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 ? 
 A: 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 $u(a_1+\cdots+a_mu^{m-1})=1$. 
A: 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.
