# Algebraic Closure of a Ring is Not a Ring?

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.

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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.
Actually I do want $R$ to be a domain. – Drew Armstrong Apr 18 '14 at 16:59
To be specific: Let $A$ be a finitely generated $K$-algebra (with no zero divisors), say $A=K[u_1,\ldots,u_n]$. I want to prove that every maximal algebraically independent subset of the generators is a transcendence basis in the sense that if $\{u_1,\ldots,u_d\}$ is such a set then every element of $A$ is algebraic over $K[u_1,\ldots,u_d]$. The proof I'm looking at passes to the field of fractions of $A$ and I don't see why we really need to do that. – Drew Armstrong Apr 18 '14 at 17:05
A satisfactory explanation of the proof is here: math.stackexchange.com/questions/759345/… I suppose we could prove directly that the algebraic closure of $R$ in $S$ is a ring by using resultants to construct equations, but I suppose passing to the field of fractions is quicker and easier... – Drew Armstrong Apr 18 '14 at 17:18