If $A$ is a commutative ring and $I \subset A$ an ideal, then I do know what $A / I$ means. However my professor now sometimes writes $J / I$ where $J$ is not a ring, but an ideal $I \subset J \subset A$ (for example $\sqrt{I} / I$ ). I was wondering is this is common notation and if someone could tell me how the 'quotient' of two ideals is defined.
Abstract Algebra – Defining Quotient of Two Ideals
abstract-algebraidealsring-theory
Related Solutions
There is no mistake; the one-to-one, inclusion preserving correspondence between ideals of a ring $R$ that contain the ideal $I$ and ideals of the quotient ring $R/I$ also gives a correspondence between prime ideals.
For the commutative-with-unity case, your proof works.
If you want a proof for the more general case, and to prove a bit more:
Definition. Let $R$ be a ring; an ideal $I$ of $R$ is completely prime if and only if whenever $ab\in I$, either $a\in I$ or $b\in I$.
Definition. Let $R$ be a ring; an ideal $I$ of $R$ is prime if and only if whenever $JK\subseteq I$, where $J$ and $K$ are ideals, either $J\subseteq I$ or $K\subseteq I$.
If $R$ is commutative, then an ideal is prime if and only if it is completely prime. The two notions are not equivalent for arbitrary rings, though. For example, in the ring of $2\times 2$ matrices over a field, the only ideals are the trivial ideal and the whole ring; in particular, the zero ideal is a prime ideal, but it is not completely prime because you can find two nonzero matrices whose product is the zero matrix.
Theorem. Let $R$ be a ring and let $I$ be an ideal of $R$. Then the natural correspondence between ideals of $R$ that contain $I$ and ideals of $R/I$ identifies prime ideals with prime ideals and identifies completely prime ideals with completely prime ideals.
Proof. Let $J$ be an ideal that contains $I$. If $J$ is completely prime, suppose that $a+I, b+I\in R/I$ are such that $(a+I)(b+I)\in J/I$. Then $ab+I\in J/I$, hence $ab\in J$ (since $I\subseteq J$); since $J$ is completely prime, either $a\in J$ or $b\in J$, so either $a+I\in J/I$ or $b+I\in J/I$. Thus, $J/I$ is completely prime. Conversely, suppose that $J/I$ is completely prime, and let $a,b\in R$ be such that $ab\in J$. Then $ab+I\in J/I$, hence either $a+I\in J/I$ or $b+I\in J/I$. If $a+I\in J/I$, then there exists $i\in I$ such that $a+i\in J$, hence $a\in J$ (since $I\subseteq J$); likewise, if $b+I\in J/I$, then $b\in J$. Thus, $J$ is completely prime.
Now suppose that $J$ is a prime ideal. Let $K/I$ and $L/I$ be ideals of $R/I$ such that $(K/I)(L/I)\subseteq J/I$, with $K/I$ corresponding to the ideal $K$ of $R$ that contains $I$, and the ideal $L/I$ corresponding to $L$. Then $(K/I)(L/I) = (KL)/I\subseteq J/I$, hence, by the inclusion-preserving correspondence, $KL\subseteq J$, so either $K\subseteq J$ or $L\subseteq J$, hence $K/I\subseteq J/I$ or $L/I\subseteq J/I$. Thus, $J/I$ is a prime ideal. Conversely, if $J/I$ is a prime ideal, let $K$ and $L$ be ideals such that $KL\subseteq L$. Then $K+I$ and $L+I$ are ideals that contain $I$, and $(K+I)(L+I)=KL+KI+IL+I^2\subseteq KL+I\subseteq KL+J=J$; therefore, $(K+I)/I$ and $(L+I)/I$ are ideals of $R/I$ whose product is contained in $J/I$, so either $K+I\subseteq J$ or $L+I\subseteq J$. But $K\subseteq K+I$ and $L\subseteq L+I$, so either $K\subseteq J$ or $L\subseteq J$. Thus, $J$ is prime. QED
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.
Best Answer
Your teacher is thinking to the ideal $J/I$ of $A/I$.
There is a surjective ring homomorphism $A\to A/I$, sending $a$ to $a+I$. The image of $J$ under this homomorphism is an ideal of $A/I$ and it is quite naturally denoted $J/I$, because it consists of all elements of the form $x+I$, with $x\in J$. This assumes that $J\supseteq I$.
In the general case, if $J$ is any ideal of $A$, its image in $A/I$ under the canonical homomorphism above is $(J+I)/I$.
For instance, $\sqrt{I}/I$ is the nilradical of $A/I$. Indeed, if $x\in\sqrt{I}$, then $x^n\in I$ for some $n$, so the element $x+I$ is nilpotent in $A/I$. The converse is also clear.