# Quotient Rings of Polynomials

Let $$R = \mathbb{C}[x,y]$$. Then $$R/Rx$$ is isomorphic to $$\mathbb{C}[y]$$.

My Proof

$$R = ax+by+c , a,b,c \in \mathbb{C}$$ and $$Rx = (ax+by+c)x , a,b,c \in \mathbb{C}$$

So, $$R/(R\cdot x) = (ax+by+c) + (a'x+b'y+c')x, a,b,c a',b',c' \in \mathbb{C}$$ $$= by+c + (a'x+b'y+(c'+a))x, a,b,c a',b',c' \in \mathbb{C}$$ So the set of cosets of $$R/(R\cdot x)$$ is clearly isomorphic to $$\mathbb{C}[y]$$.

This argument feels incredible handwavy. Is there a more rigorous way to prove this?

(I know in some cases defining an appropriate ring homomorphism and finding the kernel gives us the isomorphism by the first isomorphism theorem, but I couldn't find a homomorphism that worked.)

• I'd say $R=\mathbb C[x,y]=c+ax+by+dx^2+exy+fy^2+gx^3+hx^2y+ixy^2+jy^3+...$ – J. W. Tanner Nov 24 '19 at 15:37
• When you typed $r$, did you mean $R$? – J. W. Tanner Nov 24 '19 at 15:38
• But still $R \cdot x$ has no terms with just $y$ and no constant terms correct? So when we mod out $R \cdot x$, we have two cosets: identity coset, and one with all multiples of y @J.W.Tanner – Jess Nov 24 '19 at 15:59
• correct; $Rx=cx+ax^2+byx+dx^3+..$ – J. W. Tanner Nov 24 '19 at 16:02
• Thanks! Is there an approach that doesn't use the coset descriptions? or is this the most straightforward approach? – Jess Nov 24 '19 at 16:03

Hint. Can you prove that \begin{align*} \Phi:\mathbf{C}[X,Y] &\longrightarrow \mathbf{C}[Y] \\ f(X,Y) &\longmapsto f(0,Y) \end{align*} is a surjective ring homomorphism? Its kernel is exactly $$(X)\subset \mathbf{C}[X,Y]$$ (why?). Now you are ready to use the first isomorphism theorem.
• But don't I need the kernel to be ($X \cdot \mathbb{C}[X,Y]$)? – Jess Nov 24 '19 at 16:23
• This is just notation. We can denote the ideal by $(X)$ and $(X)\cdot \mathbf{C}[X,Y]$. Both mean the same thing: all polynomials in $X,Y$ with coefficients in $\mathbf{C}$ that are divisible by $X$. – rae306 Nov 24 '19 at 16:24