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A problem from Intro to Abstract Algebra from Hungerford.

  1. a) Let K be an ideal in a ring R. Prove that every ideal in the quotient ring R/K is of the form I/K for some ideal I in R.

This is what I've done.

Consider the function $f:R\rightarrow R/K$ defined as $f(r)= r+K$, $r \in R$. This is a surjective homomorphism, then for any ideal $I \in R$, $i \in I$, $$f(i)=i+K$$ which will be an ideal in $R/K$ since $f$ is homomorphic. Ie. We will show that the set $I' = \{f(i)\in R/K\ |\ i\in I\}$ is an ideal in $R/K$. Then, since $f$ is surjective every element of $R/K$ has a premap, $\forall f(b) \in R/K,\ f(b)*f(i) = f(bi) = f(i_{1}), \ f(i_{1}) \in I'$. Similarly for multiplying on the other side. Then for any ideal $I \in R$, $f$ maps it to an ideal in $R/K$ (also check for closure of subtraction).

Every ideal of $R/K$ has a premapping through $f$ whose input $\in R$ is itself an ideal. We can show this is true by considering the set $J = \{r\in R | f(r) \in I/K\}$ $$\forall a\in R,\ f(ra)=f(r)*f(a) \in I/K \rightarrow ra \in J$$ Similarly for multiplying on the other side. J is closed under subtraction so it is an ideal in $R$.

Hence, if every ideal in $R$ is an ideal in $R/K$ through $f$, and every ideal in $R/K$ has a premap which is an ideal in $R$, then it must be true that for each ideal in $R/K$, it is of the form $I/K$.

$\Box$

Is this a correct/rigorous proof? Thank you in advance, much appreciated.

zzz2991
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1 Answers1

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This looks good! I have only one criticism: when you conclude that the image of an ideal of $R$ under $f$ is an ideal, you should explicitly use the fact that $f$ is surjective. (Suppose $r+K\in R/K$ and conclude that $(r+K)f(i)\in I/K$.) This is important because in general the image of an ideal under a ring homomorphism is not an ideal: for example, the image of $\mathbb Z$ under the inclusion $\mathbb Z\hookrightarrow\mathbb Q$ is not an ideal of $\mathbb Q$.

user134824
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