## § Integral elements of a ring form a ring [TODO ]

- An integral element of a field $L$ (imagine $\mathbb C$) relative to an integral domain $A$ (imagine $\mathbb Z$) is the root of a monic polynomial in $A$.

- So for example, in the case of $\mathbb C$ over $\mathbb Z$, the element $i$ is integral as it is a root of $p(x) = x^2 + 1$.
- On the other hand, the element $1/2$ is not integral. Intuitively, if we had a polynomial of which it is a root, such a polynomial would be divisible by $2x - 1$ (which is the minimal polynomial for $1/2$). But $2x - 1$ is not monic.
- Key idea: take two element $a, b$ which are roots of polynomial $p(x), q(x) \in A[x]$.
- Create the polynomial $c(x)$ (for construction) given by $c(x) \equiv p(x)q(x) \in A[x]$. See that $c(x)$ has both $a$ and $b$as roots, and lies in $A[x]$.