Suppose the degree of a field extension $[\mathbb{Q}(\alpha):\mathbb{Q}]=n\gt 1$ and $\alpha$ is a root of a monic polynomial $f \in \mathbb{Q}[T]$ and the degree of $f$ is $n$.

Does the above imply that $f$ is irreducible and hence the minimal polynomial of the extension?

I'm wondering if that is the case.


If $\alpha$ is a root of $f$, then the minimal polynomial $p_\alpha$ of $\alpha$ divides $f$. Recall that $p_\alpha$ has degree $n=[\mathbb Q(\alpha):\mathbb Q]$, since $\mathbb Q(\alpha)\simeq \mathbb Q[x]/(p_\alpha)$.

Since $f$ is monic and a multiple of $p_\alpha$ of the same degree, it follows that $f=p_\alpha$ so that $f$ is irreducible. However, notice that we've used that $f$ is the minimal polynomial to deduce irreducibility -- I'm not sure about proving irreducibility first, without using some variant of the fact that $f$ must be the minimal polynomial.

Edit: Alternately, we can note that if $f(x)=g(x)h(x)$ for $g$, $h$ nonconstant polynomials, then $\alpha$ is a root of either $g$ or $h$. This implies that the minimal polynomial of $\alpha$ over $\mathbb Q$ has degree less than $[\mathbb Q(\alpha):\mathbb Q]$, a contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.