# Thread: maximal ideal in R[x,y]

1. ## maximal ideal in R[x,y]

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

2. Originally Posted by Stiger

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

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

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

*$\displaystyle m=<x-t,y^2+vy+w>$ for some $\displaystyle t,v,w \in \mathbb{R}$ with $\displaystyle v^2-4w<0$.
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 $\displaystyle k$ is a field, then any maximal ideal of $\displaystyle k[x_1, \cdots, x_n]$ can be generated by $\displaystyle n$ polynomials

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

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

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

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

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

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

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

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

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