# Polynomial ring and extension field

Let $$K$$ be a field and $$p(x) \in K[x]$$ a monic irreducible polynomial of degree $$n$$. Suppose $$F/K$$ is a field extension, and there exists $$u \in F$$ which is a root of $$p(x)$$.

1) Let $$K(u)$$ be the smallest subfield of $$F$$ containing both $$K$$ and $$u$$. Prove that $$K(u) \cong K[x]/(p)$$.

2) Deduce that $$K(u)=\{a_0+a_1u+\cdots+a_{n-1}u^{n-1}:a_0,a_1,...a_{n-1}\in K \}$$

3) Compute $$u^{-1} \in K(u)$$.

I guess that first one can be shown by fundamental theorem of homomorphism, and the second one by division algorithm, but how to do the third one ?

Hint: Let $p(x)=a_0+a_1x+\cdots+a_mx^m$. Since $a_0\ne 0$, we can assume that $a_0=-1$. Then $u(a_1+\cdots+a_mu^{m-1})=1$.
• Remark that such inversion by rearrangement of a minimal polynomial can be viewed as a generalization of computing $\,\alpha^{-1} = \dfrac{1}{\alpha}\,$ by rationalizing the denominator - see my answer. – Bill Dubuque May 5 '14 at 15:49
Hint $\$ A root $\alpha$ of a polynomial $\,f(x)\,$ divides its constant term $\,\alpha\mid f(0)$. Over a field, if $\,\alpha\ne 0\,$ then by cancelling $\,\alpha^k$ we may assume that $\,f(0)\ne 0\,$ so $\,f(0)\,$ is a unit, thus so is its factor $\,\alpha.$
Remark $\$ The above rearrangement of a minimal polynomial to show that a root divides the constant term is simply a generalization of rationalizing denominators. Indeed a quadratic $\,\alpha\,$ has minimal polynomial $\,f(x) = (x-\alpha)(x-\alpha') = x^2-tx + n,\,$ $\, t = \alpha+\alpha',\,$ $\, n = \alpha\alpha',\,$ hence rearranging $\,\alpha^2 - t\alpha + n = 0\,$ yields $\,\alpha(t-\alpha) = n,\,$ hence $\,\alpha((t-\alpha)/n) = 1,\,$ therefore $\,\alpha^{-1} = \dfrac{t-\alpha}n = \dfrac{\alpha'}{\alpha\alpha'},\,$ precisely the result of rationalizing the denominator of $\,\dfrac{1}{\alpha}.\,$
The same method works for higher degree algebraic numbers by multiplying the denominator by all other conjugates, i.e. using $\,\alpha\mid \alpha\alpha'\cdots \alpha^{(n)} = N(\alpha),\,$ for $\,N =$ norm. This works because the norm is a nonzero element of a field, hence is invertible. Essentially this exploits the norm homomorphism in order to reduce the problem of division in $\,K(\alpha)\,$ to the simpler problem of division in $\,K.$ The key idea of the hint is that we can find such a simpler multiple of $\,\alpha\,$ already in the constant term of the minimal polynomial.