# Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic

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.

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 $$F[x]$$?
• 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. – DonAntonio Nov 20 '18 at 10:41
• @DonAntonio Oh I see. Is the "monic" perhaps for extensions to non-fields like $\mathbb Z$? Thank you! – user198044 Nov 20 '18 at 11:10

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:
• $$p(x)+[x-1]=p(1)+[x-1]$$;
• $$\bigl(p(1)+[x-1]\bigr)\times\left(\frac1{p(1)}+[x-1]\right)=\bigl(1+[x-1]\bigr)$$.
• $\overline {\frac{1}{p(1)}}$ is so compact! Thank you! – user198044 Nov 20 '18 at 11:07
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 $$\frac{1}{9}+(x-2)$$.
• Oh I see. Is the "monic" perhaps for extensions to non-fields like $\mathbb Z$? Thank you! – user198044 Nov 20 '18 at 11:10