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The question I am asking is very elementary I think. Suppose $R$ is an integral domain. Let $I$ be an ideal of $R$. Then we can consider the quotient ring $R/I= I_R$. Now consider an ideal $J'$ of $R/I$ , then from the fourth isomorphism theorem of ring we can say that there is an ideal $J$ of $R$ containing $I$ such that $J' = J/I $ in $R/I$.

Now if we consider the quotient ring $I_R$ , then we can again quotient it by the ideal $J'$ i.e by $J/I$, so we have a quotient ring $I_R/J'$ which is actually $(R/I)/(J/I)$ and from third isomorphism theorem we have $(R/I)/(J/I)$ i.e $I_R/J'$ isomorphic to $R/J$.

Now while viewing some questions and answers about quotient ring on this site, I found this answer (question is attached with it) in which the very 1st line is confusing to me. Why the ordering in quotient doesn't matter ?

( Regarding to the question if we see, then we have $R = \mathbb Z[x] , I =((x^2+x+1)(x^3 +x+1)) , I_R = \mathbb Z[x]/I, J' = (2) $. So for $J'$ we have an ideal $J$ corresponding to it in $\mathbb Z[x] $ which contains $I$ . So we have $(R/I)/(J/I) $ isomorphic to $R/J$. I understand atmost to this extent. ) If I accept the 1st line, then I am ok with the proof.

Next in this answer how $ \mathbb Z[√3] /(1+2√3) $ is isomorphic to $\mathbb Z[x]/((x^2-3),(1+2x))$ ? I am ok that $\mathbb Z[√3]$ is isomorphic to the ring $\mathbb Z[x]/(x^2-3)$, and I think $\mathbb Z[√3]/(1+2√3) $ is isomorphic to $(\mathbb Z[x]/(x^2 - 3))/(1+2√3)$. So is it permissible to write $(\mathbb Z[x]/(x^2 - 3))/(1+2√3)$ as $\mathbb Z[x] /( x^2-3, 1+ 2√3) $ ? If yes then why ? Here $1+2√3$ is actually $1+2x$ because here $x^2=3$, so $x=√3 $, am I right ?

All of my questions are very easy I think, but due to a bizzare view about the Polynomial Rings and there quotients I am unable to understand these. Also my question is large , but I need to understand the concept, so I am looking for a simple explanation. Thank you.

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I think the source of your anxiety is that many people are abusing notation by using the same labels for elements of a ring and the image in the quotient ring. For example, let $a,b\in R$, and $S=R/(a)$ and $\phi:R\to S$ be the projection to the quotient. Then we could take the quotient $S/(\phi(b))$, but people will very often just write $S/b$, or use $b$ as an element of $S$. Regarding the isomorphism theorems, we have that there exists an ideal $I\subset R$ such that $R/I=S/(\phi(b))$, but we can be explicit about what that ideal is: it's $(a,b)$. This is straightforward to check.

In the second example you give, they actually ARE using a different symbol for the image of an element of $R$. $\sqrt{3}$ is the image of $x$ under the quotient morphism. Then it's really the same theorem $R/(a,b)=(R/(a))/(\phi(b))$

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  • $\begingroup$ At first I am very much fortunate that you have understood my problem. Thank you. Give some time to digest your answer. @Callus $\endgroup$ – hiren_garai Dec 15 '17 at 13:32
  • $\begingroup$ Regarding to your answer, I have a query, if $I & J $ are two ideals then the ideal generated by $I$ and $J$ i.e $(I,J)$ is same as the ideal $(I+J)$, ?@Callus $\endgroup$ – hiren_garai Dec 15 '17 at 13:48
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    $\begingroup$ That's correct. $\endgroup$ – Bernard Dec 15 '17 at 13:56
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To add some details for your first question, we have this situation (where $J'=(2)$): $$(R/I)\!\bigm/\!(J'\cdot R/I)=(R/I)\!\bigm/\!((J'+I)/I)\simeq R/(J'+I).$$

You also can see that with tensor products: since for any $R$-module $M$ and ideal $I\subset I$, we have $M\otimes_R R/I\simeq M/IM$, have here $$(R/I)\!\bigm/\!(J'\cdot R/I)\simeq(R/I)\otimes_R(R/J')\simeq R/(J'+I).$$

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  • $\begingroup$ I don't understand the $(J' \cdot R/I)$ thing . Why are we considering this ? We have $(R/I)/J'$ . Sorry but, I didn't get you. @Bernard. Will you please explain ? $\endgroup$ – hiren_garai Dec 15 '17 at 14:14
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    $\begingroup$ As I said, $J'$ is the ideal generated by $2$ in $R/I$. This is nothing else than $(2R+I)/I$. Isn't it clear? $\endgroup$ – Bernard Dec 15 '17 at 14:58
  • $\begingroup$ Yes , that's clear to me. Then why the $J' \cdot R/I$ ? $\endgroup$ – hiren_garai Dec 15 '17 at 15:04
  • $\begingroup$ I didn't explain quite exactly in my comment: J' is the ideal generated by the image of $2$ in $R/I$, whence the dot so there's confusion due to the quotient. $\endgroup$ – Bernard Dec 15 '17 at 15:07

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