# Is the definition of integral extension, why we use monic polynomial?

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!?

• Sep 10, 2019 at 12:03

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.
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.