Why do elements in $\mathbb R[x]/(x^2)$ look like at most degree one polynomials?

Question

What do elements in $\mathbb R[x]/(x^2)$ look like? Intuitively, I know they are polynomials of degree one but if we stick with the definition of quotient ring, the elements look like $f(x)+(x^2)g(x)$.

How does this show that the elements are polynomials of degree at most $1$?

Answer 1

Because every polynomial $P(x)$ can be written in one and only one way as $x^2Q(x)+R(x)$, where $\deg R(x)\leqslant1$. So, it is natural to identify $P(x)+x^2\mathbb{R}[x]$ with $R(X)$, whose degree is $0$ or $1$.

Answer 2

A nicer way of looking at the quotient $\mathbf{R}[x]/(x^2)$ is the following: $\mathbf{R}[x]$ is just a free $\mathbf{R}-$algebra (with the usual multiplication) on elements $\{1,x,x^2,\ldots\}$. Now, by quotienting by $(x^2)$, we are declaring that any element of our quotient ring from which $x^2$ can be factored out is trivial. You can think about this as imposing an algebraic relation of $x^2=0$. Then, the polynomials of $\mathbf{R}[x]$ are of the form $$ a_nx^n+\cdots+a_1x^{1}+a_0$$ After quotienting, we have that all of the monomial terms with degree $\ge2$ are killed, so that we have a resulting polynomial of $a_1x^1+a_0$. Constructing the isomorphism between these two rings makes this formal.

Answer 3

Because in a polynomial modulo $x^2$, all terms of degree $\ge 2$ are killed, just as in the representation of a element in $\mathbf Z/100\mathbf Z$, you keep only the last two digits.