@freseh_42, and
@mathwonk sorry if I seem to disappeared for the last few days. You asked about
What do you think of this reasoning? Did I make a mistake? [By the way I agree with fresh_42 that the point about this question requiring thought for me, is that ##X^5+11X^2+3## is irreducible in ##Z[X]##. How did you prove this? (I checked it by reduction ##pmod 2##.)]
I had to digged up a lot of deleted posts from MSE. I will answer your questions by providing contexts from those deleted posts. It will answer
this other post, I still have technical questions in that other post. But I will focus on providing the background here.
So it begun with encountering the following quoted example from Stephen Lovett's text
Abstract Algebra: A First Course in the section fields of fractions/localization chapter. ( By the way, the topic of localization is making an appearance in newer abstract algebra texts. )
Let ##M## be the quotient ring ##M=\frac{\mathbb{Z}[x]}{(x^2-1)}##. (For simplicity, we omit the over-line ##\bar{ax+b}##notation.) We see that ##x-1## and ##x+1## are zero divisors. Consider the multiplicatively closed set ##D=\{(1+x)^k\mid k\geq 0\}##. In ##M##, we have ##x^2-1=0##, so ##x^2=1##. Then
$$(1+x)^2=x^2+2x+1=1+2x+1=2(x+1)$$
Consequently, ##(1+x)^k=2^{k-1}(x+1)## for ##k\geq 1##.
Then the ring of fractions ##D^{-1}M## consists of expression ##\frac{a+bx}{1}## or ##\frac{a+bx}{2^n(x+1)}## for ##n\geq 0##.
I went on MSE and ask what if I am given a single variable quotient polynomial ring that is not an integral domain also, and I would like to localize the thing at a prime ideal. At the time of my asking, I did not know that I was asking about
local rings. I wanted an example that is as simple as possible. By simple, I want it in the land of one variable polynomial so that I don't have to relieve the unpleasantness of feeling like an ignoramus for having zero knowledge about algebraic geometry or advanced commutative algebra where I have to figure out terms like what sheafs or working with spectrum of a ring, etc,etc. So I tried to cook up an example:
##R=\frac{\mathbb{Z}[x]}{((x-6)(x-10))}## along with trying to describe its multiplicatively closed set:
$$R_P=(R\setminus P)^{-1}R=\left\{\frac{\bar{f(x)}+((x-6)(x-10))}{\bar{g(x)}+((x-6)(x-10))}\in R\mid \bar{f(x)},\bar{g(x)}\in R, \bar{g(x)}\not\in (P)=(x^2+1)\right\}$$
And also
Is ##P=(x^5+11)## a prime ideal of the quotient ring ##R=\frac{\mathbb{Z}[x]}{((x^2+1)(x^5+11))}## where ##M=\mathbb{Z}[x], I=(x^2+1)(x^5+11)##? am thinking that it is. Since by
theorem 2 in the below section, ##I=(x^2+1)(x^5+11)\subset (x^5+11)\subset R##.
A patient soul there, discussing over chat wrote me a few things:
1)
I will also say that ##P\subset R## is not prime, so you can't localize at ##[x]^2+1##.
Consider ##a=[x]## and ##b=-61[x]-16##, then ##ab=-61[x]^2-16[x]## so ##ab-(-60([x]^2-1))=[x]^2-16[x]+1=0##, it follows that ##ab\in \langle [x]^2+1\rangle## but neither ##a## nor ##b## are in ##\langle [x]^2+1\rangle##.
2)
I'll point out that Lovett is not localizing at a prime ideal, he is localizing at an element by the multiplicatively closed set generated by ##1+x##. In particular, you can play the same game with ##x-6## to get the fractions are all like ##(a[x]+b)/4^n([x]-6)##. If ##\mathfrak p## is a prime of ##A/I## then it corresponds uniquely to a prime ideal ##\mathfrak q## containing ##I##. In this case ##(A/I)_\mathfrak p\cong (A_\mathfrak p/I_{\mathfrak p})## where ##I_{\mathfrak p}## is the ideal generated by the image of ##I## under the localization map.
Sorry I mean ##A_\mathfrak q/I_\mathfrak q##.
I told him I did not know in the case of quotient ring, how an ideal is prime relative to a particular quotient ring. He wrote back:
3)
Let ##A## be a ring and ##I## be an ideal, then ##A/I## is the set of equivalence classes ##[a]## given by the equivalence relations ##a\sim a'## if and only if ##a-a'\in I##. If we have some prime ideal ##\mathfrak{p}## of ##A## such that ##I\not\subset \mathfrak{p}##, then there exists some ##i\in I## such that ##i\notin \mathfrak{p}##, but \pi(i)\in \pi(\mathfrak{p}) as = 0. It follows that \pi^{-1}(\pi(\mathfrak{p}))=\mathfrak{p} + I,
but this need not be prime.
(I can't get the last bit of latex to render properly in the above quote)
I was super miffed, dejected and disappointed. I thought I should find out the answer just in case the professors at my university in their twisted Hunger Games style sense of humor would ever ask me such question on a test or an exam and that I would need to know the answer then because I had the foresight to find out now.
I went back to pestering Doctor Google, and Doctor Google showed me two theorems from the volume two of a two volumes texts Abstract Algebra with Applications, vol 2, Rings and Fields by Karlheinz Spindler
Theorem 1 An ideal ##(f(x))## in ##k[x]## (single variable commutative polynomial ring over ##k##) is a prime ideal if and only if ##f(x)## is an irreducible polynomial over ##k##.
Theorem 2 Let ##R## be a commutative ring and let ##I\trianglelefteq R## be an ideal of ##R##. Then an ideal ##J## of ##R## with ##I\subset J\subset R## is prime (radical, primary, maximal) if and only if ##J/I## is, where ##J/I\subset R/I.##
I created another post asking from the above two theorem, how would I determine if an ideal is prime with respect to a quotient ring. The answer is the following:
You cannot localise ##F[X]/(g)## at ##(p)##, because ##(p)## is not an ideal of ##F[X]/(g)##. In fact, it does not even live in this ring. What you can do is considering the image of ##(p)## under the canonical projection.
This projection is in fact ##((p)+(g))/(g)##.
If you really want to consider ##(p)/(g)##, you need to assume that ##(g)\subset (p)##.
Now, if ##(g)\subset(p)##, then ##(p)/(g)## is an ideal of ##F[X]/(g)##.
But in this case, you have the ring isomorphism
##(F[X]/(g))/((p)/(g))\simeq F[X]/(p)##. In particular, ##(p)/(g)## is a prime ideal of ##F[X]/(g)## if and only if ##(p)## is a prime ideal of ##F[X]## if and only if ##p## is irreducible.
If ##(g)\not\subset (p)##, you are forced to consider ##((p)+(g))/(g)=(d)/(g)##, where ##d=gcd (p,g)## (and then you can apply the previous case).
After that, I looked over how the multiplcatively closed set is suppose to be written out in the case of quotient polynomial rings, and also for local rings, based on the notations I got from this post, I was able to figure out how to determine whether an ideal is prime with respect to a quotient polynomial ring in one variable. Sorry I hope I am not sounding like some old bag in my explanations.
