About the definition of an integral element in commutative rings

Question

Let $R \subset S$ be commutative rings with unity. The usual definition for an element $x \in S$ to be integral over $R$ is to require the existence of a monic polynomial with coefficients in $R$ such that $x$ is a root of that polynomial.

Why is it that we define it by the use of monic polynomials? Clearly, if an element is integral within this definition, it would also be integral if we would not require the polynomial to be monic. On the other hand, if an element would be the root of a polynomial with leading coefficient not a unit, it wouldn't be integral.

Furthermore, if we denote by $K$ the integral closure of $R$ according to the definition given above, and by $\tilde{K}$ we denote the "alternative integral closure" of $R$ (i.e. the set of all elements $x \in S$ s.t. there exists a polynomial - not necessarily monic - with coefficients in $R$ that vanishes at $x$) we get the relation

$$R\subset K \subset \tilde{K} \subset S$$

But why is $\tilde{K}$ apparently of less interest than $K$? After all, $\tilde{K}$ is not always equal to $S$, take e.g. $k\left[x\right] \subset k\left[x,y\right]$ for a field $k$ and consider the element $y \in k\left[x,y\right]$. Clearly, when talking about field extensions (i.e. about algebraic elements instead of integral ones) the sets $K$ and $\tilde{K}$ become equivalent. But it seems to me that in the case of general commutative rings the study of $\tilde{K}$ could also be of interest and lead to results worthwhile in their own rights.

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EDIT

I just realised that in the case of finite field extensions we might lose uniqueness of the minimal polynomial. Is this indeed the case? My assumption comes from the fact that when proving by contradiction, assuming there exist $p \neq q \in K\left[x\right] \subset L\left[x\right]$ both minimal of degree $n$, then $p-q$ is of strictly less degree but still vanishes at the element in question. This proof wouldn't work if the polynomials aren't required to be monic (the degree would not decrease necessarily). Maybe there is something similar to rings that we would loose?

Answer 1

This may be more an extended comment than a full answer, but maybe it helps.

First, consider the case $R=\Bbb Z$ and $S=\Bbb Z[\frac13]$. The extension should not be considered "integral" because, well, you added a fraction. And indeed, the minimal polynomial of $\tfrac13$ would be $3X-1\in\Bbb Z[X]$. As you can see, it is not monic.

I honestly feel that this example actually says it all: If you do not require the polynomial relations to be monic, then you will be adding something that somehow requires dividing by an element of $R$ which was not invertible. And that would simply not be "integral". You are free to add roots and stuff like that, but not invert.

If you have some context of algebraic geometry, the following might help, but otherwise feel free to ignore it as it will only be confusing:

From a geometric perspective, it means that you are allowed to bend, but you are not allowed to cut: The integral ring extension $\Bbb C[y]\subseteq \Bbb C[x,y]/(x^2-y)$ corresponds to projecting from a parabola to a line (the line was bent), while the extension $\Bbb C[y]\subseteq \Bbb C[y,y^{-1}]$ corresponds to the inclusion $\Bbb C^\times\subseteq \Bbb C$ (the origin was removed).