maximal ideal in R[x,y]

**Stiger** · January 14th 2009, 02:35 AM

Show that every maximal ideal $m \subset \mathbb{R}[x,y]$ is one of the following:
* $m=<x-a,y-b>$ for some $a,b \in \mathbb{R}$ ;
* $m=<y-rx-s,x^2+vx+w>$ for some $r,s,v,w \in \mathbb{R}$ with $v^2-4w<0$ ;
* $m=<x-t,y^2+vy+w>$ for some $t,v,w \in \mathbb{R}$ with $v^2-4w<0$ .

**NonCommAlg** · January 14th 2009, 10:45 PM

Originally Posted by Stiger

Show that every maximal ideal $m \subset \mathbb{R}[x,y]$ is one of the following:

* $m=<x-a,y-b>$ for some $a,b \in \mathbb{R}$ ;

* $m=<y-rx-s,x^2+vx+w>$ for some $r,s,v,w \in \mathbb{R}$ with $v^2-4w<0$ ;

* $m=<x-t,y^2+vy+w>$ for some $t,v,w \in \mathbb{R}$ with $v^2-4w<0$ .

this is a nice problem! i don't know what you know and what you don't, so i'll assume that you know that if $k$ is a field, then any maximal ideal of $k[x_1, \cdots, x_n]$ can be generated by $n$ polynomials

$f_1, \cdots , f_n$ such that $f_i$ is a polynomial in $k[x_1, \cdots , x_i].$ so if $m$ is a maximal ideal of $\mathbb{R}[x,y],$ then $m=<f(x), g(x,y)>.$ now we may assume that $f(x) \in \mathbb{R}[x]$ is irreducible, because if

$f(x)=f_1(x)f_2(x),$ then since $m$ is prime, either $f_1(x) \in m$ or $f_2(x) \in m.$ if $f_1(x) \in m,$ then $<f_1(x),g(x,y)>=<f(x),g(x,y)>=m.$ this proves our claim that we may assume that $f(x)$ is

irreducible. thus either $\deg f = 1,$ or $\deg f$ is an even number. [note that $f$ cannot be a constant because then $m$ wouldn't be maximal] now if $\deg f$ is even, then $f$ is a product of some

irreducible quadratic polynomials in $\mathbb{R}[x].$ thus in this case, we may assume that $f$ is an irreducible quadratic polynomial in $\mathbb{R}[x].$ so you have two cases:

case 1: $\deg f = 1.$ we may assume that $f(x)=x-a.$ now write $g(x,y)=\sum_{j=0}^n u_j(x)y^j,$ where $u_j(x) \in \mathbb{R}[x].$ then dividing each $u_j(x)$ by $x-a$ gives us $g(x,y)=(x-a)h(x,y) + p(y),$ for some

$h(x,y) \in \mathbb{R}[x,y], \ p(y) \in \mathbb{R}[y].$ hence $m=<x-a,g(x,y)>=<x-a,p(y)>.$ again, as we already showed, we may assume that either $\deg p(y)=1,$ or $p(y)$ is an irreducible quadratic

polynomial in $\mathbb{R}[y].$ if $\deg p(y)=1,$ then we'll get $m=<x-a,y-b>$ and if $p(y)$ is an irreducible quadratic polynomial, we'll get $m=<x-a, y^2 + cy + d>,$ where of course $c^2 - 4d < 0,$

because otherwise $p(y)$ would have a real root and hence it wouldn't be irreducible anymore.

case 2: $f(x)$ is an irreducible quadratic polynomial. this case is left for you to try!

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