# Is $F[x,y]$ a Euclidean Domain?

I was wondering if this is just common knowledge. So far for a field $F$ and transcendental $x$ and $y$, I know one can define the degree by

$1) \deg c =0$, for any $c \in F-\{0\}$

$2) \deg x^{n_1}y^{n_2}=n_1+n_2$, for any $n_1, n_2 \in \mathbb{N}$

$3)\deg (f+g)=\max\{\deg f , \deg g\}$, for any $f, g \in F[x,y]$

but I have yet to prove this is a Euclidean measure on $F[x,y]$, so my suspicion is that it is not. I've also been having a hard time finding this particular problem online. If anyone can either solve this or point me to an online resource with the answer, it would be greatly appreciated.

In general, I'm interested in knowing whether $F[x_1, x_2, \dots ,x_k]$ is a Euclidean Domain.

• Euclidean domains are always principal ideal rings. Can you find a non-principal ideal in $F[x,y]$?
– Hoot
Commented Jun 23, 2016 at 20:27
• @Hoot Ah, so it was trivial. I should have known. Commented Jun 23, 2016 at 20:31

## 5 Answers

Every Euclidean domain is a principal ideal domain, but the ideal $(x,y)$ in $F[x,y]$ is not principal.

No. As the other answers point out, it's not a P.I.D. However, it's a U.F.D.; i.e., any polynomial in two or more indeterminates has a factorisation into a product of irreducible polynomials, which is unique up to a unit and the order of the factors.

Furthermore, if we fix a monomial order on $F[x_1,\dots,x_n]$ (i.e. a total order on monomials compatible with multiplication), there is a division algorithm by a set of polynomials, with a unique remainder if the set of polynomials is a so-called reduced Gröbner basis of the ideal they generate.

Every Euclidean domain is a principal ideal domain. Yet $F[x,y]$ is not a principal ideal domain, and thus cannot be Euclidean.

More generally, if $R$ is a principal ideal domain (in particular Euclidean) and not a field, then $R[x]$ is not a principal ideal domain (so it is not Euclidean).

Sketch of proof. Let $(a)$ be a proper nonzero ideal in $R$ (take $a\ne0$ and not invertible). Then $(a,x)$ is not a principal ideal in $R[x]$.

• Thank you for this answer. It really helps me to understand something in generality. Commented Jul 14, 2016 at 5:15

Besides using Euclidean $\Rightarrow$ PID, we can give a direct proof that your total degree function $d$ does not work. If so $\,y = q\, x + r\,$ for $\,d(r) < d(x) = 1,\,$ do $\,d(r) = 0,\,$ so $\, r\in F.\,$ Evaluating the division at $\,y = 0 = x\,$ $\,\Rightarrow\,r = 0,\,$ so evaluating at $\,x=0\,$ $\,\Rightarrow\,y=0,\,$ contradiction.

• Oh I see. I tried something like this, but I was confused about the results. It makes sense now, thanks. Commented Jun 24, 2016 at 5:03