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Artin Algebra:

Definition of maximal ideals:

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Maximal ideals of $F[x]$:

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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.

  1. 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.

  1. 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 $F[x]$?
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    $\begingroup$ No need of the "monic" requirement. I think that what's needed here the most is to understand that an irreducible polynomial generates a maximal ideal in the polynomial ring, and also that a commutative unitary ring quotiented by an ideal is a field iff the ideal is maximal. After this you can work out inverses in the quotient, which is not a trivial problem...but it is not so hard, either. $\endgroup$ – DonAntonio Nov 20 '18 at 10:41
  • $\begingroup$ @DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $\mathbb Z$? Thank you! $\endgroup$ – user198044 Nov 20 '18 at 11:10
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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:

  • $p(x)+[x-1]=p(1)+[x-1]$;
  • $\bigl(p(1)+[x-1]\bigr)\times\left(\frac1{p(1)}+[x-1]\right)=\bigl(1+[x-1]\bigr)$.
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  • $\begingroup$ $\overline {\frac{1}{p(1)}}$ is so compact! Thank you! $\endgroup$ – user198044 Nov 20 '18 at 11:07
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  1. You are right that you have to use $(x-2)$ in some way. If $f$ is a polynomial then $$f+(x-2)=c+(x-2)$$ where $c\in\Bbb R$ is some constant. The reason for this is that you can write $$f=q(x)(x-2)+r$$ where $\deg(r)<1$. This forces $r\in\Bbb R$, in particular $r=f(2)$. With $f=2x^2+1$ we find that $2x^2+1=(2x+4)(x-2)+9$, which means that in $\Bbb R[x]/(x-2)$ we have $$2x^2+1+(x-2)=9+(x-2)$$ since $(2x+4)(x-2)\in (x-2)$. The inverse is therefore $\frac{1}{9}+(x-2)$.

  2. It would work for irreducible polynomials that are not monic also since the leading coefficient is necessarily invertible.

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  • $\begingroup$ Oh I see. Is the "monic" perhaps for extensions to non-fields like $\mathbb Z$? Thank you! $\endgroup$ – user198044 Nov 20 '18 at 11:10

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