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Let $R$ be a commutative ring with $1$. Prove that a polynomial ring in more than one variable over $R$ is not a P.I.D.. In order to show this is not a P.I.D. what do i need to show.

user146269
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  • What have you tried so far, and what are you stuck on? Can you, for example, write down the definition of a PID? – anomaly Apr 08 '15 at 18:22
  • shoe that (x,y) is not Principal – user 1 Apr 08 '15 at 18:22
  • The definition is that is an integral domain in which every ideal is principal. I know that but not every ring is an integral domain, so because if that i am having trouble with this problem. if i was working with the quotient ring then i know how to go about it. – user146269 Apr 08 '15 at 18:28
  • i got it thanks i see what i need to do. Sorry for even post this – user146269 Apr 08 '15 at 18:30
  • Do you mean "R]x,y] is not a principal ideal ring" or did you mean to hypothesize that $R$ is a domain (vs. commutative ring)? – Bill Dubuque Apr 08 '15 at 20:07
  • i mean in general i wanted to take to things like x and y but it can be more than just x, y. i post my answer can you check it out so see if i was in the right direction please thanks – user146269 Apr 08 '15 at 20:34

2 Answers2

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Let's show that if $R[x]$ is a PID, then $R$ is a field. Note first that $R$ is a subring of $R[x]$, so it is a domain.

If $a\in R$, $a\ne0$, the ideal $(a,x)$ is principal, that is, we find $f(x)\in R[x]$ with $$ a=f(x)g(x),\qquad x=f(x)h(x) $$ for some $g(x),h(x)\in R[x]$. Since $R$ is a domain, we have that $$\def\deg{\operatorname{deg}} 0=\deg a=\deg f+\deg g, \qquad 1=\deg x=\deg f+\deg h, $$ so $\deg f=0$ ($\deg f$ means, of course, the degree of $f(x)$). Therefore $f(x)=b\ne0$ is constant. Also $\deg h=1$, so $h=cx+d$ with $c\ne0$. From $$ b(cx+d)=x $$ we conclude that $bc=1$, so $b$ is invertible in $R$, hence in $R[x]$. But then $$ (a,x)=(f(x))=(b)=R[x]. $$ In particular $1=ar(x)+xs(x)$ for some $r(x),s(x)\in R[x]$: evaluating at $0$ we have $$ 1=ar(0) $$ and so $a$ is invertible in $R$.

Now, if $R[x_1,\dots,x_{n-1},x_n]=R[x_1,\dots,x_{n-1}][x_n]$ is a PID, we have that $R[x_1,\dots,x_{n-1}]$ is a field, which is not true if $n>2$, because $x_1$ is not invertible.

egreg
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Proof: Suppose that $R[x,...,x_{n}]$ is a P.I.D. that means $R[x](x,..x_{n-1})$ has to be a field since we know that if $R[x]$ is a field if and only if $R$ is a field. But $R[x](x,..x_{n-1}$ is not a field, which is a contradiction. Hence $R$ is not a P.I.D.

user146269
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  • can anyone check if this is correct or not. Please because i was reading somewhere and they mention this idea about field then sort of use it in this problem so i do not know if that is the correct approach. – user146269 Apr 08 '15 at 20:34
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    If $R[x]$ is a field, then $R$ is a field. But the converse not true. I think the rest should be correct. – Wang Kah Lun Nov 26 '17 at 01:40