A problem in my textbook asks me to show that $\mathbb{R}[X,Y]/(X^2+1, Y^2+1) \cong \mathbb{C} \times \mathbb{C}$. After I've tried solving this for a long time, I looked at the textbook's hint and it said that I should consider the homomorphism $\phi : \mathbb{R}[X, Y] \to \mathbb{C} \times \mathbb{C}$, $\phi(f)=(f(i,i), f(i, -i))$ and this would be a surjective ring homomorphism whose kernel is the ideal $(X^2+1, Y^2+1)$, so I would be done.
The fact that this is a ring homomorphism is pretty obvious to me. However, I am really struggling to show that this function is surjective and to find the kernel.
I think that this all stems from the fact that I am not used to the arithmetic properties of polynomial rings in two indeterminates. Anyway, back to the problem. For the surjectivity I tried playing arround with a few polynomials and I hoped to kind of guess the right one, but no succes. Showing that $(X^2+1, Y^2+1)\subset \operatorname{Ker}\phi$ is really straightforward, but the other inclusion is really killing me. So, how should I go about doing this?
Best Answer
Kernel: Doing polynomial division where you have two polynomials in multiple indeterminates is in general a tricky business where you first have to fix an ordering. In this case, we can kind of hack our way through.
Let $f(X,Y)$ be the polynomial we want to divide by $X^2+1$ and $Y^2+1$. We can write $f(X,Y)=(X^2+1)(...)+(Y^2+1)(...)+(c_1XY+c_2X+c_3Y+c_4)$.
When we divide by $X^2+1$, we are effectively saying replace $X^2$ with $-1$ and likewise with $Y^2$ and $-1$. So you can get a "remainder" where all the $X$ terms have degree $1$ and $Y$ terms have degree $1$.
Now, evaluate at $(i,i)$ and $(i,-i)$ to see
$0=f(i,i)=0(...)+0(...)+c_1(-1)+c_2i+c_3i+c_4=(-c_1+c_4)+i(c_2+c_3)$
$0=f(i,-i)=0(...)+0(...)+c_1(1)+c_2i-c_3i+c_4=(c_1+c_4)+i(c_2-c_3)$
You can solve the simultaneous equations in $c_i$ to see that all $c_i$ have to be $0$.
Surjective: Let $(a+bi,c+di)\in\mathbb{C}^2$.
We look at the "remainder" term from before $f(X,Y)=c_1XY+c_2X+c_3Y+c_4$. We want $f(i,i)=a+bi, f(i,-i)=c+di$.
You get $a+bi=(-c_1+c_4)+(c_2+c_3)i$ and $c+di=(c_1+c_4)+i(c_2-c_3)$. You can again repeat the same thing as before and solve for the simultaneous equations.