Let $K$ be a field and $E$ an extension of $K$. Suppose that $a \in E$ is algebraic over $K$. Show that $K[a]$ is a field. (You can assume that $K[a]$ is an integral domain).

Since $a \in E$ is algebraic over $K$ there exists a polynomial $f \in K[x]$ such that$f(a)=0$. Now I think that if $f$ is irreducible, then $K[a] \cong K[x]/\langle f(x) \rangle$ so $K[a]$ is a field since $\langle f(x) \rangle$ is maximal?

This happens only if $f$ is irreducible which I don't know it is and $a \in E$ being algberaic doesn't imply that $f$ need to be irreducible.

Is there a missing condition here or can this be shown for $f$ not irreducible also?

  • $\begingroup$ I believe you're looking for the "minimal polynomial". $\endgroup$
    – usc phd
    Aug 10, 2022 at 12:33

1 Answer 1


$a\in E$ algebraic over $K$.

Not all all anhaliting polynomial of $a$ i.e $f(x) \in K[x]$ with $f(a) =0$ works. You need minimal polynomial (i.e the least degree monic polynomial $m_a(x) \in K[x]$ with $m_a(a) =0$ )

Let $m_a(x) \in K[x]$ is the minimal polynomial of $a$ over $K$.

Then ( Try to prove)

  1. $ m_a(x) $ is irreducible polynomial in $K[x]$

  2. $K[a] \cong K[x]/\langle m_a(x) \rangle$

  1. $[K[a]:K]=\deg m_a(x) $

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