The unique proper ideal in $\mathbb R[x]/(x^2)$ is maximal

Question

The ring $\mathbb R[x]/(x^2)$ has a unique proper ideal, namely the ideal generated by $x+(x^2)$. How does it follow that this ideal is maximal? I know that the unique proper ideal is maximal provided it contains all non-units of the ring. Am I supposed to use this? I'm not sure how exactly.

Answer 1

Note that if $P(X)=a_nX^n+...+a_1X+a_0$ then $$P(X)+(X^2)=a_0+a_1X+ (X^2)$$

Moreover, if $a_0 \neq 0$ then $$\left( a_0+a_1X+ (X^2) \right)\left( a_0-a_1X+ (X^2) \right)=a_0^2 +(X^2)$$ is an unit in $\mathbb R[X]/(X^2)$.

It follows from here that all the non-invertible elements are in the ideal generated by $X+(X^2)$.

P.S. You can use this idea to prove the stronger statement that the only ideals in $\mathbb R[X]/(X^2)$ are $0+(X^2), X+(X^2)$ and $\mathbb R[X]/(X^2)$. If these are the three ideals, it is trivial to conclude that the only non-trivial one is maximal. This is what the problem is hinting towards.

Answer 2

$\mathbb R[x]/(x^2)/(x/x^2) \cong \mathbb R[x]/(x) \cong \mathbb R$, where the first isomorphism is by the third isomorphism theorem.

This can be proven with your bare hands by taking $\mathbb R[x]/(x^2) \to \mathbb R[x]/(x)$ by $x \mapsto x$. consider the kernel of this map and apply the first isomorphism theorem? The latter isomorphism is clear (I hope!)

Can you see why this is enough to conclude the result?