# Polynomial ring is not PID

I was trying to think of a proof of the following proposition:

Let $K$ be a field. Then $R=K[x_1,...,x_n]$ is not a PID for $n>1$.

So here's what I've written so far:

Suppose $R$ is a PID and consider the ideal $I=\langle x_1,x_2 \rangle$. Then $I=\langle f \rangle$ for some $f \in R$. In particular, we have $$(1) \space x_1=gf$$ $$(2) \space x_2=hf,$$

for $g,h \in R$. But then, from (1), we must have $\deg_{x_2}(f)=0$ and $\deg_{x_1}(f) \in \{0,1\}$. Analogously, from (2) it follows that $\deg_{x_1}(f)=0$ and $\deg_{x_2}(f) \in \{0,1\}$. In conclusion, $\deg_{x_1}(f)=0=\deg_{x_2}(f)$.

So the polynomial $f$ must be a non zero constant in $K$ or it must be a polynomial which doesn't have $x_1$ or $x_2$ as one of its variables. For the first case, I know how to arrive to a contradiction, for if $f=k \in K \setminus \{0\}$, then $1 \in \langle x_1,x_2 \rangle$, so $$1=hx_1+hx_2$$

Evaluating at $(x_1,...,x_n)=(0,...,0)$, we conclude $1=0$, which is absurd.

I am not sure what to do in the other case, I don't even know a notation to express an arbitrary polynomial of several variables. I would appreciate some help to conclude the proof. Thanks in advance.

• – KCd Mar 15 '16 at 4:09
• – user26857 Mar 15 '16 at 22:23

In the "the other case" use one of the two equations to get $\deg_{x_i}f=0$ for $i>2$.