# degree of minimal polynomial and degree of field extension

Mathematics Asked by Ton910 on September 20, 2020

Let $$K$$ be a field and $$a in K$$ algebraic.

Then we have the minimal polynomial $$m_a in K[X]$$ with $$deg(m_a) = n = [K(a):K]$$

This implies that $$A={1,a,a^2,…,a^{n-1}}$$ is linearly independent, because otherwise $$m_a$$ would not be the minimal polynomial. So these are $$n$$ linearly independent vectors in a vector space of dimension $$n$$ over $$K$$, therefore $$A$$ is a basis.

But intuitively i don’t see how this is a basis. I see that you can build $$a^n$$ from elements of $$A$$ because we have $$m_a$$ but in general how can we conclude that we can get $$a^k$$ for some $$k > n$$ from a linear combination of elements in $$A$$?

In other words is it possible to proof that $$A$$ is a basis of $$K(a)$$ without knowing that $$[K(a):K] = n$$? And if so how?

Given a polynomial $$p(x)in K[x]$$, we may divide by the minimal polynomial $$m_a$$ and take the remainder $$r(x)$$. Then $$p(a)=r(a)$$, which is a linear combination of elements of $$A$$.