On the kernel of the map $k[X,Y]\to k[T],X\mapsto T, Y\mapsto T$ The following exercise, with a slightly more interesting choice of $\varphi$, is probably a standard exercise in, say, an introductory course in commutative algebra.

Let $k$ be a field. Show that the kernel of the homomorhpsim 
  $$\varphi:k[X,Y]\to k[T],X\mapsto T, Y\mapsto T$$
  is exactly the ideal generated by $X-Y$. 

A very straightforward solution is to use the division algorithm in the ring $(k[Y])[X]$ of polynomials in $X$ with coefficients in $k[Y]$. I wonder, however, whether this (longer and much more complicated) approach can be made rigorous: for the nontrivial inclusion, consider a polynomial $p(X,Y)$ which is in the kernel. If we write
$$p(X,Y)=\sum_{i,j\leq n}a_{ij}X^iY^j$$
then this conditions means that
$$0=P(T,T)=\sum_{i,j}a_{ij}T^{i+j}=\sum_{r}T^r\sum_{i+j=r}a_{ij}$$
i.e., 
$$A_r:=\sum_{i+j=r}a_{ij}=0\qquad\text{for all }r.$$
Now, we want to show that $p(X,Y)$ is divisible by $X-Y$, or equivalently the rational function $p(X,Y)/(X-Y)$ is a polynomial. But 
$$
\begin{equation}
\frac1{X-Y}
=\frac1X
\frac1{1-Y/X}
=\sum_{k\geq0}
Y^k
X^{-1-k}\qquad\qquad(*)
\end{equation}$$
so
$$\frac{p(X,Y)}{X-Y}
=\sum_{i,j,k}a_{ij}Y^{j+k}X^{i-1-k}
=\sum_{r,s}X^rY^s\sum_{\substack{j+k=s\\i-1-k=r\\k\geq0}}a_{ij}
=:\sum_{r,s}B_{rs}X^rY^s$$
and if we want this to be a polynomial we need that $B_{rs}=0$ if either $r<0$ or $s$ is large enough. In such cases it is not difficult to show that $B_{rs}=A_{rs}=0$, as wanted. 
Since everything seems to work, I guess the argument is fine as long as we know in which ring the equality $(*)$ holds. At first I though it was enough to consider the quotient field of the ring of formal power series in two variables $k[X,Y]$, but in $(*)$ there are coefficients in $X^{-k}$ with $k$ arbitrarily large. 
 A: We can justify $(*)$ by working in the ring $k(X)[[Y/X]]$, the ring of formal power series in one variable (which we think of as representing $Y/X$) with coefficients in $k(X)$.  We can then make sense of your equation $$\frac{p(X,Y)}{X-Y}=\sum_{r,s}B_{rs}X^rY^s$$ in this ring by interpreting $X^rY^s$ as $X^{r+s}(Y/X)^s$ and observing that we only need terms with $s\geq 0$ and that for fixed $s$ there are only finitely many $r$ with $B_{rs}\neq 0$.  To justify reaching conclusions about $k[X,Y]$, we then just need to observe that there is an injective homomorphism $k[X,Y]\to k(X)[[Y/X]]$ mapping $X$ to $X$ and $Y$ to $X\cdot Y/X$.
A: This is not what you asked, but for the record (and because you have already accepted an answer) here's a more conceptual proof of the claim : 
From the map $k[X,Y]\to k[T]$, it is clear that we get a map $k[X,Y]/(X-Y)\to k[T]$ (one inclusion is obvious), and we wish to show that it is an isomorphism. 
Note that this map is a $k$-algebra map. Now let $\{*\}$ be any one element set and consider $\{*\}\to k[X,Y]/(X-Y), *\mapsto [X]=[Y]$. Note that with this map, the obvious triangle commutes, where we have also $\{*\}\to k[T], *\mapsto T$. 
Finally, let $A$ be a (commutative unital associative) $ k$-algebra and $\{*\}\to A$ a map, $*\mapsto a$. Then by the universal property of $k[X,Y]$, there is a unique map $k[X,Y]\to A$ with $X\mapsto a, Y\mapsto a$. 
Again it is clear that this map factors through $k[X,Y]/(X-Y)$, and it sends $[X]$ to $a$, and it's clearly unique in doing so. 
Therefore, our map $\{*\}\to k[X,Y]/(X-Y)$ realizes this algebra as the free $k$-algebra on one generator; but this characterizes $k[T]$ up to unique commuting isomorphism; and actually there is a unique commuting morphism, which is an isomorphism. But wait, we found a commuting morphism, it's our map from the beginning : it must be an isomorphism. In particular it is injective and $(X-Y)$ is the kernel of the very first map.
(Note : I make it sound long, but you do't actually need to prove the whole universal property and you can actually define the converse quite easily, this us just more general)
