Theorem 1

— Let $A$ be an algebra over a field $K$, $x$ an element of $A$ algebraic over $K$, $n$ the degree and $f$ the minimal polynomial of $x$ over $K$.

a) For a polynomial $g \in K[X]$ to satisfy $g(x) = 0$ it is necessary and sufficient that $g$ should be a multiple of $f$.

b) The mapping $g \mapsto g(x)$ defines by passage to quotients an isomorphism of the quotient algebra $K[X]/(f)$ onto the algebra $K[x]$, and the elements $1,\ x,\ \ldots,\ x^{n-1}$ form a basis of $K[x]$ over $K$. In particular, $[K[x]:K] = n$.

c) Assume that $A$ is an integral domain. Then $K[x]$ is a field and $f$ the unique monic irreducible polynomial in $K[X]$ such that $f(x) = 0$.

d) For $x$ to be invertible in $A$ it is necessary and sufficient that $f(0) \neq 0$; then we have $x^{-1} \in K[x]$.

There exists a unique algebra homomorphism $\varphi : K[X] \to A$ such that $\varphi(X) = x$; we have $\varphi(P) = P(x)$ for every $P \in K[X]$ and the image of $\varphi$ equals $K[x]$. Let $a$ be the kernel of $\varphi$: by construction, the minimal polynomial $f$ of $x$ over $K$ belongs to $a$ and it is the monic polynomial of least degree in $a$. Therefore ($IV$, p. 11, Prop. 11) we have $a = (f)$, whence $a$. Assertion $b$) follows at once from a) and the Cor. of Prop. 10 of $IV$, p. 11.

Suppose that $A$ is an integral domain. The algebra $K[x]$ is then an integral domain and of finite degree over $K$, hence it is a field ($V$, p. 10, Cor.). The ideal $(f)$ of $K[x]$ is thus maximal, that means that $f$ is irreducible in $K[X]$ ($IV$, p. 13). Finally, let $g$ be a monic irreducible polynomial in $K[X]$ such that $g(x) = 0$; by $a$) it is a multiple of $f$, hence $g = f$, and this proves $c$).

It remains to prove $d$). There exists a polynomial $g \in K[X]$ of degree $n-1$ and an element $a$ of $K$ such that $f(X) = Xg(X) + a$, whence $f(0) = a$. If $a = 0$, we have $xg(x) = f(x) = 0$, and $g(x) \neq 0$, so then $x$ is not invertible in $A$. If on the other hand $a \neq 0$, then we have $x \cdot [-a^{-1}g(x)] = 1$, so $x$ is then invertible in $A$ and $x^{-1} = -a^{-1}g(x)$.