The ring $R = F[X]/(X^2)$, and $(X^2)=0$?

Question

Let $F[X]$ be a polynomial ring over a field $F$ and let $R = F[X]/(X^2)$, and consider the ideal $(X) = (X)/(X^2)$

why the ideal $(X)^2 = 0$ in this ring?

thanks your help

Answer 1

That is the definition of a quotient ring.

Namely in $R/I$, the elements $0$ and $i$ (for $i \in I$) both belong to the coset $I$.

Answer 2

In general, suppose you have a ring $R$ and ideals $I = (r_1, \dots, r_n)$ and $J = (s_1, \dots, s_m)$ of $R$. Then $IJ = (r_i s_j \mid i=1,\dots,n; j =1,\dots,m)$, i.e., the product ideal is generated by the pairwise product of the generators. This can be proven rather straightforwardly using the definition of an ideal given by generators.

Conceptually, a quotient ring $R/I$ can be thought of as 'the ring $R$ with the additional property that all elements in $I$ are equal to $0$'. Using this idea, you can immediately say that $X^2 = 0$ in $F[X]/(X^2)$: it is because $X^2 \in (X^2)$. Of course, formally you have to define the quotient ring $R/I$ and define a homomorpism $R \to R/I$ and only then can you prove that the image of every element of $I$ under that map is $0$. Again, the proof is straightforward using the definition.

In your case, this means that the ideal $(\overline X)^2$ of the quotient ring $F[X]/(X^2)$ is equal to $({\overline X}^2)$. Since ${\overline X}^2 = \overline{X^2} = 0$, this means that $(\overline X)^2 = (0)$.