Determine the ideal in $\mathbb{R}[x]$

Question

Consider the ring $\mathbb{R}[x]$, determine the ideal $$(x^3-5x^2+6x-2,x^4-4x^3+3x^2-4x+2)$$

Is it principal? Is it prime? Is it maximal?

Attempt:

We have that $x^3-5x^2+6x-2=0$ in this ideal. Thus, we have that $(x-1)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))=0$.

If $x=1$ the ideal reduces to $$(x^3-5x^2+6x-2,-2)$$

which contains a unit. Similarly, if $x=2+\sqrt{2},2-\sqrt{2}$, the ideal contains a unit. So the ideal is the whole ring. Therefore, it is principal. The ideal is neither prime nor maximal, since such ideals are proper by definition.

Is any of this right? I am wondering if $(x-1)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))=0$ in the ideal means we can have $x$ being equal to any one of these roots. Or is just equal to one of these, and we do not know which one?

Answer 1

1. "The ideal is neither prime nor maximal, since such ideals are proper by definition." 100% true.

2. Since the ideal (call it $I$) equals the entire ring, you have $\mathbb{R}[x] / I \cong \{0\}$, and so $a + I = b + I$, for any $a,b \in \mathbb{R}[x]$. In particular, you have $$x + I = 0 + I = 1 + I = 2 + \sqrt{2} + I = 2 - \sqrt{2} + I.$$ In other words, $$ x = 0 = 1 = 2 + \sqrt{2} = 2 - \sqrt{2} $$ in $\mathbb{R}[x]/I$.

Answer 2

$\mathbf R[x]$ is a P.I.D., as all polynomial rings (in a single variable) over a field.

By definition, the ideal $(x^3-5x^2+6x-2, x^4-4x^3+3x^2-4x+2)$ is generated by the g.c.d. of its generators.

Performing the Euclidean algorithm, you'll find this g.c.d. is the quadratic polynomial $\;\color{red}{x^2-4x+2}\;$ (up to a constant nonzero factor), and that \begin{cases} x^3-5x^2+6x-2=(x-1)(x^2-4x+2),\\ x^4-4x^3+3x^2-4x+2=(x^2+1)(x^2-4x+2). \end{cases}