Let \(F\) be a field and \(p(x)\in F[x]\) an irreducible polynomial.
Prove that there exists a field extension \(K\) of \(F\) in which \(p(x)\) has a root.
Determine the dimension of \(K\) as a vector space over \(F\) and exhibit a vector space basis for \(K\).
If \(\theta\in K\) denotes a root of \(p(x)\), express \(\inverseof{\theta}\) in terms of the basis found in part (b).
Suppose \(p(x)=x^3+9x+6\). Show \(p(x)\) is irreducible over \(\mathbb Q\). If \(\theta\) is a root of \(p(x)\), compute the inverse of \((1+\theta)\) in \(\mathbb Q(\theta)\).
Solution
1. There exists an extension field \(K/F\) in which \(p\) has a root.
Let \[ K=F[x]/(p(x)). \] Because \(p\) is irreducible, the ideal \((p)\) is maximal in the PID \(F[x]\), so \(K\) is a field. If \[ \theta=x+(p)\in K, \] then by construction \(p(\theta)=0\).
2. If \(n=\deg p\), then\[ [K:F]=n, \qquad 1,\theta,\theta^2,\ldots,\theta^{n-1} \]is an \(F\)-basis of \(K\).
Every class in \(F[x]/(p)\) has a unique representative of degree \(<n\) by Euclidean division by \(p\). Hence the displayed powers span. If \[ a_0+a_1\theta+\cdots+a_{n-1}\theta^{n-1}=0, \] then the polynomial \(a_0+a_1x+\cdots+a_{n-1}x^{n-1}\) lies in \((p)\). Its degree is \(<\deg p\), so it must be zero. Thus the displayed powers are linearly independent.
3. Assume \(p(0)\ne0\).Write\[ p(x)=a_0+a_1x+\cdots+a_nx^n, \qquad a_0\ne0. \]Then\[ \theta^{-1} =-\frac{a_1+a_2\theta+\cdots+a_n\theta^{n-1}}{a_0}. \]
Since \(p(\theta)=0\), \[ a_0+\theta(a_1+a_2\theta+\cdots+a_n\theta^{n-1})=0. \] Because \(a_0\ne0\), necessarily \(\theta\ne0\), so division by \(\theta\) gives the formula.
As the source problem is written, the hypothesis \(p(0)\ne0\) is necessary: if \(p(x)=x\), then \(p\) is irreducible but its root \(\theta=0\) has no inverse.
4. The polynomial\[ p(x)=x^3+9x+6 \]is irreducible over \(\mathbb Q\).
A reducible cubic over a field has a root in that field. Since \(p\) is monic with integer coefficients, any rational root must be an integer divisor of \(6\). Direct substitution of \[ \pm1,\ \pm2,\ \pm3,\ \pm6 \] shows that none is a root. Hence \(p\) has no rational root and is irreducible over \(\mathbb Q\).
5. If \(\theta^3+9\theta+6=0\), then\[ (1+\theta)^{-1}=\frac{\theta^2-\theta+10}{4}. \]
Using \(\theta^3=-9\theta-6\), \[ \begin{aligned} (1+\theta)(\theta^2-\theta+10) &=\theta^3+9\theta+10\\ &=(-9\theta-6)+9\theta+10\\ &=4. \end{aligned} \] Therefore dividing by \(4\) gives the inverse.