Pair notation in multivariate polynomial rings: ideal vs. tuple

Solution 1:

Yes, this is what the map means. The second map you mention \begin{align*} \alpha : \Bbb{C}[x,y]&\to\Bbb{C}[x,y]^2\\ f&\mapsto (-yf,xf) \end{align*} is a completely different map. It is related to your first map $\beta : \Bbb{C}[x,y]^2\to(x,y)$ by the following formula: $\beta\circ\alpha = 0.$ We can check this explicitly: \begin{align*} \beta\circ\alpha(f) &= \beta(-yf(x,y),xf(x,y))\\ &= x(-yf(x,y)) + y(xf(x,y))\\ &= -xy f(x,y) + xy f(x,y) = 0. \end{align*} In fact, the sequence $$ 0\to\Bbb{C}[x,y]\xrightarrow{\alpha}\Bbb{C}[x,y]^2\xrightarrow{\beta}(x,y)\to 0 $$ is a short exact sequence, meaning that the kernel of $\beta$ is precisely the image of $\alpha$ (and that $\alpha$ is injective and $\beta$ is surjective). To check this, you need to show that if $(f,g)\in\ker(\beta),$ then $(f,g) = (-yh, xh)$ for some $h\in\Bbb{C}[x,y].$ First, note that $(f,g)\in\ker(\beta)$ implies that $xf(x,y) = -yg(x,y).$

Since $\Bbb{C}[x,y]$ is a unique factorization domain and $x$ and $y$ are prime elements, it follows that $x\mid g$ and $y\mid f$. So, we may write $f(x,y) = y \tilde{f}(x,y)$ and $g(x,y) = x\tilde{g}(x,y)$ for some polynomials $\tilde{f},\tilde{g}\in\Bbb{C}[x,y].$

But now we have $xy\tilde{f}(x,y) = -xy\tilde{g}(x,y),$ and since $\Bbb{C}[x,y]$ is a domain, it follows that $\tilde{f}(x,y) = -\tilde{g}(x,y).$ Hence, we have shown that $\ker(\beta) = \operatorname{im}(\alpha).$

What the author has done is construct a free resolution of $(x,y)$ as a $\Bbb{C}[x,y]$-module.

TL;DR: Your first map has domain $\Bbb{C}[x,y]^2$ and codomain $(x,y),$ where the former is ordered pairs of polynomials in two variables $x$ and $y$ and coefficients in the complex numbers, and the latter is the ideal of $\Bbb{C}[x,y]$ generated by $x$ and $y.$ Remember, $(x,y)$ is simply shorthand for the following set: $$ \{f(x,y)\in\Bbb{C}[x,y]\mid f(x,y) = x p(x,y) + y q(x,y)\textrm{ for some }p(x,y),q(x,y)\in\Bbb{C}[x,y]\}. $$ So the first map takes an ordered pair $(f(x,y),g(x,y))$ (not an ideal!) and sends it to the element $xf(x,y) + yg(x,y),$ which is in the ideal $(x,y).$

Your second map has domain $\Bbb{C}[x,y]$ and codomain $\Bbb{C}[x,y]^2,$ and sends a polynomial $f(x,y)$ to the ordered pair of polynomials $(-y f(x,y), xf(x,y)),$ which is an element of $\Bbb{C}[x,y]^2$ (not an ideal!).