Artin Algebra:
Definition of maximal ideals:
Maximal ideals of $F[x]$:
In connection with these, I think $x-2$ is monic irreducible in $\mathbb R[x]$ (and other fields if it's absolutely irreducible), so we should have the quotient ring $\mathbb R[x]/(x-2)$ to be a field. If we were to prove this from definition of a field, then we satisfy all properties besides multiplicative inverse and now must show that a non-zero element of $\mathbb R[x]/(x-2)$ has a multiplicative inverse. Here is what I have tried:
An element of $\mathbb R[x]/(x-2)$ has the form $[a+(x-2)], a \in \mathbb R[x]$, and the multiplicative identity of $\mathbb R[x]/(x-2)$ is $[1+(x-2)]$ because $$[a+(x-2)][1+(x-2)] = [(a)(1)+(x-2)] = [a+(x-2)]$$
If $a$ is constant, then $[a+(x-2)]$'s inverse is $[\frac 1 a+(x-2)]$.
And now I am stuck.
- For non-constant polynomials like $a=2x^2+1$, what's the multiplicative inverse of $$[2x^2+1+(x-2)]$$?
I think the adding relations become relevant like we introduce the relation $x-2=0$ or something. I sort of forgot, but I think we just replace $x=2$ so $\overline{2x^2+1}=[2x^2+1+(x-2)]=[9+(x-2)]=\overline{9}$. I think the inverse of $\overline{2x^2+1}$ is $\overline{\frac19}$ then.
- So then the claim in Proposition 11.8.4a is that $F[x]/(p)$ is a field if and only if $p$ is monic irreducible in $F[x]$?
Best Answer
If $p(x)\in\mathbb{R}[x]$ is such that $p(1)\neq0$, then the multiplicative inverse of $p(x)+[x-1]$ is $\frac1{p(1)}+[x-1]$. That's so because: