# The Ring extension isomorphic to the field extension

Let $\alpha$ be algebraic over $F$, with $F(\alpha)$ the smallest field containing both $F$ and $\alpha$, and with $F[\alpha]$ the smallest ring containing both $F$ and $\alpha$. I want to show $F(\alpha)=F[\alpha]$ which doesn't hold when $\alpha$ is transcendental (I think). The ring is clearly contained in the field, but given some $\sum a_k\alpha_k\in F(\alpha)$, I have no idea how to show it also exists in $F[\alpha]$.

## 1 Answer

Hint: All you have to prove is that, if α is algebraic over $F$, the $F[\alpha]$ is a field. Let $x=p(\alpha)\ne 0$. Consider multiplication by $x$ on $F[\alpha]$ and note 1) it is an $F$-linear map, 2) $F[\alpha]$ is a finite-dimensional $F$-vector space.

• oh, so you're saying if I define $T_x:a\mapsto ax$, then this map is bijective, hence must be 1 at some place? – serge hective Apr 7 '16 at 17:51
• It's exactly that. – Bernard Apr 7 '16 at 18:27
• You could also say that, if g is the minimal polynomial of a over F, then the ring F[a] is isomorphic in a standard way to the quotient ring F[t]/(g), where (g) is the principal ideal generated by g. But in the PID ring F[t], g is irreducuble iff (g) is maximal, hence F[a] is a field, and F[a] = F(a) , as desired . – nguyen quang do Apr 8 '16 at 13:57