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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?

One Answer

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 $A$.

Correct answer by tkf on September 20, 2020

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