# Maximal ideals of C[x,y]

• Feb 9th 2009, 09:45 PM
Different
Maximal ideals of C[x,y]
Why maximal ideals in a ring of two-variable polynomials with complex coefficients are those generated by x - c, and y - d, for some complex c and d?

Thanks
• Feb 9th 2009, 11:58 PM
NonCommAlg
Quote:

Originally Posted by Different

Why maximal ideals in a ring of two-variable polynomials with complex coefficients are those generated by x - c, and y - d, for some complex c and d?
Thanks

this is a special case of weak Nullstellensatz*. one side is trivial because $\displaystyle \frac{\mathbb{C}[x,y]}{<x-c,y-d>} \simeq \mathbb{C}.$ the other side is much deeper: let $\displaystyle \mathfrak{m}$ be a maximal ideal of $\displaystyle R=\mathbb{C}[x,y].$ then $\displaystyle R/\mathfrak{m}$ is clearly a

finitely generated $\displaystyle \mathbb{C}$ algebra, which is also a field. a well-known result in commutative algebra says that a finitely generated domain over a field F is a field iff it's algebraic over F**. so $\displaystyle R/\mathfrak{m}$

must be an algebraic extension of $\displaystyle \mathbb{C},$ which is possible only if $\displaystyle R/\mathfrak{m}=\mathbb{C}$ because $\displaystyle \mathbb{C}$ is algebraically closed. now let $\displaystyle c,d \in \mathbb{C}$ be the image of $\displaystyle x,y$ under the natural projection $\displaystyle R \longrightarrow R/\mathfrak{m}=\mathbb{C}.$

then clearly $\displaystyle x-c \in \mathfrak{m}$ and $\displaystyle y-d \in \mathfrak{m}.$ thus $\displaystyle <x-c,y-d> \subseteq \mathfrak{m}.$ but we already proved that $\displaystyle <x-c,y-d>$ is a maximal ideal of $\displaystyle R.$ therefore $\displaystyle \mathfrak{m}=<x-c,y-d>. \ \Box$

* if $\displaystyle F$ is any algebraically closed field, then maximal ideals of polynomial ring $\displaystyle F[x_1,x_2, \cdots, x_n]$ are exactly the ideals $\displaystyle <x_1-a_1, x_2-a_2, \cdots , x_n - a_n>, \ \ a_j \in F.$

** see for example page 162 of the book "Graduate Algebra: Commutative View". the author is Louis Halle Rowen.
• Feb 10th 2009, 03:33 AM
Different
Thanks a lot. Funny enough, I understand the other direction, but could you explain why C[x,y] / <x - c, y - d> is isomorphic to C?

I assume the isomorphism sends x to c and y to d....so, why does a polynomial f(x,y) s.t. f(c,d)=0 belongs to the ideal <x - c, y - d> ?
• Feb 10th 2009, 01:30 PM
NonCommAlg
Quote:

Originally Posted by Different
Thanks a lot. Funny enough, I understand the other direction, but could you explain why C[x,y] / <x - c, y - d> is isomorphic to C?

I assume the isomorphism sends x to c and y to d....so, why does a polynomial f(x,y) s.t. f(c,d)=0 belongs to the ideal <x - c, y - d> ?

right! you define $\displaystyle f:\mathbb{C}[x,y] \longrightarrow \mathbb{C}$ by $\displaystyle f(g(x,y))=g(c,d),$ which is obviously a surjective homomorphism. let $\displaystyle I=<x-c,y-d>.$ it's clear that $\displaystyle I \subseteq \ker f.$ now let $\displaystyle g \in \ker f.$ then:

$\displaystyle g(x,y)=\sum a_{ij}x^iy^j=\sum a_{ij}(x-c+c)^i(y-d+d)^j \equiv \sum a_{ij}c^id^j =g(c,d)=0 \mod I.$ thus $\displaystyle g \in I,$ i.e. $\displaystyle \ker f \subseteq I.$
• Feb 10th 2009, 03:49 PM
Different
All clear now. Thanks!