Show that the polynomial ring in $n-1$ variables is isomorphic to the polynomial ring in one variable $x_{n}$ For a ring $R$ and for $n \geq 1$, define $ S := R[x_{1},...,x_{n-1}]$ for the polynomial ring in $n-1$ variables with coefficients in $R$. Show that $R[x_{1},...,x_{n}]$ is isomorphic to the polynomial ring $S[x_{n}]$ in one variable $x_{n}$ with coefficients in $S$The hint I have been given is to write down a map between two rings and prove it is an isomorphism. I'm not sure how to even begin however.
 A: Here's the intuition for this problem: a polynomial in $n$ variables (something in $R[x_1,\dots,x_n]$) is exactly the same as a polynomial in one variable whose coefficients are polynomials in $n-1$ variables (something in $S[x_n]$). For example, $x_1x_2^3x_3 + 4x_1x_2 + x_1x_3^2 + x_3$ is the same as $x_1\cdot x_3^2 +(x_1x_2^3+1)\cdot x_3 + 4x_1x_2$. In any case, the isomorphism is (I hope!) so obvious that there's hardly anything to write down. Formally, you could define an isomorphism $\phi:S[x_n]\to R[x_1,\dots,x_n]$ by $$
\phi\big(p_d\,x_n^d + \cdots p_1\,x_n + p_0\big) = p_d\,x_n^d + \cdots p_1\,x_n + p_0,
$$
where $p_0,p_1,\dots,p_d$ are polynomials in $n-1$ variables with coefficients in $R$. It's clear that $\phi$ is an isomorphism: both injectivity and surjectivity are obvious.
A: If you have a polynomial in $n$ variables you can factor out the last variable $x_n$ from every term and collect terms with like powers of $x_n$.  This way you can view a polynomial in $n$ variables as a polynomial in $x_n$ over polynomials in $n-1$ variables.  If you want to be formal you can use this idea to define the isomorphism.  
A: More generally, $ R [A][B]\cong R [A\sqcup B] $ for sets $ A $ and $ B $. This follows from the universal property of the polynomial ring.
