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I want to show that any ideal in $\mathbb{C}[x,y]$ containing $y$ can be generated by 2 elements.

Here is my proof so far. I would like verification that what I've done is correct, and to see whether there are any neater solutions.

Let $I$ be such an ideal. I know that any ideal in $\mathbb{C}[x,y]$ is finitely generated, and without loss $y$ can be one of the generators, so we have $I=\langle y,f_1,...,f_m\rangle$ for some $f_i \in \mathbb{C}[x,y]$.

If each of the $f_i$ are divisible by $y$ then $I\subset \langle y \rangle$ and as $y\in I$ we have $\langle y \rangle \subset I$ so then $I=\langle y \rangle$.

Otherwise, there exists some $f \in I$ with $\deg_y f=0$, i.e. $f = f(x) \in \mathbb{C}[x]$. Without loss suppose $f$ is of minimal degree among all polynomials in $I$ of the form $f(x)$. Then, $\langle y, f\rangle \subset I$. Conversely, take some $g\in I$. If $g \in \langle y \rangle$ then we're done. Else, $g$ contains some term not divisible by $y$, i.e. of the form $h(x)$ (where we write $g$ as a polynomial in $y$ with coefficients in $\mathbb{C}[x]$). Then as $\langle y \rangle \subset I$ and all other terms are divisible by $y$ we can rearrange to find $h \in I$. Then dividing $h$ by $f$ in $\mathbb{C}[x]$ we have $h(x)=f(x)q(x)+r(x)$, where $r=0$ or $\deg(r)<\deg(f)$. Then $r=0$ else it contradicts the minimality of $f$ (as rearranging gives $r \in I$). So $h \in \langle f \rangle$ and thus $g\in \langle y, f \rangle$. So $I= \langle y, f \rangle$ is generated by 2 elements.

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Martin
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  • It is not true for $y^n$, for $n>1$. Order monomials $y^ax^b$ by the lexicographic order of $(a,b)$. For each element $f$ of $I$, mark a point $a,b$ in the plane $\mathbb{N}^2$ for the largest monomial $y^ax^b$ of $f$ and all points $(c,d)$ with simultaneously $c\geq a$ and $d\geq b$. You will need a generator for each of the pointy ends in the picture that results. In the case $y=1$ you can get two generators because there is only space for two pointy peaks, the one at $(1,0)$ and perhaps one at some $(0,a)$. – MoonLightSyzygy Jan 18 '20 at 11:46
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    @MoonLightSyzygy But any ideal I can think of containing $y^n$ can be generated by $n+1$ elements; I have no idea how one would come up with a counterexample. – Martin Jan 18 '20 at 11:50
  • Wait what did you want to prove? Two generators or $n+1$? – MoonLightSyzygy Jan 18 '20 at 11:51
  • If you can allow $n+1$ generators for $y^n$ in the ideal, then the same argument helps. A Groebner basis will have initial monomial producing one of the pointy peaks in the diagram above. Then note how you can't fit more than $n+1$ of those when you have $(n,0)$ in it. – MoonLightSyzygy Jan 18 '20 at 11:53
  • @MoonLightSyzygy $n+1$ generators - I appreciate it might not be clear in the first paragraph! I have no idea what a Groebner basis is, or the lexicographic order, we haven't covered that yet, so need more of an 'elementary' argument. – Martin Jan 18 '20 at 11:56
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    That things have names doesn't mean that they are not elementary. Once you study them you will see that it is nothing else than polynomial long division. That is what is involved in the proof. First: consider $(a,b)>(c,d)$ when $a>c$ or $a=c$ and $b>d$. That is lexicographic order. Next, for each polynomial in $I$ place a point on the plane for the exponent tuple of its largest monomial. Since $y^n$ is assumed in the ideal, you will put the point $(0,n)$. Note that because it is an ideal all the points $(a,n+b)$ are also exponents of the largest monomial of some element $x^ay^{n+B}$ of $I$ – MoonLightSyzygy Jan 18 '20 at 12:01
  • So, the diagram looks like a union of a bunch of translated quadrants. Finitely many quadrants actually, that are not one inside the other. Now, let's take an element of $I$ for each of the vertices of those finitely many quadrants, such that their largest monomial has exponent corresponding to that vertex. Let's show that those generate $I$. Let $f$ be an arbitrary element of $I$. Its largest monomial is necessarily above one of the vertices. Therefore, by doing long division with one of our to-be-generators we can kill the largest term of $f$. Each elimination drops the terms down ... – MoonLightSyzygy Jan 18 '20 at 12:04
  • ... in the lexicographic order, but still inside the set of quadrants above. Noting that the lexicographic order is a well-order, this process cannot go on forever. Therefore, at some point you reduce $f$ all the way to zero. Hence it is generated by the elements chosen. The proof ends by noting that if $(0,n)$ is one of the vertices of the quadrants, then you can only fit $n$ more quadrants in the picture. – MoonLightSyzygy Jan 18 '20 at 12:10
  • By the way, compare this with the argument that you gave for $n=1$. You took $y$ as one of the generators. Note that its largest terms gives you the point $(0,1)$. Then you searched for an element $f=f(x)$ of minimum degree. Note how that gives you a vertex $(m,0)$ of the quadrant $(m+c,d)$ for $c,d\in\mathbb{N}$. The reason for searching for those vertices is because $(1,0)$ can kill all monomials $(1+c,d)$ and $(m,0)$ can kill all monomials $(m+c,d)$, for $c,d\in\mathbb{N}$, in the polynomial long division. Note how you did the division in order, first by $y$ and then by $f(x)$ ... – MoonLightSyzygy Jan 18 '20 at 12:30
  • ... This corresponds to $y=y^1x^0$ being a larger monomial than $x^m=y^0x^m$ in the lexicographic order of the exponents $(1,0)>(0,m)$. Likewise, in the $y^m$ you will cancel monomials using the candidate generators starting from the largest monomial in the lexicographic order. – MoonLightSyzygy Jan 18 '20 at 12:30

1 Answers1

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Your proof has the right idea, but the case you give is not correct - so let's figure out how to fix it, and then generalize. You, at various points in the argument, basically say that if $f_i$ is not divisible by $y$, then $\deg_y f_i=0$, which is false; for instance, the ideal $$\langle y,x+y\rangle$$ has a generator $x+y$ which is neither divisibly by $y$ nor in $\mathbb C[x]$. A more correct observation would be to note that every element $f\in\mathbb C[x,y]$ can be written as $f=a+by$ where $a\in \mathbb C[x]$ and $b \in \mathbb C[x,y]$. This is analogous to pulling off the constant term treating $\mathbb C[x,y]$ as the space of polynomials in $y$ over $\mathbb C[x]$. Then, you can observe that $$I=\langle y,a_1+b_1y,a_2+b_2y,\ldots\rangle=\langle y,a_1,a_2,\ldots\rangle$$ since the generators on the right are linear combinations of those on the left and vice versa.

At this point you can use your argument to find some polynomial $f\in I \cap \mathbb C[x]$ of minimal degree and to find out that $I=\langle y,f\rangle$.

However, this correct better illuminates what is going on here: when we write an element as $f=a+by$ we're really looking at which coset of $\langle y\rangle$ contains $f$, taking $a\in\mathbb C[x]$ as a representative. This hints to us that the proof can be made much more compact:

Let $I$ be an ideal of $\mathbb C[x,y]$ containing $y$. Consider the quotient map $q:\mathbb C[x,y]\rightarrow \mathbb C[x,y]/\langle y\rangle$ and note that $\mathbb C[x,y]/\langle y\rangle \cong \mathbb C[x]$. Since $q$ is surjective, $q(I)$ is an ideal of $\mathbb C[x]$ and hence, as $\mathbb C[x]$ is a PID, $q(I)$ is generated by a single element $f$. Since $\ker q \subseteq I$ it must be that $f\in I$ if and only if $q(f) \in q(I)$.

Let $\bar f$ be any generator of $q(I)$ and $f$ be any element of $\mathbb C[x,y]$ such that $q(f)=\bar f$. The ideal $I$, by the above facts, must be equal to $\ker q + \langle f\rangle = \langle f,y\rangle$.

where we avoid repeating the argument that $\mathbb C[x]$ is a PID as you do and use that fact directly. This essentially uses the Fourth Isomorphism Theorem for Rings, if you are familiar with that terminology. You can also see that this generalizes: what we really need to do is to show that every ideal of $\mathbb C[x,y]/\langle y^n\rangle$ has at most $n$ generators, and then just take those generators and add $y^n$ to generate those ideals of $\mathbb C[x,y]$ containing $y^n$.

At this point, we do need another ingredient; you proof won't generalize immediately. We can observe that every element of $\mathbb C[x,y]/\langle y^n\rangle$ can be written uniquely as $a_0+a_1y+a_2y^2+\ldots+a_{n-1}y^{n-1}$ for $a_i\in \mathbb C[x]$. Essentially, $\mathbb C[x,y]/\langle y^n\rangle$ looks like the sets of coefficients $\mathbb C[x]^n$ with some strange multiplication rule and we don't really need to worry about.

We can then use your basic idea: Let $\bar I$ be an ideal of $\mathbb C[x,y]/\langle y^n\rangle$. Let $A_0$ be the set of constant terms $a_0\in \mathbb C[x]$ that appear in the expansions of elements of $\bar I$. Note that $A_0$ is an ideal of $\mathbb C[x]$, since it is closed under multiplication by $\mathbb C[x]$ and addition, thus there is some generator $g_0$ of this module. Let $\tilde g_0$ be any element of $\bar I$ whose constant term (in $y$) is $g_0$.

Then, we move on to the second coordinate and have to be a bit more clever - note that we have no control over the terms of $\tilde g_0$ except the first one. Thus, we would like our next generator not to include a constant term, since this is the only way we can maintain control over the first term. Formally, let's let $A_1$ be the set of linear terms $a_1\in\mathbb C[x]$ that appear in the expansion of elements of $\bar I$ whose constant term is $0$. Again, we can find an element $g_1$ generating the ideal $A_1$ in $\mathbb C[x]$ and a $\tilde g_1$ lifting this in $\bar I$ with no constant term. We can then proceed similarly to define $A_k$ to be the set of terms $a_k$ appearing in elements of the form $a_ky^k+a_{k+1}y^{k+1}+\ldots+a_{n-1}y^n\in\bar I$ and will this way get a sequence of generators $\langle \tilde g_0, \tilde g_1,\ldots, \tilde g_{n-1}\rangle$. You can then inductively show that every element of $\bar I$ may be written as a sum of these generators with coefficients in $\mathbb C[x]$, since we can match the first coefficient using $\tilde g_0$, then the second using $\tilde g_1$ and so on. This proves the lemma in a manner which is about as close as one can get to your original proof.

Note that we could also abstract this away: $\mathbb C[x,y]/\langle y^n\rangle$ is a free module over $\mathbb C[x]$ of rank $n$. Our above argument can be repeated to show that if $R$ is a PID, then any submodule of a free module of rank $n$ is a free module of rank $r\leq n$ - which, using that an ideal of $\mathbb C[x,y]/\langle y^n\rangle$ is also a submodule of that same set viewed as a module of $\mathbb C[x]$, gives the result we want. Abstracting this way would separate the proof a bit better: First, we reduce the given problem to a problem about $\mathbb C[x,y]/\langle y^n\rangle$ via the Fourth Isomorphism Theorem for Rings. Then, we reduce this to a common property of free modules. We could also, conversely, roll everything into one by immediately considering coefficients of elements of an ideal of $\mathbb C[x,y]$, finding $n$ generators to match the first $n$ terms of any element as in the proof, then using multiples of $y^n$ to handle the rest.

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