Maximal ideals of $\mathbb R[x] / \langle (x-a)(x-b) \rangle$ , where $a,b$ are reals and of $\mathbb R[x] / \langle x-(a+\bar a)x+a\bar a \rangle$?

Question

How to determine maximal ideals of $\mathbb R[x] / \langle (x-a)(x-b) \rangle$ , where $a,b$ are reals ? I know it has only four ideals , the ring itself cannot be maximal . Also the zero ideal i.e. $\langle x^2 \rangle /\langle x^2 \rangle$ cannot be maximal as $\mathbb R[x] / \langle (x-a)(x-b) \rangle$ is not a field . So which one of $\langle x-a \rangle/ \langle (x-a)(x-b) \rangle$ , $\langle x-b \rangle/ \langle (x-a)(x-b) \rangle$ is maximal ? ( Ithink both , but cannot actually prove it ) . Please help . And what happens if $a$ is not real and we want to determine maximal ideals of $\mathbb R[x] / \langle x-(a+\bar a)x+a\bar a \rangle$ ?

Answer 1

In general, the maximal ideals of $R/I$ where $R$ is a ring are the images of the maximal ideals of $R$ that contain $I$.

$R=\mathbb R[x]$ is a principle ideal domain, so its maximal ideals are ideals $\langle p(x)\rangle>$ where $p(x)$ is irreducible. The only irreducibles in $\mathbb R[x]$ are of the form $x-\alpha$ for some $\alpha\in \mathbb R$ and $x-2\alpha x + \alpha^2+\beta^2$, which is just another way of writing $(x-(\alpha+\beta i))(x-\overline{\alpha+\beta i})$.

Answer 2

Let $R$ be a commutative ring with identity and let $I$ be an ideal of $R.$ Then the ideals of $R/I$ are the ideals of $R$ containing $I.$ Also the prime ideals of $R/I$ are the prime ideals of $R$ containing $I$ and the maximal ideals of $R/I$ are the maximal ideals of $R$ containing $I.$

In this case $R=\mathbb R[x], I=\langle (x-a)(x-b)\rangle.$ And the ideals $\langle x-a \rangle, \langle x-b\rangle$ are maximal ideals of $R$ containing $I.$ So the image of these two ideals in $R/I$ are maximal ideals.