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Background

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.

Question

Suppose we have some particular/specific single variable polynomial quotient ring $F[x]/(g(x))$ over the commutative ring $F$, and the ideal $(g(x))$ is generated by the polynomial $g(x)\in F[x]$. I would like to localize this polynomial quotient ring $F[x]/(g(x))$ at the prime ideal $(p(x))$ generated by the polynomial $p(x)\in F[x]$. The issue one has to settle is whether $(p(x))$ is actually a prime ideal of $F[x]/(p(x))$. I would like to know how to decide if an ideal $(p(x))$ is prime in the quotient polynomial ring $F[x]/(g(x))$? Specifically, are there are theorems that one can derive easy computational methods or criteria? From Theorem 1 above, we know how to decide whether an ideal of a single variable polynomial ring is prime but I am not sure if it applies to polynomial quotient ring. As for Theorem 2 above, I am not sure if it is sufficient to know that if $g(x)\in (p(x))$ or equivalently $(g(x))\subset (p(x))$, then we can conclude that the ideal $(p(x))$ is a prime ideal of the quotient ring $F[x]/(g(x))$? Similarly, I am not sure if there is an easy way to show that $(p(x))/(g(x))$ is a prime ideal. Also it is not difficult to come up with cases where $F[x]/(p(x))$ is an integral domain, and hence $(p(x))$ is a prime ideal of $F[x]$ but $(p(x))$ is not a prime ideal of the quotient ring $F[x]/(g(x))$.

I am concentrating on single variable polynomial ring because I understand that for even two variable cases, more advanced techniques from commutative algebra/algebraic geometry might be required.

I also try to do a search about this topic through existing textbooks, online lecture notes and on here that doesn't involve advance notions and ideas from commutative algebra, basically staying within the level of an introductory abstract algebra course but was not able to locate the needed information.

Thank you in advance

Seth
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1 Answers1

2

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).

GreginGre
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  • when you say "What you can do is considering the image of $(p)$ under the canonical projection". Do you mean a map $f:(p)\to ((p)+(g))/(g)$? Also where does $((p)+(g))/(g)=(d)/(g)$ come from? – Seth Feb 21 '25 at 11:28
  • First question: i mean the image of (p) under the canonical projection $F[X]\to F[X]/(g)$. Second question: in a PID, we have $(a)+(b)=(gcd(a,b))$. – GreginGre Feb 21 '25 at 11:56
  • sorry i had to delete my previous comment due to a mistake. The canonical projection map $F[x]\to F[x]/g(x)$ is given as $p\mapsto p+(g)$? Also, I am assuming when you say $\frac{((p)+(g))}{(g)}=\frac{(d)}{(g)}$, that is an ideal to be used in the third isomorphism theorem: $\frac{\frac{F[X]}{(g)}}{\frac{((p)+(g))}{(g)}}=\frac{\frac{F[X]}{(g)}}{\frac{(d)}{(g)}}\simeq \frac{F[X]}{(p)}$. But why would one have to be forced to consider that $F[x]$ is a PID? I tried find theorem or results from standard abstract algebra texts,... – Seth Feb 21 '25 at 17:46
  • ....but I was not having any success. How i did you figured it out can I ask? Also, is there a source/reference I can look at? – Seth Feb 21 '25 at 17:46