Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2006 / 3

The evaluation homomorphism $\mathbb{R}[X]\to \mathbb{R}$, $p\mapsto p(0)$, is surjective and has kernel of a ring homomorphism $J_1=(X)$. Hence $J_1$ is an ideal, and the first isomorphism theorem for rings gives $\boxed{\mathbb{R}[X]/J_1\cong \mathbb{R}}$. Since this quotient is a field, $\boxed{J_1\text{ is maximal}}$.

It is not an ideal. The constant polynomial one belongs to $J_2$, but its product with $X$ does not: the derivative of $X$ at zero is one. Thus $J_2$ fails closure under multiplication by arbitrary elements of $\mathbb{R}[X]$, a required ideal property. Closure under addition alone would not suffice.

The conditions set the constant and linear coefficients to zero, so $\boxed{J_3=(X^2)}$. This is a principal ideal, but it is not maximal: there is a strict chain $(X^2)⊊(X)⊊\mathbb{R}[X]$. In the quotient ring, the nonzero class of $X$ has square zero, so the quotient is not a field.

Here $J_4=(X^2+1)$ is a principal ideal. The evaluation homomorphism $p\mapsto p(i)$ from $\mathbb{R}[X]$ onto $\mathbb{C}$ has this kernel of a ring homomorphism: division by $X^2+1$ leaves a remainder $a+bX$, and $a+bi=0$ with real $a,b$ forces both to vanish. Surjectivity follows because every complex number is $a+bi$. ConsequentlyEquivalently, $X^2+1$ is an irreducible polynomial over the real numbers.