# maximal ideal in $\mathbb{R}$[x]

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.

• What about the ideal generated by $x^2$? – MooS Jun 12 '17 at 12:44
• What about the ideal $(1)$? – Arthur Jun 12 '17 at 12:44
• Non constant irreducible polynomial... – Martín-Blas Pérez Pinilla Jun 12 '17 at 19:58
• $x^2=x.x$ 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 – dipali mali Jun 13 '17 at 3:55
• But , I is maximal ideal then there exists a non constant polynomial f(x) $\in$I of degree $\le$2 is always true. – dipali mali Jun 13 '17 at 4:04