Describe the multiplication in the ring $F[x]/(x^2)$.
Is this a field?
What type of element is $[x]$.
I'm learning about field extensions and I have a general idea of what is happening but I'm having trouble working with the elements and the notation.
So for the first part, I would describe this as, we take $a,b \in F[x]/(x^2)$. Then $a \times b$, I think would be $a^2 \times b^2 = (ab)^2 \in F[x]/(x^2)$. I'm not entirely sure what is being asked here.
For the second part, we already know $F[x]/(x^2)$ is a ring and all we have to do is show multiplicative commutativity, identity, and inverse. But I don't know how to do this because as stated before, I don't know what elements of this should look like like.
Finally I don't understand what is being asked by type of element.
Any help proceeding further is appreciated.
No. Modulo $(x^2)$, the ideal generated by $x^2$, all polynomials are represented by some (unique) linear $a+bx$ since $x^k\equiv 0$ for $k\ge 2$. In particular the product of two such elements is given by $$(a+bx)(a'+b'x)=aa'+(ab'+a'b)x.$$ In this ring the inverse of an element $a+bx$, if it has one, is the element $t+ux$ such that $$at+(au+bt)x=1\iff \begin{cases}at=1\\au+bt=0\end{cases}$$ This shows $a+bx$ is invertible if and only if $a\ne 0$. It's easy to check that necessarily, $$t=a^{-1},\quad u=-a^{-2}b,\quad\text{so that}\quad (a+bx)^{-1}= a^{-1}-a^{-2}bx.$$