If $a$ is algebraic, prove that there is a minimal polynomial $p(x)$ in $Q[x]$ such $p(a)$ = $0$.

Question

If $f_a$: $Q[x]$ -> $C$ is the evaluation at $a$ map, then a polynomial $q(x)$ in $ker(f_a)$ is irreducible iff it generates $ker(f_a)$.

Let $ker(f_a)$ = $h(x)$ so that $h(x)$ is irreducible and $f_a(a)$ = $0$. Dividing $h(x)$ by its leading coefficient, we get a monic irreducible polynomial that vanishes at $a$.For uniqueness, if there are polynomials $p(x)$, $P(x)$ both monic and irreducible such that $p(x)$ = $P(x)$ = 0, then how do we conclude that $ker(f_a)$ is generated by $p(x)$ and $P(x)$? And that $p(x)$, $P(x)$ are associates? Does this show that $p(x)$ = $P(x)$?

Answer 1

[Supposing $Q$ is a field] Take the ideal $I=\{p(x)\in Q[x]:p(a)=0\}$. Because $Q$ is a field you have a division algorithm so the ideal $I$ is principal; i.e. exists a $p(x)\in Q[x]$ so that $I=(p(x))$. Now if $a_n$ is the coefficient of the greatest power of $p(x)$ take the polynomial $q(x)=p(x)/a_n$. Obviously $q(x)$ vanishes at $a$, it is monic and has to be irreducible over $Q$ because otherwise you can find a polynomial in $I$ that it is not in $(q(x))=(p(x))$. Furthermore, uniqueness follows from the fact that if $q_1(x)$ and $q_2(x)$ are such that $I=(q_1(x))=(q_2(x))$ then they must be the same up to a constant multiplication (otherwise you could divide one by the other and...)

Answer 2

you can do it easily by using that the minimal polynomial has lowest degree.

if there is another q(x) which is minimal and monic and deg(q(x)) = deg(p(x)) then p(x) - q(x) has degree one less than p(x) and p(a) - q(a) = 0

Answer 3

Since the polynomial ring over a field is a principal ideal domain, every ideal in $\Bbb Q[x]$ is principal; in particular, there is some $p(x)\in\Bbb Q[x]$ such that $\ker f_a=(p).$
We can choose $p$ to be monic, by multiplying by a scalar if necessary.
Now all that is left is to show that $p$ is irreducible and is the unique monic $p$ such that $\ker f_a=(p);$ and I shall leave this to you. :P

Hope this helps.
If some point is unclear or inappropriate, please point it out, thanks.