# 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.

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?

• I believe you're looking for the "minimal polynomial". Aug 10, 2022 at 12:33

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