Why is the polynomial ring $\Bbb R[x]$ a PID but $\Bbb Z[x]$ is not?

Question

Why is the polynomial ring $\Bbb R[x]$ a PID but $\Bbb Z[x]$ is not? The question is asking me to prove that $\Bbb R[x]$ is a PID. I'm assuming you would go about this knowing that every field is a PID, but I'm stuck on the differences between $\Bbb R[x]$ and $\Bbb Z[x]$.

Answer 1

There are many ways to answer this question, at different levels, but I would say that I think the right reason is because $\mathbb{Z}$ is not a field.

I will give you 3 supporting evidences for that.

1. The proof that $\mathbb{R}[X]$ is a PID relies on long division of polynomials, and the fact that you can divide a polynomial by any non zero polynomial, and get a quotient and a remainder in $\mathbb{R}[X]$. This relies on the fact that any nonzero real number is invertible.

This does not work in $\mathbb{Z}[X]$: for example, if you divide $X$ by $2$, you will get a quotient which does not lie in $\mathbb{Z}[X]$. What goes wrong here is that the invertible elements of $\mathbb{Z}$ are $\pm 1$.

OK, this may be not satisfactory, because what this really proves is that $\mathbb{Z}[X]$ is not Euclidean for the degree function. It does not prove that $\mathbb{Z}[X]$ is not Euclidean for another function. Worse, there exists PID's which are not Euclidean.

2. Another reason (see Bernard's answer): in a PID, any non zero prime ideal is maximal. This is not the case for $\mathbb{Z}[X]$, since $p\mathbb{Z}[X]$ is a prime ideal but not maximal, for any prime number $p$.

Note that this cannot happen in $\mathbb{R}[X]$ ( $p\mathbb{R}[X]=\mathbb{R}[X]$, since $p$ is invertible in $\mathbb{R}$).

Once again, the main difference is that you have a lot of non invertible elements in $\mathbb{Z}$. Ok, but this is still too vague. SO let's go for :

3. Thm. Let $A$ be a commutative ring with $1$. Then the following properties are equivalent:

i) $A$ is a field

ii) $A[X]$ is an Euclidean domain

iii) $A[X]$ is a PID.

Only iii)$\Rightarrow$ i) needs a proof: in a PID, any nonzero prime ideal is maximal. Since $A[X]$ is a PID, it is an integral domain, and thus so is $A$ (this is a subring of $A[X]$)

Since evaluation at $0$ induces a ring isomorphism $A[X]/(X)\simeq A$, $(X)$ is a nonzero prime ideal. Since $A[X]$ is a PID, $(X)$ is thus a maximal ideal. Consequently, $A[X]/(X)$ is a field, meaning that $A$ is a field.

Answer 2

For any field $F$, the polynomial ring $FX]$ is a P.I.D. because it is a Euclidean domain, with the degree of a polynomial as a Euclidean function.

On the other hand, $\mathbf Z[X]$ has no general Euclidean division (how would you divide $X+1$ by $2X+1$? Furthermore, a P.I.D. has Krull dimension $1$ (every nonzero prime ideal is maximal), and $\mathbf Z[X]$ has Krull dimension $2$: if $p$ is a prime number, $ (0)\subsetneq (p)\subsetneq (p,X)$ is a chain of prime ideals of length $2$.

Answer 3

You can show that $K[X]$ is an euclidean domain via the degree- map for any field K. For $\mathbb{Z}[X]$ consider the ideal $(2,X).$ If this ideal were an principal ideal there were an $f\in\mathbb{Z}[X]$ such that $(2,X)=(f).$ For this $f$ it holds $f| 2$ and $f|X.$ So there is $r$ s.t. $2=fr$ an $q$ s.t. $X=fq.$ Hence we get $$ 1=\deg(X)= \deg(f)+\deg(q) $$ and $$0= \deg(2)=\deg(f)+\deg(r).$$ So the only possibilitiy is $\deg(f)=0.$ Hence $f\in\mathbb{Z}.$ Because of $X=1*X$ we see that $1=f.$ Hence $(f)=(1)=\mathbb{Z}[X].$ But $(2,X)$ is not equal to $\mathbb{Z}[X].$