Why Is Inverse of a Polynomial Still a Polynomial? The fourth paragraph of the wikipedia article algebraic extension states that $K[a]$ is a field, which means every polynomial has an inverse. The inverse has to be a polynomial over $K$ as well. It seems it requires $K$ to be a non-integral domain. How do we resolve the confusion? How do we prove this paragraph? 
 A: Polynomials over a field $K$, that is, elements of $K[x]$, don't generally (unless they have degree zero) have an inverse.
$K[a]$ is a different object, it is isomorphic to the quotient ring $K[x]/(m)$ where $m$ is the minimal polynomial of $a$ (you can see this via the first isomorphism theorem using the morphism $\varphi:K[x]\to K[a]$ which sends $\sum k_i x^i\mapsto\sum k_i a^i$), assuming $a$ lives inside a field extension of $K$ its minimal polynomial will be irreducible, so the ideal $(m)$ it generates will be maximal and the quotient $K[x]/(m)$ will be a field.
Inverses in this field can be computed via the usual euclidean algorithm, take a polynomial $f\in K[x]$ with $\gcd(f,m)=1$, then by Bezout's theorem there are $h$ and $g$ such that $fh+gm=1$, but in the quotient ring $gm=0$ so $h$ is the inverse of $f$
A: First, $\;f(a)\;$ is not a polynomial but the evaluation of a polynomial at $\;a\;$ , meaning: a number. 
The complete claim is: If $\;K\;$ is a field and $\;F\;$ is some extension of $\;K\;$ , and $\;a\in F\;$ is algebraic over $\;K\;$ , then
$$K[a]:=\left\{\,f(a)\;/\;f(x)\in K[x]\,\right\}\;\;\text{is a field, which means}\;\;K[a]=K(a)$$
The proof: since $\;a\;$ is algebraic over $\;K\;$ there exists $\;0\not\equiv m_a(x)\in K[x]\;$ of minimal degree s.t. $\;m_a(a)=0\;$ , but then:
$$m_a(x)=\sum_{k=0}^{n-1}b_kx^k+x^n\implies b_0=-a^n-\sum_{k=1}^{n-1}b_ka^k\;\;\color{red}{(*)}$$
(...Observe the claim is trivial if $\;a\in K\;$ , and we can assume $\;b_0\neq0\;$ (why?) ...)
Yet $\;0\neq a\in F\;$ and $\;F\;$ is a field, so $\;a^{-1}\in F\;$ , and then multiplying $\;\color{red}{(*)}\;$ by $\;a^{-1}\;$:
$$a^{-1}=-\frac1{b_0}\left(a^{n-1}+\sum_{k=1}^{n-1}b_ka^{k-1}\right)\in K[a]$$
Deduce now that $\;K(a)\subset K[a]\;$ , and since the other contention is trivial if we remember $\;K(a)\;$ is the fraction field of the integral domain $\;K[a]\;$ , we're done.
