Is the ideal $(X^2-3)$ proper in $\mathbb{F}[[X]]$?

Question

Let $\mathbb{F}$ be a field and $R=\mathbb{F}[[X]]$ be the ring of formal power series over $\mathbb{F}$. Is the ideal $(X^2-3)$ proper in $R$? Does the answer depend upon $\mathbb{F}$?

Clearly $X^2-3=(X+\sqrt3)(X-\sqrt3)$ and hence $X^2-3$ is not zero.

I have no idea whether the ideal is proper or not.

So far, I didn't learn any theorem to prove an ideal is proper. Perhaps I should start with definition of proper ideal? Find one element in $R$ but not in the ideal ?

Answer 1

The element $\sum _{i \geq 0} a_i X^i \in R$ is invertible in $R$ if and only if $a_0\neq 0$.
The key to the proof of that relatively easy result is the identity $(1-X)^{-1}=\sum_{i \geq 0} 1. X^i \in R $
In your question $a_0=-3$ so that the element $X^2-3$ is invertible in $R=F[[X]]$ (which is equivalent to the ideal $(X^2-3)\subset R$ being proper) if and only if the characteristic of the field $F$ is not $3$: $$ (X^2-3)\subset R \;\text {proper} \iff \operatorname{char} F\neq 3.$$

Answer 2

HINT: $$\frac1{x^2-3}=-\frac13\cdot\frac1{1-\frac{x^2}3}=-\frac13\sum_{x\ge 0}\frac{x^{2n}}{3^n}\in R\text{ provided that}\dots ?$$

Answer 3

Since $\mathbb{F}$ is a field then R as you have defined it is an integral domain. Therefore if you can prove that the ideal does not contain 1 then the ideal is proper.