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Let $I$ be an ideal of a ring (commutative with unity) $R$ and let $q:R\to R/I$ be the quotient map. Then there is a well known correspondence between ideals of $R$ containing $I$ and ideals of $R/I.$

Let $J$ be an ideal of $R$ containing $I$ and let $J'$ be the corresponding ideal in the quotient ring. I had to show that $J$ is radical/prime/maximal iff $J'$ is radical/prime/maximal.

Showing $J$ is radical/prime/maximal if $J'$ is was simple enough by direct element manipulation arguments.

For the other direction, I did it by using $J'=J/I,$ the fact that ideals are radical/prime/maximal iff their quotient rings are reduced/domains/fields and the second isomorphism theorem. However, I tried unsuccessfully to find a "direct" proof that perhaps manipulates elements of the ring in a similar way to the other direction. Is there a more direct proof? I'm wondering this especially for the radical ideals, because this is Exercise 1.22 (page 7) in Fulton's Algebraic Curves and he has never even mentioned reduced rings or nilradicals yet, and doesn't expect you to know it already because he introduced what radical ideals were.

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up vote 2 down vote accepted

You actually already proved the difficult part.

If $J$ is maximal, then by the correspondence of ideals you mention $J'$ has to be maximal, too. If it weren't, there would be a proper ideal $T'\supset J'$, which would give a proper ideal $T\supset J$.

Now assume $J$ prime and let $a'b'\in J'$, where $a',b'$ are the classes in $R/I$ of some $a,b\in R$. This means that there are $i_a,i_b\in I$ such that $(a+i_a)(b+i_b)=ab+ai_b+bi_a+i_ai_b\in J+I=J$, as $I\subseteq J$. Thus $ab\in J$ and, say, $a\in J$. Hence $a'\in J'$, so $J'$ is prime.

Finally, suppose $J$ radical and let $(a')^r\in J'$ for some $a\in R$, $r\in\Bbb N$. Then there are some $i_a,i\in I$ such that, by binomial expansion and since $I$ is an ideal, $(a+i_a)^r=a^r+i\in J+I=J$. Hence $a^r\in J$, so $a\in J$. Therefore $a'\in J'$ and $J'$ is radical.

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