Chapter 31

If \(K\) is the root field of some polynomial \(a(x)\) over \(F, K\) is also called a normal extension of \(F .\) There are other possible ways of defining normal extension, which are equivalent to the above. We consider the two most common ones here: they are precisely the properties expressed in Theorems 7 and \(6 .\) Let \(K\) be a finite extension of \(F .\) Prove the following: Suppose that for every irreducible polynomial \(p(x)\) in \(F[x]\), if \(p(x)\) has one root in \(K\), then \(p(x)\) must have all its roots in \(K\). Prove that \(K\) is a normal extension of \(F .\)

Let \(\mathrm{h}: F_{1} \rightarrow F_{2}\) be an isomorphism. If \(a(x) \in F_{1}[x]\), let \(K_{1}\) be the root field of \(a(x)\) over \(F_{1}\), and \(K_{2}\) the root field of \(h a(x)\) over \(F_{2}\). Prove the following : If \(p(x)\) is an irreducible factor of \(a(x), u \in K_{1}\) is a root of \(p(x)\), and \(v \in K_{2}\) is a root of \(h p(x)\), then \(F_{1}(u) \cong F_{2}(v)\).

Let \(F\) be a field. An irreducible polynomial \(p(x)\) in \(F[x]\) is said to be separable over \(F\) if it has no multiple roots in any extension of \(F\). If \(p(x)\) does have a multiple root in some extension, it is inseparable over \(F\). Prove the following : If \(F\) has characteristic 0 , every irreducible polynomial in \(F[x]\) is separable. Thus, for characteristic 0, there is no question whether an irreducible polynomial is separable or not. However, for characteristic \(p \neq 0\), it is different. This case is treated next. In the following problems, let \(F\) be a field of characteristic \(p \neq 0\)

Find \(c\) such that \(Q(\sqrt{2}, \sqrt{-3})=Q(c)\). Do the same for \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})\).

Find the root field of \(\mathrm{x}^{2}+1\) over \(\mathbb{Z}_{3} .\) ANSWER By the basic theorem of field extensions, $$ \mathbb{Z}_{3}[x] /\left\langle x^{2}+1\right\rangle \cong \mathbb{Z}_{3}(u) $$ where \(u\) is a root of \(x^{2}+1 .\) In \(\mathbb{Z}_{3}(u), x^{2}+1=(x+u)(x-u)\), because \(u^{2}+1=0\). Since \(\mathbb{Z}_{3}(u)\) contains \(\pm u\), it is the root field of \(x^{2}+1\) over \(\mathbb{Z}_{3} .\) Note that \(\mathbb{Z}_{3}(u)\) has nine elements, and its addition and multiplication tables are easy to construct. (See Chapter 27, Exercise C.) Show that, in any extension of \(\mathbb{Z}_{3}\) which contains a root \(u\) of $$ a(x)=x^{3}+2 x+1 \in \mathbb{Z}_{3}[x] $$ it happens that \(u+1\) and \(u+2\) are the remaining two roots of \(a(x)\). Use this fact to find the root field of \(x^{3}+2 x+1\) over \(\mathbb{Z}_{3}\). Write its addition and multiplication tables

If \(\mathbb{Z}_{p}[y]\) is the domain of polynomials (in the letter \(y\) ) over \(\mathbb{Z}_{p}\), let \(E=\mathbb{Z}_{p}(y)\) be the field of quotients of \(\mathbb{Z}_{p}[y]\). Let \(K\) denote the subfield \(\mathbb{Z}_{p}\left(y^{p}\right)\) of \(\mathbb{Z}_{p}(y)\). Prove that \(a(x)=x^{p}-y^{p}\) has the factorization \(x^{p}-y^{p}=(x-y)^{p}\) in \(E[x]\), but is irreducible in \(K[x]\). Conclude that there is an irreducible polynomial \(a(x)\) in \(K[x]\) with a root whose multiplicity is \(p\). Thus, over an infinite field of nonzero characteristic, an irreducible polynomial may have multiple roots. Even these fields, however, have a remarkable property: all the roots of any irreducible polynomial have the same multiplicity. The details follow: Let \(F\) be any field, \(p(x)\) irreducible in \(F[x], a\) and \(b\) two distinct roots of \(p(x)\), and \(K\). the root field of \(p(x)\) over \(F\). Let \(i: K \rightarrow i(K)=K^{\prime}\) be the isomorphism of Theorem 4, and \(\bar{i}: K[x] \rightarrow K^{\prime}[x]\) the isomorphism described immediately preceding Theorem \(3 .\)

If \(F \subseteq I \subseteq K\) and \(K\) is a root field of \(a(x)\) over \(F\), then \(K\) is a root field of \(a(x)\) over \(I .\)

Find the root field of \(\mathrm{x}^{2}+1\) over \(\mathbb{Z}_{3} .\) ANSWER By the basic theorem of field extensions, $$ \mathbb{Z}_{3}[x] /\left\langle x^{2}+1\right\rangle \cong \mathbb{Z}_{3}(u) $$ where \(u\) is a root of \(x^{2}+1 .\) In \(\mathbb{Z}_{3}(u), x^{2}+1=(x+u)(x-u)\), because \(u^{2}+1=0\). Since \(\mathbb{Z}_{3}(u)\) contains \(\pm u\), it is the root field of \(x^{2}+1\) over \(\mathbb{Z}_{3} .\) Note that \(\mathbb{Z}_{3}(u)\) has nine elements, and its addition and multiplication tables are easy to construct. (See Chapter 27, Exercise C.) 2 Find the root field of \(x^{3}+x^{2}+x+2\) over \(\mathbb{Z}_{3}\), and write its addition and multiplication tables.

Find the root field of \(a(x)=\left(x^{2}-3\right)\left(x^{3}-1\right)\) over \(\mathbf{Q} .\) ANSWER The complex roots of \(a(x)\) are \(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\), so the root field is \(\mathbb{Q}\left(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\right)\). The same field can be written more simply as \(\mathbb{Q}(\sqrt{3}, i)\). Prove that \(x^{2}-3\) and \(x^{2}-2 x-2\) are both irreducible over \(\mathbb{Q}\). Then find their root fields over \(\mathbf{Q}\) and show they are the same.

In the following, let \(F\) be a subfield of \(\mathbb{C}\). An injective homomorphism \(h: F \rightarrow \mathrm{C}\) is called a monomorphism; it is obviously an isomorphism \(F \rightarrow h(F)\). Let \(F \subseteq K\), with \([K: F]=n\). If \(h: F \rightarrow \mathbb{C}\) is a monomorphism, prove that there are exactly \(n\) monomorphisms \(K \rightarrow \mathbb{C}\) which are extensions of \(h\)

Find \(c\) such that \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{-5})=Q(c)\).

Find the root field of \(a(x)=\left(x^{2}-3\right)\left(x^{3}-1\right)\) over \(\mathbf{Q} .\) ANSWER The complex roots of \(a(x)\) are \(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\), so the root field is \(\mathbb{Q}\left(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\right)\). The same field can be written more simply as \(\mathbb{Q}(\sqrt{3}, i)\). Find the root field of \(x^{4}-2\), first over \(Q\), then over \(\mathbb{R}\).

In the following, let \(F\) be a subfield of \(\mathbb{C}\). An injective homomorphism \(h: F \rightarrow \mathrm{C}\) is called a monomorphism; it is obviously an isomorphism \(F \rightarrow h(F)\). Prove: The only possible monomorphism \(h: \mathbf{Q} \rightarrow \mathrm{C}\) is \(h(x)=x\). Thus, any monomorphism \(h: \mathbb{Q}(a) \rightarrow C\) necessarily fixes \(\mathbb{Q}\).

Find the root field of \(a(x)=\left(x^{2}-3\right)\left(x^{3}-1\right)\) over \(\mathbf{Q} .\) ANSWER The complex roots of \(a(x)\) are \(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\), so the root field is \(\mathbb{Q}\left(\pm \sqrt{3}, 1, \frac{1}{2}(-1 \pm \sqrt{3} i)\right)\). The same field can be written more simply as \(\mathbb{Q}(\sqrt{3}, i)\). Find irreducible polynomials \(a(x)\) over \(\mathbb{Q}\), and \(b(x)\) over \(\mathbb{( i )}\), such that \(Q(i, \sqrt{3})\) is the root field of \(a(x)\) over \(\mathbb{Q}\), and is the root field of \(b(x)\) over \(Q(i)\). Then do the same for \(Q(\sqrt{-2}, \sqrt{-3})\).

Let \(a(x)\) be a polynomial of degree \(n\) in \(F[x]\), and let \(K\) be the root field of \(a(x)\) over \(F\). Prove that \([K: F]\) divides \(n !\)