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The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal.

I don't know how to consider it. Any suggestions?

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    $\begingroup$ Suppose it's generated by $P_0$. Then $x=PP_0$ and $y=QP_0$ for some $P,Q\in k[x,y]$. For each $y$, $P(x,y)P_0(x,y)$ is a polynomial of degree $1$ with respect to $x$ so the maximal degree with respect to $x$ of $P_0(x,y)$ is at most $1$. Do the same for $y$ and get a contradiction. $\endgroup$ – Davide Giraudo Feb 18 '12 at 21:11
  • $\begingroup$ @DavideGiraudo That looks like an answer to me. I'd say you should post it as one so it can be accepted. $\endgroup$ – Alex Becker Feb 18 '12 at 21:17
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    $\begingroup$ Sledgehammer proof: the height of $I$ is at least 2, but the height of any principal ideal in any noetherian ring is at most 1 by Krull's Hauptidealsatz. $\endgroup$ – Zhen Lin Feb 18 '12 at 21:17
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A generator $g$ of $\langle x,y \rangle$ would divide $x$ and $y$, which are irreducible and non-associated. Thus $g$ would be a unit, and $\langle x,y \rangle$ would be the unit ideal, contradiction.

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  • $\begingroup$ Why is $g$ is a unit? I've tried like mad to prove this but I can't seem to scare up a proof. $\endgroup$ – user193319 Feb 21 '18 at 17:13
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    $\begingroup$ @user193319 - I considered that it was implicitly assumed that $k$ is a field and $x$ and $y$ are indeterminates. Then $k[x,y]$ is a unique factorization domain, $x$ and $y$ are irreducible and non-associated, and, in any unique factorization domain, an element which divides two non-associated irreducible elements is a unit. $\endgroup$ – Pierre-Yves Gaillard Feb 21 '18 at 19:11
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Suppose it would be principal. Then there would exist $p(x,y)$ such that for all polynomials $g(x,y)$ and $h(x,y)$ in $k[x,y]$, we would have some $i(x,y)$ such that $$ xg + yh = ip. $$ In particular, $x = i_1 p$ and $y = i_2 p$, so that the degree of $ p$ is at most $1$. It can't be zero, because that would mean p is a unit or zero and we would lose the fact that $\langle p \rangle = \langle x,y \rangle$. So $\deg p = 1$, hence $p=ax+by+c$ for $a,b,c \in k$. But then $x = i_1(ax+by+c)$ means $b= 0$ and $ y =i_2 (ax+c)$ means $a=0$, hence $p$ has not degree $1$, contradiction.

Hope that helps,

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  • $\begingroup$ I see all the comments now and I think im a little too late, but this is the first time I post an answer from my cellphone. >.> $\endgroup$ – Patrick Da Silva Feb 18 '12 at 21:25
  • $\begingroup$ I was wondering why "the degree of $p$ is at most 1". $\endgroup$ – John Conn Feb 18 '12 at 21:47
  • $\begingroup$ Thats because deg p + deg i = dex (x) = 1. $\endgroup$ – Patrick Da Silva Feb 18 '12 at 22:02
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$k[x,y]$ modulo a proper principal ideal is infinite dimensional, yet your ideal is of codimension $1$.

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Suppose it's generated by $P_0$. Then $x=PP_0$ and $y=QP_0$ for some $P,Q\in k[x,y]$. For each $y$, $P(x,y)P_0(x,y)$ is a polynomial of degree $1$ with respect to $x$ so the maximal degree with respect to $x$ of $P_0(x,y)$ is at most $1$, so $P_0(x,y)=a_0(y)+a_1(y)x$ where $a_0,a_1\in k[y]$.

Doing the same for $y$ we get that the maximal degree with respect to $y$ of $P_0(x,y)$ is at most $1$ for each $x$ so the maximal degree of $a_0$ and $a_1$ is $1$. We have $P_0(x,y)=a_{0,0}+a_{0,1}y+a_{1,0}x+a_{1,1}xy$ where $a_{0,0},a_{1,0}, a_{0,1},a_{1,1}\in k$. Identifying the term of highest degree in the equality $x=P(x,y)(a_{0,0}+a_{0,1}y+a_{1,0}x+a_{1,1}xy)$ we get that $a_{1,1}=0$ and since $a_{0,1}y+a_{1,0}x\in\langle x,y\rangle$, $a_{0,0}=0$. We finally get that $a_{1,0}=a_{0,1}=0$, which is a contradiction.

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  • $\begingroup$ Why a1, 0 and a0, 1 is zero $\endgroup$ – Pradip Aug 20 '19 at 4:17
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HINT $\ $ Among principal ideals: contains $\equiv$ divides; thus having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible). Thus the only principal ideal containing $(x)$ is $(1)$. But $(x,y)\ne (1)$ by evaluating at $x=0=y.\:$ QED. Below is more detail

Observe the ideals $\: (x) \subsetneq (x,y) \subsetneq (1)\:$ are distinct $\: $ primes $\rm\: [\:or\ (1)\:]\:$ since
their residue rings $\ k[y]\ \supsetneq\ k\:\ \supsetneq\:\ (0)\:$ are distinct domains $\rm[\:or\ (0)\:].$

So $(x,y)$ isn't principal since principal primes $\ne (0)$ are maximal among principal ideals.

This maximality is familiar from PIDs, e.g. in $\mathbb Z$ or $k[x]$.
It simply says that $\:\gcd(x,p) = p\ \ {\rm or}\:\ 1\:$ if $\:p\:$ is prime,
or, in ideal language, $(x,p) = (p)\ {\rm or}\ (1)\:$ so $(p)$ is maximal.

More generally, in any domain where ideals satisfy contains $\equiv$ divides (e.g. Dedekind domains), prime ideals $\ne 0$ are maximal. This characterizes PIDs, i.e. PIDs are precisely those UFDs where every prime ideal $\ne 0$ is maximal (i.e. Krull dimension $\le 1$). For a handful of other useful characterizations of when a UFD is a PID see this post.

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Here's a complete simple proof. Put $\:\rm c = y,\ D = k[y]\:$ below, using $\rm\: y\:g(y) \ne 1\:$ by eval at $\rm\: y=0$.

THEOREM $\ $ Let $\rm\: 0\ne c\in D\:\!$ a domain. In $\rm\: E = D[x]:\ (c,x) = (f)\ \Rightarrow\ c\ |\ 1\: $ in $\rm D.\ \ $ Proof:

$\rm f\in (c,x)\: \Rightarrow\: f = cg + xh,\:$ some $\rm g,h\in E$. Eval at $\rm\: x=0\: \Rightarrow\: f(0) = cg(0) = cd$, some $\rm d\in D$.

$\rm c\in (f)\ \Rightarrow\ c = fg,\:$ some $\rm g\in E,\:$ so $\rm\:0\ne c\in D\:\!$ a domain $\rm\: \Rightarrow\ deg\ f = 0\ \Rightarrow\ f = f(0) = cd$.

$\rm x\in (f)\: \Rightarrow\: x = fh = cdh,$ some $\rm h\in E.\:$ Eval at $\rm\: x=1\: \Rightarrow\: 1 = cdh(1)\:\Rightarrow\:c\ |\ 1\:$ in $\rm D.\ \ $ QED

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    $\begingroup$ If you are going to write complete proofs you could also write complete words! $\endgroup$ – Mariano Suárez-Álvarez Feb 19 '12 at 1:23

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