TITLE: Showing $K[u]$ is a field when $u$ is algebraic over $K$.
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: 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$~K[u]$.