Determine maximal ideals of $\mathbb R[x]/(x^2)$

Question

Artin Algebra Chapter 11

For (b), a solution can be found here, which I think is the same as Takumi Murayama here. I have a question about the solution of Brian Bi here.

1. What does it mean that "Obviously $I$ can't contain any nonzero constant without being the entire ring." ?

My confusion is that the elements of $R$ are of this form $$[p(x)+(x^2)]$$ rather than real numbers. So constants in $R$ are $[p(x)+(x^2)]$ for $p(x)=x^2+k$ for some $k \in \mathbb R$?

I think it means that for $a \in \mathbb R$, if $[a+(x^2)] \in I$, then $I = R$. But I think we can also have $p(x) \in \mathbb R[x]$ such that $[p(x)+(x^2)]=[b+(x^2)]$ for some $b \in \mathbb R$ such as $p(x)=x^2-17$ and $b=-17$. If I'm right, then is this correct?

$$[a+(x^2)] \in I \implies [a+(x^2)][\frac1a+(x^2)] = [1+(x^2)] \in I \implies I = R$$

2. Why exactly do we have

$$\forall c \in I \setminus (\alpha), \exists a,b \in R: c = a\alpha + b, a \ne 0 \ne b?$$

I think I understand intuitively, for finitely generated $I=(\alpha, \beta_1, ..., \beta_n)$, we have $$c=a \alpha + \sum_{i=1}^{n} b_i \beta_i$$

where at least one of the $b_i$'s is not zero, but

• Why is $a \ne 0$?

• What if $I$ is not finitely generated?

Answer 1

The simplest way to approach these questions is by using the isomorphism theorem, in the form

The ideals of $\mathbb R[x]/(f(x))$ correspond to the ideals of $\mathbb R[x]$ that contain $(f(x))$.

Then use that $\mathbb R[x]$ is a PID to find the ideals of $\mathbb R[x]$ that contain $(f(x))$.