I have to construct a ring isomorphism $\mathbb{C}[x,y]/\langle x^2+2xy+y^2-529,x+3y-1\rangle \cong \mathbb{C}\times \mathbb C$ and its inverse.
I tried to find a function with $\langle x^2+2xy+y^2-529,x+3y-1\rangle$ as a Kernel but couldn't figure it out. I also don't quite understand what $\mathbb{C}\times\mathbb{C}$ actually is other than it means the cartesian product of $\mathbb C$ with itself
It feels like a fairly simple question with a simple solution but I'm getting nowhere. Any help would be greatly appreciated as I'm currently tearing my hair out trying to do anything with it.
Let $A$ be $ℂ[x, y]$ and let $I$ be the given ideal in $A$. Let $J ≔ ⟨x + 3y - 1⟩$, which is an ideal in $A$ with $A ⊇ I ⊇ J$. By the third isomorphism theorem we have $$ A / I ≅ (A / J) / (I / J) \,. $$
The quotient $A / J$ is easy to determine: modding out the polynomial $x + 3y - 1$ identifies $x$ with $1 - 3y$. More explicitly, we have the isomorphism $$ φ \colon A / J ≅ ℂ[y] \,, \quad [x] \mapsto 1 - 3y \, \quad [y] \mapsto y \,. \tag{1} $$ Under the isomorphism $φ$ the ideal $I / J$ of $A / J$ corresponds to an ideal $K$ in $ℂ[y]$, and the isomorphism $φ$ descends to an isomorphism $$ (A / J) / (I / J) ≅ ℂ[y] / K \,. $$ The ideal $I / J$ is generated by the two residue classes $[x^2 + 2xy + y^2 - 529]$ and $[x + 3y - 1]$, whence the ideal $K$ is generated by $φ([x^2 + 2xy + y^2 - 529])$ and $φ([x + 3y - 1])$. The second generator of $K$ is $0$ because already $[x + 3y - 1] = 0$ in $A / J$ (after all, we modded out this generator when we formed $A / J$). The first generator of $K$ is $$ (1 - 3y)^2 + 2(1 - 3y)y + y^2 - 529 = 4 y^2 - 4 y - 528 = 4 ⋅ (y^2 - y - 132) \,. $$ The factor $4$ is invertible in $ℂ[y]$, whence $K$ is generated by the single polynomial $y^2 - y - 132$. We have the decomposition $$ y^2 - y - 132 = (y - 12) (y + 11) \,. $$ It now follows from the Chinese Remainder Theorem (for principal ideal domains) that \begin{align*} ℂ[y] / K &= ℂ[y] / ⟨y^2 - y - 132⟩ \\ &= ℂ[y] / ⟨(y - 12)(y + 11)⟩ \\ &≅ ℂ[y] / ⟨y - 12⟩ × ℂ[y] / ⟨y + 11⟩ \tag{2} \\ &≅ ℂ × ℂ \,. \end{align*} We use here that $$ ℂ[x] / ⟨x - a⟩ ≅ ℂ \,, \quad [p] \mapsto p(a) \tag{3} $$ for every $a ∈ ℂ$.
If you wish to write down an explicit isomorphism between $A/I$ and $ℂ × ℂ$ then you’ll need to trace the images of $[x]$ and $[y]$ under the above sequence of isomorphisms \begin{align*} A / I &≅ (A / J) / (I / J) \\ &≅ ℂ[y] / K \\ &≅ ℂ[y] / ⟨y - 12⟩ × ℂ[y] / ⟨y + 11⟩ \\ &≅ ℂ × ℂ \,. \end{align*} We already have explicit formulas for the steps $(1)$ and $(3)$, and the isomorphism $(2)$ is given by $[p] \mapsto ([p], [p])$. We thus find that $$ [x] \mapsto [[x]] \mapsto [1 - 3y] \mapsto ([1 - 3y], [1 - 3y]) \mapsto (1 - 3 ⋅ 12, 1 - 3 ⋅ (-11)) = (-35, 34) $$ and $$ [y] \mapsto [[y]] \mapsto [y] \mapsto ([y], [y]) \mapsto (12, -11) \,. $$ The overall isomorphism between $A / I$ and $ℂ × ℂ$ is thus given by sending $[x]$ to $(-35, 34)$ and $[y]$ to $(12, -11)$.