This is the definition of integral over $R$:

Let $S$ be an extension ring of $R$ and $s\in S$. If there exists a monic polynomial $f(X)\in R[X]$ such that $s$ is a root of $f$ (i.e., $f(s)=0$), then $s$ is said to be integral over $R$.

My question is: why do we use monic polynomial!?


2 Answers 2


The ring of integers $\mathcal O_K$ of a number field $K$ is defined as the set of roots of monic polynomials with integer coefficients to mimic the relationship of $\Bbb Z$ to $\Bbb Q$: $a \in \Bbb Q$ is in $\Bbb Z$ iff it is the root of a (monic) polynomial of the form $x - b$ with $b \in \Bbb Z$.

If we relaxed the condition to allow arbitrary polynomials in $\Bbb Z[x]$, then any rational number $r/s$, being a root of $sx - r$, would be an "integer" and the ring of integers would coincide with the field it lies in (since $K = \operatorname{Frac}(\mathcal O_K$)), making the definition useless.

This motivates the definition of an integral element for arbitrary ring extensions too.


Well, if you have monic polynomial $f(x)\in R[x]$ of minimal degree with root $s$ and there is another polynomial $h(x)\in R[x]$ which has root $s$, then you can divide $h(x)$ by $f(x)$ as the leading coefficient of $f(x)$ is $1$ (or a unit in $R$). Otherwise, this would not be possible.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .