Let $K$ be a field and $\alpha$ be an algebraic element. Consider the following ring homomorphism by which $\phi:K[X]\to K(\alpha)$ by which $\phi(X)=\alpha.$ I was wondering how would one show such $\phi$ must be a surjection directly? ($K(\alpha)$ denotes the smallest field to contain both $K$ and $\alpha$.)
I wanted to show this is a surjection thus the minimal polynomial of $\alpha$ must be irreducible. (I know other ways of proving this statement but I thought it was interesting trying out this way)
I know that we can use FIT and thus $\frac{K[X]}{\langle m_{\alpha}(X)\rangle}\cong K(\alpha)$; however, via this route we need to use $m_\alpha$, the minimal polynomial, is irreducible and thus I do not want to end up in a circular argument. Hence I was wondering if there exists other clever way of showing such map must be surjective.
Many thanks in advance!
Best Answer
I think this is somewhat of an inefficient way of proving $m_\alpha(x)$ is irreducible (since you can just use the fact that $K[\alpha]$ is an integral domain). However, this is a proof using what you said.
This boils down to showing that $K[\alpha] = K(\alpha)$ since you already know that the image is $K[\alpha]$. Notice that $K[\alpha]$ naturally has the structure of a $K$-vector space and using the minimal polynomial you can see that this is finite dimensional. Hence, it is algebraic over $K[\alpha]$ and for $\beta \in K[\alpha]$ you can use that it has an inverse by using an annihilating polynomial. If $\sum_{i=0}^k \lambda_i \beta^i = 0$ for $\lambda_0,..., \lambda_k \in \mathbb{Q}$ we can write
$$\beta \left(\frac{1}{\lambda_0} \sum_{i=1}^k \lambda_i \beta^{i-1}\right) = 1$$
Note that we can assume $\lambda_0 \neq 0$ since we could just divide the polynomial by $x$ if $0$ was a root.
Hence, every element has an inverse and $K[\alpha] = K(\alpha)$ so the map is surjective.