-5

F is a Field implies F[X] is a Principal Ideal Domain. But is the converse true? Can we also always say that if F[x] is a PID then F is a Field?

I follow Gallian's Contemporary Abstract Algebra wherein is a Theorem stating F is a Field implies F[x] is a PID. The converse isn't given. I had come across the converse in a few answers to the question asked here: Is $\mathbb{Z}[x]$ a principal ideal domain?. Hope you understand the context.

Krishan
  • 178
  • Now that you have detailed answers to this question, perhaps you can undo your downvotes in https://math.stackexchange.com/questions/500254/is-mathbbzx-a-principal-ideal-domain – lhf Feb 27 '21 at 18:41
  • Proved in this answer in the linked dupe (and iirc at least a handful of other places here). – Bill Dubuque Mar 01 '21 at 10:24

2 Answers2

3

Yes.

Take $\alpha\in F\setminus\{0\}$ and take the ideal $\langle \alpha, x\rangle$ of $F[x]$. Being a principal ideal, it is generated by a single polynomial $f\in F[x]$, which divides both $\alpha$ and $x$. $f$ dividing $\alpha\ne 0$ means that $f$ is of degree $0$, i.e. $f\in F$, and $f$ dividing $x$ means that its "leading coefficient" (i.e. $f$ itself) is a unit. This means that, WLOG, we can assume that $f=1$. Thus $1\in\langle\alpha, x\rangle$, i.e. $1=\alpha p(x)+xq(x)$ for some polynomials $p,q\in F[x]$. Now for the final blow: set $x=0$ to obtain $1=\alpha p(0)$, i.e. $\alpha$ is invertible.

3

If $F[x]$ is a PID then so is F. Consider the ideal $(a, x)$ where $a\in F$ and $a\neq 0$. By hypothesis there is a polynomial $g$ that generates $(a, x)$. Thus for some $f\in F[x]$ we have $g f(x)= a$. The only way this can happen is if g and f are in $F$. Thus $g \in F$. Since $x \in (g)$ then there is a polynomial $h(x)$ such that $g h(x) = x$. Thus $h(x) = c x$ for some $c \in F$ and $g c x = x$. Hence $g$ is a unit and thus $a$ is a unit since the ideal $(a, x) = F[x]$. Therefore $F$ is a field.

  • (1) Have you assumed already that if F[x] is a PID then F is a PID too?(If yes, can you please prove it). (2) You have said there is a polynomial 'g' that generates (a,x), and so there exists an f(x) in F[x] such that "g.f(x)=a", so f has to be a constant polynomial (an element in F) and so also 'g' too. is this how you have concluded that f and more importantly g are elements of F? – Krishan Feb 28 '21 at 05:40
  • (3) This line is very beautifully architected and I am very clear until "Hence g is a unit", Have you concluded 'a' is a unit because 'g' is a unit means <a,x>==F[x] and so 'a' must also be a unit because <a,x>={a.m(x)+x.n(x)}=F[x] must contain all the zero degree polynomials of F[x] which is possible only if it contains the constant polynomial '1', therefore making the existence of a^-1 in F a necessary condition for [x]. – Krishan Feb 28 '21 at 05:40
  • @Krishan: A simpler argument to the whole problem is to first prove that in a PID prime ideals are maximal. The ideal (x) is thus maximal and hence F[x]/(x) is a field. I didn’t actually use that F is a PID in my post. I really just needed that F is an integral domain. What you say in (3) is correct and to (2) yes. –  Feb 28 '21 at 18:54