\sqrt{2})\) (b) \(Q(1+\sqrt{2})\) (c) \(Q(\sqrt{1+\sqrt{2})}\)

If \(p(x)\) has degree 2 , then \(\mathbb{Q}[x] /\langle p(x)\rangle\) contains both roots of \(p(x)\).