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Given a ring $R$ and a prime ideal $P$, then we can define localized ring denoted $R_p$ effectively giving all elements not in the ideal the property of unit. I understand that $R_p/I_p\cong (R/I)_p$ for any ideal $I$ of $R$. But I'm having trouble understanding what an element in $(R/I)_p$ look like.

For instance, let $R=\mathbb{R}[x]$ and let $P=(x)$ and $I=(x^3)$. We know $R/I$ contains element of the form $f(x)+(x^3)$, or more precisely $r_0+r_1x^1+x_2x^2+(x^3)$ where $r_i\in\mathbb{R}$. But what does an element in $(R/I)$ localized at the prime ideal $(x)$ look like? What exactly is this entity? Is it simply $\dfrac{r_0+r_1x^1+x_2x^2}{g(x)}+(x^3)$ where $\gcd(g(x),x)=1$?

Thanks.

Student Lv 1
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  • It should really be $(R/I)_{P/I}$, as $P$ is not a prime ideal of $R/I$. So it looks "the usual way": they are (equivalencee classes) of fractions of the form $\frac{f(x)+I}{q(x)+I}$, where $q(0)\neq 0$. – Arturo Magidin Apr 28 '22 at 17:54
  • @Arturo I reread your comment like 20 times and now I'm more confused lol. So $\frac{x^2+I}{x^2+1+I}$ would be an element? – Student Lv 1 Apr 28 '22 at 18:10
  • It would represent an equivalence class of elements, yes. – Arturo Magidin Apr 28 '22 at 18:13
  • $\newcommand{\p}{\mathfrak{p}}$On a similar-ish note, you may want to try the following: Let $R$ be a nonzero commutative ring and $\p$ a prime ideal of $R$. Then, $R/\p$ is a domain and you can talk about its field of fractions. On the other hand, $R_{\p}$ is a local ring with maximal ideal $\p R_\p$. Thus, you can talk about the field $R_\p/\p R_\p$. Show that the two fields are isomorphic. – Aryaman Maithani Apr 29 '22 at 10:19

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As $P$ is not an ideal of $R/I$, you should really have here $$\frac{R_p}{I_p} \cong \left(\frac{R}{I}\right)_{P/I}.$$

For your specific example: viewed as elements of the right hand side, they have the "usual" form: elements of $R/I$ are cosets of the form $f(x)+I$, and you may pick your representative $f(x)$ to be of the form $a_0+a_1x+a_2x^2$. So the localization consists of equivalence class of expressions of the form $$\frac{f(x)+I}{g(x)+I} = \frac{a_0+a_1x+a_2x^2 + I}{b_0+b_1x+b_2x^2 + I}$$ with $b_0\neq 0$ (so $g(x)+I\notin P/I$). Two such fractions $\frac{f(x)+I}{g(x)+I}$ and $\frac{m(x)+I}{n(x)+I}$ are equivalent if and only if there exists $p(x)\notin P$ such that $p(x)(f(x)n(x)-g(x)m(x))\in I$. But because $I=(x^3)$ and $p(x)$ is not a multiple of $x$, this happens if and only if $f(x)n(x)+I = g(x)m(x)+I$, i.e., the usual "cross multiplications agree".

To view elements of the left hand side, on the other hand, elements of $R_p$ are (equivalence classes of) fractions of the form $\frac{f(x)}{g(x)}$, with $g(0)\neq 0$. Because we have an integral domain and our multiplicative set does not include $0$, we know that $\frac{f(x)}{g(x)}$ is equivalent to $\frac{m(x)}{n(x)}$ if and only if $f(x)n(x)=g(x)m(x)$.

A fraction $\frac{f(x)}{g(x)}$ lies in $I_p$ if and only if there exists $h(x)\notin P$ such that $h(x)f(x)\in I$: indeed, if this happens, then $\frac{f(x)}{g(x)}=\frac{h(x)f(x)}{h(x)g(x)}$ lies in $I_p$. And if $\frac{f(x)}{g(x)} = \frac{m(x)}{n(x)}$ with $m(x)\in I$, then $n(x)f(x)=m(x)g(x)\in I$, with $n(x)\notin P$. So every element of $I_p$ can be expressed in the form $\frac{m(x)}{g(x)}$, where $g(0)\neq 0$ and $x^3\mid m(x)$.

Thus, given an element $\frac{f(x)}{g(x)} = \frac{a_0+a_1x+\cdots +a_nx^n}{b_0+b_1x+\cdots+b_mx^m}\in R_p$, with $b_0\neq 0$, we have that $$\frac{f(x)}{g(x)} + I_p = \frac{a_0+a_1x+a_2x^2}{g(x)} + I_p$$ Thus, elements of $\frac{R_p}{I_p}$ are equivalence classes of fractions of the form $$\frac{a_0+a_1x+a_2x^2}{g(x)} + I_p$$ where $g(0)\neq 0$.

So you can see that in general, the isomorphism ${R_p/I_p}\to (R/I)_{P/I}$ is going to take $\frac{f(x)}{g(x)} + I_p$ to $\frac{f(x)+I}{g(x)+I}$.

Arturo Magidin
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  • Thanks for your time explaining and typing up all these details. Now I understand how to view this quotient! Much appreciated! – Student Lv 1 Apr 28 '22 at 18:17
  • Note that even $P/I$ is not an ideal of $R/I$. It should be something more like $(P + I)/I$, unless we are assuming $I \subset P$. – Aryaman Maithani Apr 29 '22 at 10:20
  • @AryamanMaithani If $I$ is not contained in $P$, it'll contain units in the localization, and both sides collapse. The correspondence is between ideals disjoint from the multiplicative set (so in this case, contained in $P$) and ideals of the localization. – Arturo Magidin Apr 29 '22 at 12:53