Why does $A_s = k[U,T,S]/(UT-S) \otimes_{k[S]} k[S]/(S-s)$ simplify to $ A_s = k[U,T] / (UT-s)$? When reading algebraic geometry (on the technique of base change) in the book Algebraic Geometry 1 - Schemes by Ulrich Gortz, et.al, I came up with the following tensor product:
$$
A_s = k[U,T,S]/(UT-S) \otimes_{k[S]} k[S]/(S-s), \quad \quad (\star)
$$
and the author claimed that
$$
A_s = k[U,T] / (UT-s).      \quad \quad (\star\star)
$$
My question is: how to simplify $(\star)$ to $(\star\star)$?
More on this question:
I have read Atiyah and MacDonald's Comm. Algebra and know what a tensor product is. Yet I have not been familiar with the concrete calculation of tensor products (though I know the universal property of tensor products, its relation with localization, its exactness and etc.)
So beside the above question, I hope to know that what is going on in your mind when calculating the tensor product?
For example, when calculating the quotient ring $k[x,y]/(y-x^2)$, we can imagine that $y-x^2 = 0$ and hence $y=x^2$, then in the ring $k[x,y]$, we can make $y$ be $x^2$ and the quotient ring is isomorphic to $k[x]$.
For example, when calculating the quotient ring $k[x,y]/(1-xy)$, we can imagine that $xy=1$ and hence $y=1/x$, then in the ring $k[x,y]$, we can make $y$ be $1/x$ and the quotient ring is isomorphic to $k[x, 1/x]$, or the localization $k[x]_{x}$.
Then, when calculating the tensor product, is there a way like these above in mind to help us calculate these?
Thank you all! :)

An arbitrary element of $A_s$ is a $k$-linear combination of elements of the form $p(U,T)S^i\otimes 1$, where $p$ is a polynomial over $k$, in two variables.  As we are tensoring over $k[S]$, we have the equality:
\begin{eqnarray*}p(U,T)S^i\otimes 1&=&p(U,T)\otimes S^i\\&=&p(U,T)\otimes s^i\\&=&p(U,T)s^i\otimes 1\end{eqnarray*}
Thus in effect we substitute in $s$ for $S$ in the ring $k[U,T,S]/(UT-S)$, resulting in $k[U,T]/(UT-s)$.
