Localizing single variable quotient polynomial ring at a prime ideal

[elias001]

TL;DR

I would like to know if the following example localizing a single variable quotient polynomial ring at a prime ideal is correct.

Background

I have questions about localizing the following quotient polynomial ring $R=\frac{\mathbb{Z}[x]}{(x^2+1)(x^5+11x^2+3)}$ at the prime ideal $M=(x^5+11x^2+3)$


Question:

For the question above, my attempted solution is as follows:

Let $I=((x^2+1)(x^5+11x^2+3))$ and let $S$ be the mulitplicative closed set where $$S=R\setminus M=\{f(x)+I\in R\mid f(x)\in \mathbb{Z}[x], \mathrm{gcd}(f(x), M)=1 \}.$$ Then the ring of fractions is the quotient ring $R$ localized at the prime ideal M, consisting of coset elements from $R$: $$p(x)+I\in R, p(x)\in \mathbb{Z}[x], f(x)+I\in R, f(x)\in \mathbb{Z}[x],$$ so that $$R_M=\left\{\frac{p(x)+I}{f(x)+I}\mid f(x)\not\in M\right\}.$$

I would like to know if the expression for the multiplicative closed set $S$ and the ring of fractions $R_M$ is correct?

Thank you in advance

 

[fresh_42]

TL;DR Summary: I would like to know if the following example localizing a single variable quotient polynomial ring at a prime ideal is correct.

Background

I have questions about localizing the following quotient polynomial ring $R=\frac{\mathbb{Z}[x]}{(x^2+1)(x^5+11x^2+3)}$ at the prime ideal $M=(x^5+11x^2+3)$


Question:

For the question above, my attempted solution is as follows:

Let $I=((x^2+1)(x^5+11x^2+3))$ and let $S$ be the mulitplicative closed set where $$S=R\setminus M=\{f(x)+I\in R\mid f(x)\in \mathbb{Z}[x], \mathrm{gcd}(f(x), M)=1 \}.$$ Then the ring of fractions is the quotient ring $R$ localized at the prime ideal M, consisting of coset elements from $R$: $$p(x)+I\in R, p(x)\in \mathbb{Z}[x], f(x)+I\in R, f(x)\in \mathbb{Z}[x],$$ so that $$R_M=\left\{\frac{p(x)+I}{f(x)+I}\mid f(x)\not\in M\right\}.$$

I would like to know if the expression for the multiplicative closed set $S$ and the ring of fractions $R_M$ is correct?

Thank you in advance

I had to check some intermediate steps such like whether $M\subseteq R$ is actually a prime ideal and why $f(x)\not\in M$ is equivalent to $\operatorname{gcd}(f(x),x^5+11x^2+3)=1,$ and I still don't see why you needed this. $R_M=\left(R\setminus M\right)^{-1}R,$ so your calculations are correct.

 

[elias001]

@martinbn thank you for checking if my expressions are correct. I replied back to your comment in my other post I used another example and asked if my understanding of the original example was correct.

Also in your replied to this other post of mine earlier. Is it possible to do exercises without resorting to reference to concepts and methods from the topic of localization? Thank you in advance.

 

[fresh_42]

@martinbn thank you for checking if my expressions are correct. I replied back to your comment in my other post I used another example and asked if my understanding of the original example was correct.

Also in your replied to this other post of mine earlier. Is it possible to do exercises without resorting to reference to concepts and methods from the topic of localization? Thank you in advance.

Looks as if I'm not the only one suffering the heat and being slower than usual ...

 

[elias001]

@martinbn where i am, it is getting to be like an oven. I posted other things too. I am wondering if you can also take a lookat them whenever you have time.

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they are questions i am trying to find answers to, hopefully I can consult someone on here before I start my abstract algebra course. Some of them has to do with basic category theory. I took a bit from learning algebra and ran into the category theory rabbit hole for a few months.


thank you in advance.

 

[mathwonk]

In reference to post #1, your description of the localization is technically correct but for me does not do much towards actually describing the structure of the ring. I have not checked these next comments by making any calculations, but it seems to me that the localization is actually isomorphic to the field Q[X]/M, where Q is the rational numbers and M is the ideal generated by f = X^5+11X^2+3. I.e. localizing at the (complement of the) prime ideal M, means that all non multiples of the polynomial f become units. In particular X^2+1 becomes a unit, so the ideal generated by (X^2+1).(X^5+11X^2+3) becomes the ideal generated by just the prime f. And since we are modding out by this ideal, and we are localizing at everything outside this ideal, we get a field, which must then be the quotient field of the domain Z[X]/(f), i.e. Q[X]/(f). 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 mod 2.)]

Note this quotient polynomial ring localization problem is partly analogous to the example in your other post, where we localized the quotient ring Z/(2.3) at powers of 3, making the 3 in the ideal (2.3) into a unit, hence changing that ideal into the ideal (2), and thus making the localized quotient ring into some localization of Z/(2), which however is already a field.

 

[elias001]

@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.