Let $k$ be a field, and consider a polynomial ring $R=k[x_1,x_2,\dots,x_n]$ with ideal $I = (x_1-x_2x_3,x_2-x_3x_4,\dots,x_{n-2}-x_{n-1}x_n)$. Now I claim that
$$R/I\cong k[x_{n-1},x_n]=:S.$$
To me, it seems obvious. Take the composition $S\hookrightarrow R\rightarrow R/I$, the first being the obvious inclusion, the second the quotient map. Then we have a well defined map $S\to R/I$ by $p\mapsto [p]$. By the iterative nature of the generators of $I$, it is clear that it is surjective, since by iteration every $[f]\in R/I$ has a representative in only the variables $x_{n-1},x_n$. In fact, you can show $[x_i]=[x_{n-1}^{F_{n-i-1}}x_n^{F_{n-i}}]$, where $F_k$ is the $k$'th Fibonacci number.
To show injectivity, however, we need to argue that every class has only one representative in only the variables $x_{n-1},x_n$, and this is where I'm having difficulties. I imagine there is an easy proof, so I won't discuss my many half-baked ideas. Can anyone see how to do it?
EDIT: Since interest in this question hasn't been high, let me present some of those half-baked ideas.
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@George has a good treatment of the problem, reducing it to proving that every polynomial has a factor in which some $x_i$ for $i\neq n-1,n$ occurs. This claim seems obvious, but arguing it formally is causing difficulties for me. This is equivalent to showing that there is no non-trivial linear combination $\sum_i^{n-2} p_i\cdot(x_i-x_{i+1}x_{i+2}) = 0$, where $p_{n-2}$ has no pure $x_{n-1},x_n$-part.
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Another method could be by induction over $n$, since for $n=2$ the case is trivial. For the step, you could think of $f\in[f]\in R/I$ as a polynomial in $x_n$ over $k[x_1,\dots,x_{n-1}]$ and use the induction hypothesis on each "coefficient", so you have unique representations in the variables $x_{n-2},x_{n-1}$. I haven't been able to use this effectively though.
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Finally, although the naïve degree function is not well defined, I thought you could be able to refine the notion as the degree of $[f]$ being the degree of (the) a representative in only the variables $x_{n-1},x_n$. You could do the degree in $x_{n-1}$, the degree in $x_n$ and the multivariable degree, and then showing these notions are well defined should probably give us uniqueness of the representative.
Best Answer
I found a way to solve it using the theory of Gröbner basis. Assume $n \geq 3$ (the nondegenerate case).
Let $f_i = X_i - X_{i+1} X_{i+2}$ for $1 \leq i \leq n-2$. By definition $I = (f_1, \dots, f_{n-2})$.
Claim: $\{f_i\}_i$ is a Gröbner basis of $I$ with respect to the lexicographic order such that $X_1 > X_2 > \cdots > X_n$. We'll prove this in the end.
This implies that, for every $g \in \mathbb{K}[X_1, \dots, X_n]$, there exist $g_i \in \mathbb{K}[X_1, \dots, X_n]$ and a unique $r \in \mathbb{K}[X_1, \dots, X_n]$ such that:
This implies the isomorphism $\mathbb{K}[X_1, \dots, X_n]/I \cong \mathbb{K}[X_{n-1}, X_n]$.
The only thing that remains to show is that $\{f_i\}_i$ is a Gröbner basis form $I$ with respect to the already defined order. For this, one needs to check that the S-polynomials $S(f_i, f_j)$ vanish modulo $\{f_i\}_i$.
Without loss of generality, suppose $i < j$. Then: $$ S(f_i, f_j) = \stackrel{\textrm{computation}}{\cdots} = X_i X_{j+1} X_{j+2} - X_j X_{i+1} X_{i+2} = X_{j+1}X_{j+2} f_i - X_{i+1} X_{i+2} f_j. $$
This shows that every $S$-polynomial vanishes module $\{f_i\}_i$ and hence that $\{f_i\}_i$ is a Gröbner basis. The claim is completed.