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

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.

• There is something strange about the claim 1, is it really talking about $c$ or $x$ being irreducible in $D$? Also in that case wouldn't $\mathbb{Z}[p] = \mathbb{Z}$ be a counterexample for any prime $p$ (and thus irreducible in $\mathbb{Z}$)? – Tob Ernack Jan 2 '18 at 17:17
• @TobErnack The statement is not taking $x$ to be irreducible in $D$. – user193319 Jan 2 '18 at 17:19
• @TobErnack $x$ is a variable. The statement says that if there is an irreducible element in $D$, then $D[x]$ is not a PID. For example, there is no irreducible element in the field $F$, so $F[x]$ being a PID does not contradict statement 1. – Pedro Jan 2 '18 at 17:21
• Statement 1 should be formulated as ‘If $D$ is an integral domain which contains an irreducible element, then …’. – Bernard Jan 2 '18 at 17:28

If $D$ is an integral domain, an irreducible element in $D$ is also an irreducible element in $D[X]$.
• Okay. But, taking $D=F[x_1,...,x_{n-1}]$, I still need to show that $D$ has an irreducible element--namely, $x_1$--which I was trying to prove using induction. – user193319 Jan 2 '18 at 23:30
• I'll complete the hint: since $x_1$is irreducible in $F[x_1]$, it is irreducible in $F[x_1,x_2]$, hence also in $F[x_1,x_2, x_3]$, &c. – Bernard Jan 2 '18 at 23:33
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 $$F[x_{1},\ldots,x_{n}]\cong F[x_{1},\ldots,x_{n-1}][x]$$