Third Isomorphism Theorem for Rings confusions

Question

I am having some confusion about the Third Isomorphism Theorem for Rings that I hope could be clarified.

Take $\mathbb{R}[x,y]$. I want to mod out by $(x^2 + 1, y^2 + y + 7)$. Therefore, I use the third isomorphism theorem to say that doing so is equivalent to $\mathbb{R}[x,y] / (x^2 +1) / (x^2 + 1, y^3 + y) / (x^2 + 1)$.

I know that from the proof of the third isomorphism theorem it must be the case that $(x^2 + 1, y^2 + y + 7) / (x^2 + 1)$ is an ideal of $\mathbb{R}[x,y] / (x^2 + 1)$. But here is my confusion:

1) is there a way to say explicitly what $(x^2 + 1, y^2 + y + 7) / (x^2 + 1)$ is (besides just saying that it is a set of cosets)?

2) $\mathbb{R[x,y]} / (x^2 + 1)$ is isomorphic to $\mathbb{C}[t]$ under the isomorphism derived as a result of the map from $\mathbb{R}[x,y]$ to $\mathbb{C}[t]$ $x \mapsto i$, $y\mapsto t$ (first isomorphism theorem). Therefore, the ideal $(x^2 + 1, y^2 + y + 7) / (x^2 + 1)$ will be mapped (under the inverse of the above isomorphism) to an ideal in $\mathbb{C}[t]$. In my notes, I have (with no explanation) that this ideal is $(t^2 + t + 7)$. Could someone explain why this is the case?

Thanks!

Answer 1

Since you have an isomorphism, you can go in both directions, so just apply the isomorphism you gave:

$$(x^2 + 1, y^2 + y + 7) / (x^2 + 1) \mapsto (i^2 + 1, t^2 + t + 7) = (0, t^2 + t + 7) = ( t^2 + t + 7).$$

Or just notice that

$$ (x^2 + 1, y^2 + y + 7) / (x^2 + 1) = (y^2 + y + 7) / (x^2 + 1).$$

Answer 2

1) One way to visualize this quotient would be by viewing it as the first ideal "minus" the other. We have a homomorphism from $(x^2+1,y^2+y+7)$ to $(x^2+1,y^2+y+7)/(x^2+1)$ which sends elements from $(x^2+1)$ to $0$ , so that, in one sense, we "eliminated" the second ideal from the other.

2) Indeed, if we map $x$ to $i$ and $y$ to $t$, $(x^2+1,y^2+y+7)\mapsto (i^2+1,t^2+t+7)=(0,t^2+t+7)=(t^2+t+7)$ and $(x^2+1)\mapsto (i^2+1)=0$. As well, $(t^2+t+7)/{0}=(t^2+t+7)$, which is the result you wanted.