is $\mathbb{R}[x]/(x^2)$ a field? My guess is no, but I cannot find a counter example. I started off by saying let $x_1,x_2 \in \mathbb{R}[x]/(x^2)$ then consider $x_1x_2 = (a+bx)(c+dx) = (ad+bc)x + ac = 0$ but I'm not sure how to get a contadiction from here
 A: $x\in\mathbb{R}[x]/(x^2)$ and $x\neq0$, but $x$ has no inverse since $x(a+bx)=ax\neq 1$ for any $a,b$.
A: Hint $\ $  In a field, $\ \, xf = 0\,\Rightarrow\,x=0\,$ or $\,f = 0,\ $ so $\, x^2 = 0\,\Rightarrow\, x = 0\, \ \Rightarrow\!\Leftarrow$
Or $\,\ xf = 1\,$ in $\,\Bbb R[x]/\color{#c00}{xg}\,\Rightarrow\, xf = 1 + \color{#c00}{xg}\, h\ $ in $\,\Bbb R[x]\,\overset{\large x\,=\,0}\Rightarrow\, 0 = 1\,$ in $\,\Bbb R\ \Rightarrow\!\Leftarrow$
Remark $\ $ This is known as the algebra of dual numbers over the ring $\Bbb R.\,$  Such rings and higher order analogs $\,\Bbb R[x]/x^n \;$  prove quite useful when studying (higher) derivations algebraically since such rings provide very convenient algebraic models of tangent / jet spaces. For example, they permit easy transfer of properties of homomorphisms to derivations -- see for example section $8.15$ in N. Jacobson, Basic Algebra II. See this post for further discussion and links.
A: We have a few facts.


*

*Let $\mathfrak a$ be an ideal in a ring $A$. Then $A/\mathfrak a$ is a field if and only if $\mathfrak a$ is a maximal ideal.

*Let $\mathfrak a$ be an ideal in a PID $A$. Then $\mathfrak a$ is a maximal ideal if and only if $\mathfrak a$ is a prime ideal.

*If $k$ is a field, then $k[x]$ is a PID.

*If $k$ is a field with $f\in k[x]$, then $\langle f\rangle$ is a prime ideal in $k[x]$ if and only if $f$ is irreducible.


Can you see how these facts combine to prove that $\Bbb R[x]/\langle x^2\rangle$ is not a field?
A: No need of complicated proofs. Let $I=(x^2)$ and consider $x+I$; then
$$
(x+I)^2=x^2+I=I
$$
Thus the ring is not a domain. We just need to ensure that $x\notin I$, but this is clear because nonzero polynomials in $I$ have degree at least $2$.
