Showing $K[u]$ is a field when $u$ is algebraic over $K$. Let $K$ be a field, and let $u$ be algebraic over $K$.  Show that $K[u]$ is a field.
Context/progress so far:  This is a generalization of a problem in the first few pages of Peterson's "Linear Algebra."  If possible, I'd like to avoid using the machinery from abstract algebra and show directly that elements of $K[u]$ are invertible.  I can show that $\{1, u, ..., u^{n-1}\}$ is a basis for $K[u]$ over $K$, where $n$ is the degree of the minimal polynomial of $u$.  Using the same polynomial, I can show that $u^{-1}$ exists.  But I'm having trouble inverting general elements of $K[u]$.  Any hints?
 A: I am not sure if this proof is what you want, but here goes :
Let $p(x)\in K[x]$ be the minimal polynomial of $u$, then
$$
K[u] = K[x]/(p(x))
$$
Consider a polynomial $f(X) \in K[X]$ such that $\overline{f}\neq \overline{0}$ in $K[u]$. Replacing $f$ by its remainder on division with $p$, we may assume without loss of generality that $deg(f) < deg(p)$.
Since $p$ is prime,
$$
gcd(f,p) = 1
$$
Thus, there exists $g(x), r(x) \in K[x]$ such that
$$
fg + rp = 1
$$
Thus
$$
\overline{f}\overline{g} = \overline{1}
$$
And hence $\overline{f}$ is invertible.
A: Suppose we have any element $x$ algebraic over $K,$ with minimal polynomial $x^n + a_1 x^{n-1} + \ldots + a_n =0.$ Then $x$ has inverse $(-a_n)^{-1} (x^{n-1}+ a_1 x^{n-2} + \cdots + a_{n-1}).$ Now the key is that sums or products of algebraic elements is algebraic. 
A: This result is only true if $u$ is supposed to live in an integral domain containing $K$ (for instance in a field extension, a context suggested by the term "algebraic"). I'll suppose this hypothesis.
That $u$ is algebraic means that $K[u]$ is a finite dimensional vector space over$~K$. Multiplication by any nonzero element $a\in K[u]$ is an injective (because $K[u]$ is an integral domain) $K$-linear map $K[u]\to K[u]$. By finite dimensionality, such $K$-linear maps are surjective as well; in particular the element$~1\in K[u]$ is in the image of multiplication by$~a$, which means that $a$ is invertible.
If you need to compute the inverse of $a$ explicitly, you can proceed as follows. Let $\mu\in K[X]$ be the minimal polynomial of$~u$ over$~K$, which is an irreducible monic polynomial of degree$~n$ (if it were reducible $\mu=PQ$, then $P[u]Q[u]=\mu[u]=0$ would contradict that $K[u]$ is an integral domain), and write $a$ as a polynomial (that can be taken to be of degree${}<n$) in$~u$, say $a=A[u]$. Then since $a\neq0$ and $\mu$ irreducible one has $\gcd(A,\mu)=1$, so if $S,T\in K[X]$ are Bezout coefficients $1=SA+T\mu$, then $S[u]$ is an inverse of$~a$ in$~K[u]$.
A: Will this work?:
Let $\phi(x) \in K[x]$ be the minimal polynomial of $u$.  Consider $f(u) \in K[u]$ where
$f(x) \in K[x]$.  I claim we can take $\deg f < \deg \phi$ without loss of generality.  Why?  If $\deg f \ge \deg \phi$, we can use the division algorithm in $K[x]$ to write $f(x) = \phi(x) q(x) + r_f(x)$, where $\deg r_f < \deg \phi$.  Then $f(u) = \phi(u) q(u) + r_f(u)$, whence, since $\phi(u) = 0$, $f(u) = r_f(u)$.  I claim that $\phi(x)$ must be irreducible in $K[x]$.  Why?  If $\phi(x) = \alpha(x) \beta(x)$ in $K[x]$, and neither $\alpha(x)$ nor $\beta(x)$ is constant, i.e. of degree $0$, then we would have $\alpha(u) \beta(u) = \phi(u) = 0$, whence either $\alpha(u) = 0$ or $\beta(u) = 0$. But $0 < \deg \alpha < \deg \phi$, $0 < \deg \beta < \deg \phi$, so this contradicts the minimality of $\phi(x)$.  Since $\phi(x)$ is irreducible, it is prime in $K[x]$, a principal ideal domain.  Since $\phi(x)$ is prime, we must have $\gcd(\phi(x), r_f(x)) = 1$, since we cannot have $\phi(x) \vert r_f(x)$ by virtue of the fact that $\deg r_f < \deg \phi$, whence there exist $g(x), h(x) \in K[x]$ such that $r_f(x)g(x) + \phi(x)h(x) = 1$.  But then $r_f(u)g(u) = 1$, since $\phi(u) = 0$.  QED.
P.S.  After the above breakdown, I must point out that the evaluation homomorphism
$\theta:K[x] \to K[u]$ has kernel precisely $(\phi(x))$ in $K[x]$, whence $K[u]$ is isomorphic to $K[x]/(\phi(x))$, a field since $\phi(x)$ is prime etc. etc. etc.  Less typing! ;)
