Proving an ideal of polynomial ring

Question

I posted this question here intended to include it as part of my writing. However, I got stuck again by something of entirely different nature: How do you prove that $(x_1), (x_1, x_2), (x_1, x_2, x_3),$ etc., are ideals of of $\mathbb C [x_1, x_2,...]$?

Especially, suppose that $f(x_1) \in (x_1)$ and $g(x_1, x_2,...) \in \mathbb C[x_1, x_2, ...]$, how should I convince myself that $(x_1)$ absorbs multiplication, i.e., $fg \in (x_1)$?

Thank you for your time and help.

Answer 1

The definition of $(x_1)$ depends on the ring it lives in: $$(x_1)\subset \mathbb C[x_1]\Rightarrow (x_1)=\{x_1f(x_1)\mid f(x_1)\in \mathbb C[x_1]\}$$ $$(x_1)\subset \mathbb C[x_1,x_2]\Rightarrow (x_1)=\{x_1f(x_1,x_2)\mid f(x_1)\in \mathbb C[x_1,x_2]\}$$ and so on. In general, $(a)\subset R$ is defined by $(a)=\{ar\mid r\in R\}$ (EDIT: well, when $R$ is commutative anyhow), so it is transparent in the definition that the ring itself matters.

Edit: In particular, in your context, $(x_1)$ (which you might denote by $(x_1)_{\mathbb C[x_1,x_2,\ldots,]}$ or $x_1\mathbb C[x_1,x_2,\ldots]$ to make this clearer) is the set $\{x_1f(x_1,x_2,\ldots)\mid f(x_1,x_2,\ldots)\in \mathbb C[x_1,x_2, \ldots]\}$ so contains, for example, $x_1x_2$, $x_1x_2x_3$, $x_1x_2x_3x_4$, and so on by definition.

Answer 2

By definition $(f_1, f_2, \ldots, f_n)$ denotes the ideal generated by $\{f_1, f_2, \ldots, f_n\}$, that is, the smallest ideal containing $\{f_1, f_2, \ldots, f_n\}$.

In particular, the typical element of $(x_1)$ is $x_1g(x_1, x_2,...)$, not $f(x_1)$.