The Polynomial Ring $F[x_1,...,x_n]$ over the field $F$ is not a PID.

Question

1. If $D$ is an integral domain containing an irreducible element, then $D[x]$ is not a principal ideal domain.

2. If $F$ is a field and $n \ge 2$, then $F[x_1,...,x_n]$ is not a principal ideal domain. [Hint: show that $x_1$ is irreducible in $F[x_1,...,x_{n-1}]$.

I have already worked on part 1, but I am having trouble proving what the hint in part 2. suggests. I imagine the statement is to be proven through induction. Here is my proof of the base case. Suppose that $x =pq$ for $p,q \in F[x]$. Then $1 =\deg(pq) = \deg(p) + \deg (q)$, which implies, WLOG, that $p = ax + b$ and $q = c$, where $a,b,c \in F$ and $c \neq 0$. Then $x = pq = acx + bc$ implies $ac=1$ and $bc=0$. Since $c \neq 0$, it must be that $b=0$, and $a$ and $c = q$ are units. Hence, $x$ is irreducible.

The inductive step is giving me a little more trouble. Suppose that $x_1$ is irreducible in $F[x_1,...,x_{n-1}]$, and suppose that $x_1 = pq$ for $p,q \in F[x_1,..,x_n]=(F[x_1,...,x_{n-1}])[x_n]= D[x_n]$ (strictly speaking, I think this is an isomorphism). Note that $x_1$ as a constant in $D[x_n]$, so that $0=\deg(x_1) = \deg(pq)= \deg(p)+\deg(q)$ implies $p = f(x_1,...,x_{n-1})$ and $q=g(x_1,...,x_{n-1})$ are constants. In particular, they satisfy $1 = fg$ and so are units in $D=F[x_1,...,x_{n-1}]$......

Isn't it true that the units of $D[x_n]$ will be the units of $D$? If so, can I conclude that $p$ and $q$ are units in $D[x_n]$ and then conclude that $x_1$ is irreducible in $F[x_1,...,x_n]$? Is that right? Obviously proving 2. is trivial at this point: $F[x_1,...,x_{n-1}]$ contains the irreducible element $x_1$, and so by part 1. $F[x_1,...,x_n]$ cannot be a PID.

Answer 1

Hint:

For step 2, you don't need really induction. You just have to prove this:

If $D$ is an integral domain, an irreducible element in $D$ is also an irreducible element in $D[X]$.

Answer 2

The argument you give for the base case, combined with similar arguments as $deg_{x_{i}}(p)=deg_{x_{i}}(q)=0$ for all $i>1$, shows at once that $x_{1}$ is irreducible in $F[x_{1},\ldots,x_{n-1}]$. Once you know this, then the statement follows immediately from part 1, since $$ F[x_{1},\ldots,x_{n}]\cong F[x_{1},\ldots,x_{n-1}][x] $$

Answer 3

Though years passed, there is still no a complete proof. So let me write a complete one.

Observation $1$: Note that $F$ is a field. In particular, it is an integral domain, and so is $F[x_1,...,x_{n-1}]$.

Observation $2$: The only units in $F[x_1]$ are elements in $F$. More generally the only units in $F[x_1,...,x_{n-1}]$ are elements in $F$. This shows that $x_1$ is not invertible (i.e. non-unit) in $F[x_1,...,x_{n-1}]$.

Observation $3$: $(F[x_1,...,x_{n-1}])[x_n]=F[x_1,...,x_n]$.

Now using the first result you have proved, i.e. If $D$ is an integral domain containing an irreducible element, then $D[x]$ is not a principal ideal domain, with $D$ being $F[x_1,...,x_{n-1}]$. We conclude that $F[x_1,...,x_n]$ is not a PID for $n\geq 2$.