1

I recently learnt that the maximal ideals of $\mathbb C[x, y]$ are precisely the ones of the form $(x-a, y-b)$ for some $a, b\in \mathbb C$.

I am unable to prove it.

So I considered an easier version of the problem.

Let $M=(p, q)$ be an ideal in $\mathbb C[x, y]$, where $p$ and $q$ are elements of $\mathbb C[x, y]$. If $M$ is maximal, then $\deg p=\deg q=1$.

I am stuck even at this.

An ideal is maximal if quotienting the parent ring with it gives a field, thus $\mathbb C[x, y]/(p, q)$ is a field. But I am having no ideas.

user26857
  • 52,094
caffeinemachine
  • 18,510
  • 2
    The way you phrased the 'easier problem', it will not work. Take any polynomial $f \in \mathbb{C}[x,y]$ with degree larger than $1$ and look at $p = x-a$ and $q = y - b + f \cdot (x - a) $ then the degree of $q$ is not $1$ but still $(p,q)$ will be a maximal ideal. The point is that there exist (!) generators with degree $1$ but it may not be the ones you started with. Even though this might be overkill for your situation you might want to look into Hilbert's Nullstellensatz for a more general statement. – Matthias Klupsch Aug 04 '15 at 08:31
  • Thank you. Do you know some easy proof of the non-easier problem? – caffeinemachine Aug 04 '15 at 08:47
  • Try to proof&use this: $\mathbb{C}$ have no countable/finite algebraic extensions. – qwenty Aug 04 '15 at 08:54
  • Then: $\mathfrak{m} \subset \mathbb{C}[x_1,\dots,x_n]$ where $\mathfrak{m}$ - maximal. $k = \mathbb{C}[x_1,\dots,x_n]/\mathfrak{m}$ - at most countable algebraic extension therefore $k=\mathbb{C}$ therefore $\mathfrak{m} = (x_1-a_1,\dots,x_n-a_n)$ (prove it). – qwenty Aug 04 '15 at 09:01

1 Answers1

0

You have to use Hilbert's Nullstellensatz that's is a polynomial say $p$ in $\mathbb C[x,y]$ have a zero that a point $(a,b)\subset \mathbb{C}^2$,where it vanish in this case it imply the ideal generated by the polynomial $(p)\subset (x-a,x-b)$. Now you can prove.......

user26857
  • 52,094
GGT
  • 1,045