Showing $K[u]$ is a field when $u$ is algebraic over $K$.

Solution 1:

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.

Solution 2:

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.

Solution 3:

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$~K[u]$.