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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.

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  • $\begingroup$ It's quite fine for me. $\endgroup$ – Bernard Nov 1 '18 at 23:28
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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.

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  • $\begingroup$ 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. $\endgroup$ – user437309 Nov 2 '18 at 1:28
  • $\begingroup$ 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. $\endgroup$ – Badam Baplan Nov 2 '18 at 1:46
  • $\begingroup$ 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... $\endgroup$ – user437309 Nov 2 '18 at 1:46
  • $\begingroup$ 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. $\endgroup$ – Badam Baplan Nov 2 '18 at 1:51
  • $\begingroup$ Oh, right, I don't need that irreducible implies prime. $\endgroup$ – user437309 Nov 2 '18 at 1:53
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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!

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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 $\mathbb{Q}[x,y]/(x,y)\cong\mathbb{Q}$.

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  • $\begingroup$ What do you mean by "identity between polynomials in $y$ over $\mathbb Q[x]$"? $\endgroup$ – user437309 Nov 3 '18 at 2:13
  • $\begingroup$ @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 $\mathbb Q[x]$. $\endgroup$ – Ennar Nov 3 '18 at 2:22
  • $\begingroup$ Ohh, for some reason I interpreted "identity" as an identity in some ring, and not as an "equality". $\endgroup$ – user437309 Nov 3 '18 at 3:16
  • $\begingroup$ @user437309 Changed the word $\endgroup$ – egreg Nov 3 '18 at 10:11

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