Maximal ideals of C[x,y]

**Different** · February 9th 2009, 09:45 PM

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

**NonCommAlg** · February 9th 2009, 11:58 PM

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 $\frac{\mathbb{C}[x,y]}{<x-c,y-d>} \simeq \mathbb{C}.$ the other side is much deeper: let $\mathfrak{m}$ be a maximal ideal of $R=\mathbb{C}[x,y].$ then $R/\mathfrak{m}$ is clearly a

finitely generated $\mathbb{C}$ 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 $R/\mathfrak{m}$

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

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

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

** see for example page 162 of the book "Graduate Algebra: Commutative View". the author is Louis Halle Rowen.

**Different** · February 10th 2009, 03:33 AM

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> ?

**NonCommAlg** · February 10th 2009, 01:30 PM

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

$g(x,y)=\sum a_{ij}x^iy^j=\sum a_{ij}(x-c+c)^i(y-d+d)^j \equiv \sum a_{ij}c^id^j =g(c,d)=0 \mod I.$ thus $g \in I,$ i.e. $\ker f \subseteq I.$

**Different** · February 10th 2009, 03:49 PM

All clear now. Thanks!

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