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I'm trying to determine what $\Bbb{C}[X,Y]/(Y-X^2)$ is equal to. The following method works for simple one variable stuff like $\Bbb{R}[X]/(X^2)$ for example, but I'm wondering if what I've done in the case of two variables here is OK?

Since we are quotienting out by the ideal $(Y-X^2)$ we have the relation $Y-X^2=0 \iff Y=X^2$ in our quotient ring. Then, working with this relation, we get $$\begin{align} \Bbb{C}[X,Y]/(Y-X^2) &=\{ a_0 + a_1X + a_2X^2 + \cdots + b_0 + b_1Y + b_2Y^2 + \cdots \mid a_i, b_i \in \Bbb{C}, Y=X^2\} \\ &= \{ a_0 + b_0 +a_1X + (a_2 + b_2)X^2 + a_3X^3 + \cdots \mid a_i, b_i \in \Bbb{C} \} \\ &\cong \Bbb{C}[X]. \end{align} $$ I'm aware that this can probably be done with a generalised form of Gauss' Lemma but I was wondering whether this way was correct? Thanks.

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  • $\begingroup$ $\mathbb C[X,Y]/(Y-X^2) \cong \mathbb C[X,X^2]\cong \mathbb C[X]$ $\endgroup$ – Mustafa Jul 8 '18 at 21:25
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It is in spirit correct, although the first equation is not strictly speaking correct. Elements of $\mathbb{C}[X,Y]/(Y-X^2)$ are sets of polynomials.

In every element $F\in\mathbb{C}[X,Y]/(Y-X^2)$ there is a unique polynomial $g_F(X)=f(X,X^2)$, for $f\in F$, for which all coefficients of the terms with positive powers of $Y$, are zero. The isomorphism that you are trying to construct sends $F\to g_F$.

$g_F$ is unique because $g_1(X),g_2(X)\in F$ implies that $g_1(X)=g_2(X)+(Y-X^2)H$, for $H\in\mathbb{C}[X,Y]$. Then, setting $Y=X^2$, gives $g_1=g_2$.

Prove that is a ring homomorphism, which is routine computation. It is clear that only $(Y-X^2)$ is sent to the zero polynomial in $\mathbb{C}[X]$. Done.


If you have been studying the isomorphism theorems, then maybe it is suitable to present your argument in this form: Show that the map $\mathbb{C}[X,Y]\to\mathbb{C}[X]$ that sends $f(X,Y)$ to $f(X,X^2)$ is a ring homomorphism. Then prove that the polynomials that are sent to zero are exactly those in $(Y-X^2)$. Finally, apply the first isomorphism theorem.

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