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Find all the maximal ideals in the ring $\mathbb{R}[x]$.

My work:

If $M$ is maximal, then $\mathbb{R}[x]/M$ is a field. Then if $M$ is of the form $(p(x))$ then $p(x)$ must be irreducible. So the principal ideals generated by irreducible polynomials in $\mathbb{R}[x]$ are maximal in $\mathbb{R}[x]$.

But $\mathbb{R}[x]$ is not a PID. Hence, there are ideals which are not principal. So how can I find all maximal ideals? Can anyone please help me to find it?

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    $\begingroup$ $\mathbb{R}[x]$ is a PID. in fact $k[x]$ is a PID whenever $k$ is a field. actually it is more than that: it is an Euclidean domain. $\endgroup$ – Krish Dec 31 '14 at 18:15
  • $\begingroup$ Oh thanks, Ok then, I have to find all irreducible polynomials in $\mathbb{R}[x]$. Then I think the maximal ideals are of the form $(x^{2n}+a)$ where $n\in \mathbb{N}, a>0$. Am I correct? $\endgroup$ – Extremal Dec 31 '14 at 18:21
  • $\begingroup$ Ok then the maximal ideals are of the form $(ax^2+bx+c)$ where $b^2-4ac<0$ or $(x+a)$ where $a\in \mathbb{R}$ $\endgroup$ – Extremal Dec 31 '14 at 18:34
  • $\begingroup$ By the way $k[x,y]$ is not always a PID when $k$ is a field. Take $k=\mathbb{C}$ and the ideal $(x,y)$. I think this ideal is not principal for any field $k$ but I'm not fully sure. $\endgroup$ – user135520 Jan 1 '15 at 22:34
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Since every odd degree real polynomial has a root in $\mathbb{R},$ an irreducible polynomial of degree $> 2$ must be of even degree. Let $p(x) \in \mathbb{R}[x]$ be an irreducuble polynomial of degree $2d, d \geq 1.$ We can also assume that it is monic. In $\mathbb{C}[x], p(x)$ will factor in linear factors. Let $p(x) = \prod_{1 \leq i \leq2d} (x - \alpha_i) \in \mathbb{C}[x].$ Since complex roots occurs in conjugate pairs, we can rename the $\alpha_i$'s to write $p(x) = \prod_{1 \leq i \leq d} (x - \alpha_i)(x - \overline{\alpha}_i)$ where $\overline{\alpha}_i$ is the complex conjugate of $\alpha_i.$ From this we can write, $p(x) = \prod_{1 \leq i \leq d}q_i(x)$ where $q_i(x) = (x - \alpha_i)(x - \overline{\alpha}_i)$ is a real quadric polynomial. This shows that $d = 1.$ So $p(x)$ is of the form $ax^2 + bx + c$ with $b^2 - 4ac < 0.$

Hence the maximal ideals of $\mathbb{R}[x]$ are of the form $(ax^2 + bx + c)$ with $b^2 - 4ac < 0$ or of the form $(x - a), a \in \mathbb{R}.$

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