I want to show that $I=(X+Y,X^2+1)\in R=\mathbb{R}[X,Y]$ is a prime ideal. Since $I$ a prime ideal is equivalent to $R/I$ an integral domain, I want to create a homomorphism $\phi: R \to \mathbb{C}$ given by $\phi: p(X,Y)\mapsto p(i,-i)$. This is a surjective homomorphism, so by the first isomorphism theorem we have that $R/\ker(\phi)$ isomorphic to $\mathbb{C}$ and since $\mathbb{C}$ is an integral domain, $R/\ker(\phi)$ is an integral domain, so $\ker(\phi)$ is a prime ideal.
So what's left to show is that $\ker(\phi)=I$.
Since $\phi(X+Y)=i+(-i)=0$, and $\phi(X^2+1)=i^2+1=-1+1=0$, we have that $I\subset\ker(\phi)$. I now want to show that no other elements but elements of $I$ are in the kernel, but I don't understand the following proof:
Suppose $p(X,Y)\in\ker(\phi)$, in other words, $p(i,-i)=0$. This implies that the polynomial $P(X,-X)$ evaluated at $X=i$ is zero and so $p(X,-X)$ is divisible by $X^2+1$. So $P(X,-X)=(X^2+1)Q(X)$ for some $Q\in\mathbb{R}[X]$. Furthermore, $p(X,Y)-p(X,-X)$ is divisble by $Y-(-X)=Y+X$ in $\mathbb{R}[X,Y]$. Hence $p(X,Y)=p(X,-X)+(Y+X)T(X,Y)=(X^2+1)Q(X)+(Y+X)T(X,Y)$. So $p(X,Y)\in(X+Y,X^2+1)$.
I don't understand why $p(i,-i)=0$ implies $p(X,-X)$ evaluated at $X=i$ equals zero (we just fill in the $i$'s?), nor why it then would be divisible by $X^2+1$.
I also don't understand the part starting from "Furthermore". Like why we use that subtraction, or what that means, and why it would be divisible by $Y-(-X)$. Also what is $T(X,Y)$, an equivalent of $Q$?