If $R$ is a commutative ring with identity then $R[x_1, x_2, ..., x_n]$ is not a P.I.D.?

Question

I am working on problem $7$, section $9.1$ in Dummit & Foote:

Let $R$ be a commutative ring with $1$. Prove that a polynomial ring in more than one variable over $R$ is not a principal ideal domain.

My Strategy:

It is enough to show it for polynomials of two variables. Then, for any polynomial ring $R[x_1, x_2, ..., x_n]$ we can view it as $R[x_1, x_2, ..., x_{n-2}] [x_{n-1}, x_n]$ (I think). And since $R[x_1, x_2, ..., x_{n-2}]$ is commutative with identity, $R[x_1, x_2, ..., x_n]$ will not be a P.I.D.

The Difficulty:

I am having trouble proving it for two variables. The issue is that there may be zero divisors in $R$. The previous problem in the book asked to prove that $(x, y)$ is not a principal ideal in $\mathbb{Q}[x, y]$. I proved it using the theorem that $\deg p \cdot q = \deg p + \deg q$ when there are no zero divisors in the coefficient ring. So I would like to also show that $(x,y)$ is not principal in $R[x, y]$ when $R$ is just commutative with identity, but I don't know how to get around the issue of possible zero divisors.

Answer 1

If $R$ has zero divisors, then so does $R[x,y]$, whence $R[x,y]$ is not a principal ideal domain, since by definition a principal ideal domain is also an integral domain.

Answer 2

Incidentally, even if we drop the integrality condition, $R[x,y]$ still fails to be a principal ideal ring: a single polynomial is just not enough to generate all the elements of the ideal $(x,y)$. For instance, assuming $(x,y)=(r(x,y))$ and using the notation $r(x,y)=\sum\,r_{ij}\,x^iy^j\in R[x,y]$ from now on, we can tinker a bit with some coefficients and see that $$\begin{array}{ccc} r(x,y)=x\,\lambda(x,y)+y\,\mu(x,y)&\Rightarrow& r_{00}=0,\,r_{10}=\lambda_{00},\,r_{01}=\mu_{00}\\ x=r(x,y)\,p(x,y)&\Rightarrow&r_{00}\,p_{10}+r_{10}\,p_{00}=r_{10}\,p_{00}=1\\ y=r(x,y)\,q(x,y)&\Rightarrow&r_{00}\,q_{01}+r_{01}\,q_{00}=r_{01}\,q_{00}=1\\ \end{array}$$ which in particular implies $r_{10}\neq 0$. Moreover $$\begin{array}{ccc} x\,\lambda=r\,p\,\lambda,\;y\,\mu=r\,q\,\mu&\Rightarrow&r=x\,\lambda+y\,\mu=r\,(p\,\lambda+q\,\mu) \end{array}$$ and thus $r\,(p\,\lambda+q\,\mu-1)=0$. Now the constant term of the second factor $s=p\,\lambda+q\,\mu-1$ turns out to be $$s_{00}=p_{00}\,\lambda_{00}+q_{00}\,\mu_{00}-1=p_{00}\,r_{10}+q_{00}\,r_{01}-1=2-1=1$$ but then $$\begin{array}{ccc} r\,(p\,\lambda+q\,\mu-1)=0&\Rightarrow&r_{00}\,s_{10}+r_{10}\,s_{00}=r_{10}=0 \end{array}$$ and we get to a contradiction.

Answer 3

This statement should be true in any commutative ring. Take the ideal $(x_1,x_2)$ - all polynomials which monomials have at least one positive power of $x_1$ or $x_2$. It is not hard to verify that this ideal is not principal. (If it is so, try to think by what it can be generated to contain both $x_1$ and $x_2$).