First, is it appropriate to write $F[x,y] = (F[x])[y] $? In particular, if $F$ is a field, then we know $F[x]$ is a Euclidean domain. Are there necessary and sufficient conditions for when $F[x,y]$ is also a Euclidean domain?
Interpreting $F[x,y]$ for $F$ a field.
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1Why did you think $F[x,y]$ was ever Euclidean? – 2012-04-10
3 Answers
In most constructions of polynomial rings (e.g., as almost null sequences with values in the ground ring), the rings $F[x,y]$, $(F[x])(y)$, and $(F[y])[x]$ are formally different objects: $F[x,y]$ is the set of all functions $m\colon \mathbb{N}\times\mathbb{N}\to F$ with $m(i,j)=0$ for almost all $(i,j)$; $(F[x])[y]$ is the set of all functons $f\colon\mathbb{N}\to F[x]$ with $f(n)=0$ for almost all $n$; and $(F[y])[x]$ is the set of all functions $g\colon\mathbb{N}\to F[y]$ with $g(m)=0$ for almost all $m$.
However, there are natural isomorphisms between them, $(F[x])[y]\cong F[x,y]\cong (F[y])[x].$ Informally, this corresponds to the fact that we can write a polynomial in $x$ and $y$ by "putting $y$'s together" or by "putting $x$'s together. So, for instance, $1+x+2y + 3x^2y - 7xy^4 + 8x^2y^2$ can be written as $\begin{align*} 1+x+2y + 3x^2y - 7xy^4 + 8x^2y^2 &= 1 + (1-7y^4)x + (3y+8y^2)x^2\\ &= (1+x) + (2+3x^2)y + (8x^2)y^2 - (7x)y^4. \end{align*}$ So, yes, you can say that $F[x,y]$ is "essentially" equal to $(F[x])[y]$.
However, $F[x,y]$ is never a Euclidean domain, because it is never a PID: $\langle x,y\rangle$ is never principal.
In fact:
Theorem. Let $D$ be a domain. Then $D[x]$ is a PID if and only if $D$ is a field.
Proof. If $D$ is a field, then $D[x]$ is a Euclidean domain, hence a PID.
If $D$ is not a field, then let $a\in D$ be an element that is a nonzero nonunit. Then $\langle a,x\rangle$ is not a principal ideal: if $\langle a,x\rangle = \langle r\rangle$, then $r|a$, hence $r$ must be a polynomial of degree $0$; since $r|x$, then $r$ must be a unit or an associate of $x$; since it must be degree $0$, $r$ must be a unit. So $\langle a,x\rangle$ is principal if and only if it is equal to the entire ring. But $D[x]/\langle a,x\rangle \cong D/\langle a\rangle\neq 0$ (since $\langle a\rangle\neq D$), so $\langle a,x\rangle\neq D[x]$. Hence $\langle a,x\rangle$ is not principal. $\Box$
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5+1, one cannot compete with Arturo. (Not that I'm trying to.) – 2012-04-09
Euclidean domains are always principal ideal domains. The ideal $(x,y)$ is never principal. Hence $F[x,y]$ is never a Euclidean domain.
Yes it is appropriate to write $F[x,y]$ as $(F[x])[y]\cong(F[y])[x]$, so there is no ambiguity. $F[x,y]$ is not a PID as it contains the non-principal ideal $(x,y)$. This means in particular that $F[x,y]$ is not euclidian.