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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 $\mathbb R[x] / \langle x-(a+\bar a)x+a\bar a \rangle$ ?

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  • $\begingroup$ For the last question, do you mean $\mathbb C[x]/<(x-a)(x-\overline a)>$? $\endgroup$ – Gregory Grant Mar 15 '15 at 15:17
  • $\begingroup$ @GregoryGrant : No I actually mean something like $\mathbb R [x] / \langle x^2+x+1 \rangle $ $\endgroup$ – user217921 Mar 15 '15 at 15:18
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    $\begingroup$ In the case of $a,\overline{a}$, the ring $\mathbb R/\langle (x-a)(x-\overline{a})\rangle\cong \mathbb C$ is a field, so the zero ideal is the only ideal and it is maximal. $\endgroup$ – Thomas Andrews Mar 15 '15 at 15:20
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    $\begingroup$ @GregoryGrant $(x-a)(x-\overline{a})\in\mathbb R[x]$. $\endgroup$ – Thomas Andrews Mar 15 '15 at 15:22
  • $\begingroup$ Anyway, both those ideals you mention are maximal. Why do you think it's only one of them? $\endgroup$ – Gregory Grant Mar 15 '15 at 15:22
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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 $(x-(\alpha+\beta i))(x-\overline{\alpha+\beta i})$.

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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.

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  • $\begingroup$ Is commutativity required for your first paragraph to be true? $\endgroup$ – Gregory Grant Mar 15 '15 at 15:31
  • $\begingroup$ @GregoryGrant: Not really! But in that case one has to consider left ideals, left prime ideals... In this particular case, since it is commutative, I assume it in the statement. $\endgroup$ – Krish Mar 15 '15 at 15:34
  • $\begingroup$ But we can't take the quotient if it's just a left or right ideal, it has to be two-sided right? So is the statement true if we take "ideal" (unqualified) to always mean two-sided? $\endgroup$ – Gregory Grant Mar 15 '15 at 16:02
  • $\begingroup$ @GregoryGrant: You are right. That was my point actually. Sorry for being sloppy. In general the homomorphisma and isomorphism theorems are stated for ideals (which means two-sided ideals), without the commutativity assumption. $\endgroup$ – Krish Mar 15 '15 at 16:11
  • $\begingroup$ I understand, thank you $\endgroup$ – Gregory Grant Mar 15 '15 at 16:15

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