# $(x,y)\subset \mathbb Q[x,y]$ is not a principal ideal

Is my proof that $$(x,y)\subset \mathbb Q[x,y]$$ is not a principal ideal correct?

Suppose $$(x,y)=(f(x,y))$$. Since $$x,y\in (x,y)$$, the degree of $$f$$ in $$x$$ and $$y$$ (separately) cannot exceed 1. So $$f(x,y)=ax+by+c$$. Since $$x\in (x,y)$$, there must exist $$g(x,y)$$ such that $$f(x,y)g(x,y)=x$$. The only event in which this happens is when $$f(x,y)=ax,\ g(x,y)=1/a$$. So $$b=c=0$$. Since $$y\in (x,y)$$, $$a=c=0$$. Thus $$f(x,y)=0$$. But $$(x,y)$$ is not the zero ideal, a contradiction.

• It's quite fine for me. Nov 1, 2018 at 23:28

I second Alex J Best's comment about a little proof fix, but also I think it's worth noting that this holds in $$R[x,y]$$ for $$R$$ any commutative ring with identity. So $$\mathbb{Q}$$ is not special in this regard. Your proof will continue to work for any integral domain. But if $$R$$ has zero divisors, you cannot immediately say that $$f$$ has degree $$1$$, and the proof is a little more complicated.

You might like to try to prove the following for a general ring $$R$$

$$f \in R[x,y]$$ divides $$x$$ and $$y$$ iff $$f$$ is a unit

This immediately implies that $$(x,y)$$ is not principal. I include a proof sketch in the spoiler below if you are interested. Like your proof, it mostly involves comparing coefficients.

Proof: Write $$fg = x$$ and $$fh = y$$. Let $$f_x, f_y$$ and $$f_0$$ denotethe $$x$$ coefficient, $$y$$ coefficient, and constant coefficient of $$f$$. First we can show that $$f_0$$ is a unit. Suppose not the case. Looking at $$1 = f_0g_x + f_xg_0$$ and $$0 = f_0 g_y + f_y g_0$$ modulo $$(f_0)$$, we deduce $$f_x \notin (f_0), f_y \in (f_0)$$. But similarly we could do the same with $$fh = y$$ to deduce $$f_y \notin (f_0), f_x \in (f_0)$$. Together these yield a contradiction, so conclude $$f_0$$ is a unit.

Now, knowing that $$f_0$$ is a unit, compare coefficients in $$fg = x$$ to see that $$g \in R[x]$$, and similarly $$h \in R[y]$$. Considering $$fg = 0 \text{ mod } x$$ we see that $$x$$ divides $$g$$, and similarly $$y$$ divides $$h$$, which gives us the equation $$yg = xh$$. Comparing coefficients in $$yg = xh$$ we finally see $$g = x$$ and $$h = y$$ up to unit multiples. It follows that $$f$$ is a unit.

• If $R=F$ is a field, is your lemma obvious? In that case, $F[x,y]$ is a UFD, so $x$ is irreducible iff $x$ is prime, and $x$ is certainly prime because $F[x,y]/(x)$ is the domain $F[y]$. Similarly, $y$ is irreducible. Moreover, they ($x$ and $y$) are not associates because they generate different ideals. So if $f$ divides the two non-associate irreducible elements $x,y$, then $f$ must be a unit. Nov 2, 2018 at 1:28
• The proof you just gave works perfectly so long as $x$ and $y$ are principal primes, which is the case exactly when $R = D$ is a domain. Nov 2, 2018 at 1:46
• What do you mean by "principal primes"? I think my proof uses that irreducible elements in $R[x,y]$ are prime, which holds in UFDs. But maybe there is a different way to prove that $x,y$ are irreducible in $R[x,y]$ when $R$ is a domain... Nov 2, 2018 at 1:46
• I just mean that the elements $x$ and $y$ are prime (the ideals they generate are prime), sorry for the confusion. Like you mentioned, you know that $(x)$ is prime because $D[x,y]/(x) = D[y]$ is a domain. Prime elements are always irreducible. Nov 2, 2018 at 1:51
• Oh, right, I don't need that irreducible implies prime. Nov 2, 2018 at 1:53

Its pretty much there, but I think a little more is needed, for example:

"The only event in which this happens is when $$f(x,y)=ax, g(x,y)=1/a$$"

we could have the other way around also right? For example $$f(x,y) = 1$$, $$g(x,y) = x$$. You should say also why this can't happen!

Your proof is almost fine, it just needs a couple of details. But you're overcomplicating things.

Consider $$f(x,y)g(x,y)=x$$ as an equality between polynomials in $$y$$ over $$\mathbb{Q}[x]$$. As this has degree $$0$$, both $$f$$ and $$g$$ must have degree $$0$$. Thus $$b=0$$. Similarly, $$a=0$$.

Therefore $$f(x,y)=c$$, which either generates the whole ring or the zero ideal. On the other hand, $$(x,y)$$ is a nonzero proper ideal, as $$\mathbb{Q}[x,y]/(x,y)\cong\mathbb{Q}$$.

• What do you mean by "identity between polynomials in $y$ over $\mathbb Q[x]$"? Nov 3, 2018 at 2:13
• @user437309, $\mathbb Q[x,y]\cong (\mathbb Q[x])[y]$, thus $f$ anf $g$ can be thought of as polynomials in single variable $y$ with coefficients in $\mathbb Q[x]$. Nov 3, 2018 at 2:22
• Ohh, for some reason I interpreted "identity" as an identity in some ring, and not as an "equality". Nov 3, 2018 at 3:16
• @user437309 Changed the word Nov 3, 2018 at 10:11