I'm trying to motivate the notion of integrality in a ring extension. It seems that the following would be a good motivation, because it would show that the notion of algebraic elements over a ring is not useful.


Here's the thing I believe is true: Let $R\subseteq S$ be a ring extension. The set of elements of $S$ that are algebraic over $R$ (i.e. satisfy a polynomial equation with coefficients in $R$) is **not** necessarily a ring.


But I can find no discussion of this point on the internet. Can someone provide a concrete example? Thanks.



Algebraic Closure of a Ring is Not a Ring?
You definition of "algebraic" is very strange when $R$ is not a domain. Due to that strangeness, it is easy to find examples: Let $k$ be a field, let $R = k \times k$ and let $S = k[t] \times k[t]$. Then $(t,0)$ obeys $(0,1) \cdot (t,0)=(0,0)$ and $(0,t)$ obeys $(1,0) \cdot (0,t) = (0,0)$.


However, I claim that $(t,t)=(t,0) + (0,t)$ is not algebraic over $R$. If 
$$\sum (a\_i, b\_i) (t,t)^i=0$$
then $\sum a\_i t^i=\sum b\_i t^i=0$ in $k[t]$, so all $a\_i$ and all $b\_i$ are zero.


A more reasonable definition would probably be to require that your element satisfy a polynomial which is not zero modulo any minimal prime of $R$. But perhaps the failure of the unreasonable definition is good enough for your purposes.


