Is $\alpha$ (an algebraic number) integral over $\mathbb{Z}$?
(a) $\alpha_1:=\frac32$
(b) $\alpha_2:=\frac{(1+\sqrt2)}{2}$
(c) $\alpha_3:=\frac{(1+\sqrt5)}{2}$
In the lecture we just had the definition:
Given commutative rings $R \subset S$, we say that $x \in S$ is integral over $R$ if there exists a monic polynomial $f \in R[X]$ such that $f(x) = 0$.
So for (c) I found that $\alpha_3$ satisfies $x^2-x-1=0$, so $f_3(x):=x^2-x-1\in\mathbb{Z}[X]$ is monic and therefore $\alpha_3$ is integral over $\mathbb{Z}$.
For (a) I did $x-\frac32=0\iff 2x-3=0$ (simplified the denominator since we're in $\mathbb{Z}$), while for (b): $x-\frac{(1+\sqrt2)}{2}=0\iff 2x-1-\sqrt2=(2x-1)-\sqrt2=0\iff $
$((2x-1)-\sqrt2)((2x-1)+\sqrt2)=0\iff 4x^2-4x-1=0$ (so I simplified the square root)
For both (a) and (b) I would say that since $\gcd(2,3)=1$ (which implies $\gcd(2^n,3^n)=1$) and $\gcd(4,1)=1$ (which implies $\gcd(4^n,1^n)=1$), there is no monic polynomial such that $f(\alpha_{1,2})=0$.
I don't know if it is right to affirm such thing. Any help would be great, thank you.
If $\alpha$ satisfies an irreducible primitive polynomial $p(X)$ over $\mathbb Z$, that polynomial must be the minimal polynomial of $\alpha$, i.e. it generates the ideal $J(\alpha) = \{f(X) \in \mathbb Z[X]: f(\alpha)=0\}$ of polynomials satisfies by $\alpha$. Every nonzero polynomial in the ideal is $p(X) q(X)$ for some $q(X) \in \mathbb Z[X]$, and the leading coefficient of such a polynomial is an integer multiple of the leading coefficient of $p(X)$. Thus if $p(X)$ is not monic, $\alpha$ is not integral.