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Let $\mathbb{R}$[x] be the polynomial ring over $\mathbb{R}$ in one variable .Let I$\subseteq$$\mathbb{R}$[x] be an ideal. Then

'I is a maximal ideal iff there exists a non constant polynomial f(x)$\in$I of degree $\le$ 2'

Is this statement is true?

I know that $\mathbb{R}$[x] is PID and hence an ideal is irreducible iff it is maximal ideal.

I know that degree of any irreducible polynomial over $\mathbb{R}$ is 1 or 2.

so according to me this statement is correct .

please correct me if i am wrong.

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  • $\begingroup$ What about the ideal generated by $x^2$? $\endgroup$ – MooS Jun 12 '17 at 12:44
  • $\begingroup$ What about the ideal $(1)$? $\endgroup$ – Arthur Jun 12 '17 at 12:44
  • $\begingroup$ Non constant irreducible polynomial... $\endgroup$ – Martín-Blas Pérez Pinilla Jun 12 '17 at 19:58
  • $\begingroup$ $x^2=x.x$ so $x^2 $ is reducible in $\mathbb{R}$ ok got it .this statement is not true other way that is $x^2 \in (x^2)$ but $(x^2) $ is not maximal ideal. But $\endgroup$ – dipali mali Jun 13 '17 at 3:55
  • $\begingroup$ But , I is maximal ideal then there exists a non constant polynomial f(x) $\in$I of degree $\le$2 is always true. $\endgroup$ – dipali mali Jun 13 '17 at 4:04

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