0
$\begingroup$

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.

$\endgroup$
2
2
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.