# finding the multiplicative inverse in a field

Let $L/K$ be a field extension. Let $a\in L$ and $K[a]=\{p(a)\;|\; p\in K[x]\}$; then $K[a]$ is clearly an integral domain. I want to show that when $a$ is algebraic over $K$, then $K[a]$ is a field.

So I want to know what is the multiplicative inverse of a non zero element $p(a)\not =0$ for some $p\in K[X]$. Since $a$ is algebraic it has a minimum polynomial $m\in K[X]$ now $m$ and $p$ are coprime so there exists $\alpha$ and $\beta$ in $K[X]$ such that $\alpha m+\beta p=1$; passing to evaluation on $a$ we get $\beta(a).p(a)=1$ hence $p(a)$ is invertible with inverse $\beta(a)$. My only doubt is about $m$ and $p$ being coprime. I think a justification is that $p$ is irreducible in $K[X]$ and that $p$ can not be a multiple of $m$ is that correct?

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Yeah, $p$ and $m$ are coprime. By definition, $m$ is irreducible over $k$. Now, if $p$ and $m$ are not coprime, then $p$ would have an associate of $m$ as a factor ($gcd(m,p) =$ some associate of $m$), which is impossible, since $p(\alpha) \neq 0$. – Rankeya Nov 5 '11 at 14:18
I think this is a nice way to think about why $k[\alpha]$ is a field, as in it is a more direct way. I would always prove it indirectly, because it is true that $k[x]/(m(x)) \cong k[\alpha]$, and $k[x]/(m(x))$ is a field. – Rankeya Nov 5 '11 at 14:24
1) what do you mean by "associate" ? 2) $k[x]/(m(x))$ is obviously a field but why $k[x]/(m(x)) \cong k[\alpha]$ is the iso induced by the evaluation $eval_{\alpha}:K[X]\rightarrow K[\alpha];\;p\mapsto p(\alpha)$? – palio Nov 5 '11 at 14:55
By associate I mean a unit times $m$. Well, $ev_{\alpha}:k[x] \rightarrow L$ has image $k[\alpha]$ and kernel $(m(x))$. – Rankeya Nov 5 '11 at 14:58

two ways are presented here to answer why $K[a]$ is a field for a given algebraic element $a$.
The direct way : Since $K[a]$ is clearly an integral domain, it is enough to show that any non zero element $p(a)\not = 0$ has a multiplicative inverse. Now since $a$ is an algebraic element then it has a minimal polynomial $m$. Now it is clear that $m$ and $p$ are coprime since $m$ is irreducible and $p$ can not be a multiple of $m$ because $p(a)\not = 0$. It follows that there exist $\alpha$ and $\beta$ in $K[X]$ such that $\alpha .m+\beta . p=1$. Now evaluate this identity on $a$ to get $\beta(a) . p(a)=1$ hence $p(a)$ is invertible with inverse $\beta(a)$.
The indirect way : consider the evaluation map $$ev_a:K[X]\rightarrow L;\; p\mapsto p(a)$$ the map $ev_a$ is clearly a ring homomorphism with image $K[a]$ and kernel the principal ideal $(m)$ generated by the minimal polynomial $m$ which is a maximal ideal because $m$ is irreducible, therefore we have a field isomorphism $k[x]/(m(x)) \cong k[a]$