# An example of how to select ideals to obtain quotient rings with desirable properties

The following self-contained post demonstrates how to construct quotient rings with desirable properties by elaborating on an example taken from chapter 19, p. 194, of the book “A book of abstract algebra” by Charles C. Pinter.

The fundamental homomorphism theorem for rings employs the kernel $K$ of a homomorphism $f:A\rightarrow B$ to construct a quotient ring $A/K$ isomorphic to ring $B$. Conversely, the homomorphic image $A/J$ of a ring $A$ is useful for factoring out unwanted features of $A$ to obtain a quotient ring $A/J$ with the desirable properties. Selecting an appropriate ideal $J$ is the key for acquiring the desired quotient ring $A/J$.

In some cases, picking a suitable ideal $J$ can be done methodically by tracing the desired property of $A/J$ back to an analogous property of $J$. To demonstrate the process of selecting $J$, the simple example of obtaining a quotient ring $A/J$ that doesn’t contain any divisors of zero will be considered.

The example under consideration is stated as follows. Let $A$ be a commutative ring with unity. An ideal $J$ of $A$ is prime if and only if $A/J$ is an integral domain (see chapter 19, p. 194, in Charles Pinter’s book “A book of abstract algebra”).

Briefly recall the associated definitions. An integral domain is a commutative ring with unity that doesn’t have any divisors of zero. An ideal $J$ of a commutative ring $A$ is prime if $\forall a,b\in A$ with $ab\in J$ it follows that $a\in J$ or $b \in J$.

It is straightforward to prove the equivalence of this example. To prove that the integral domain has no divisors, it suffices to observe that $a\in J \Leftrightarrow J=J+a$. Emphasis will be shifted towards understanding why one would think of using a prime ideal to retain the cancellation property in $A/J$, i.e. to ensure that there are no divisors of zero in $A/J$.

Think first how the lack of divisors of zero is stated for $A/J$. The product of two cosets $J+a$ and $J+b$ in $A/J$ is $(J+a)(J+b)=J+ab$. Moreover, the zero element of $A/J$ is the coset $J=J+0$, so to say that $A/J$ has no divisors of zero implies that $J+ab=J\Rightarrow J+a=J$ or $J+b=J$.

Recall also that for any $x\in A$, $x\in J \Leftrightarrow J+x=J$. It then becomes obvious that the property $J+ab=J\Rightarrow J+a=J$ or $J+b=J$ in $A/J$ corresponds to the property $ab\in J\Rightarrow a\in J$ or $b\in J$, which is the defining property of a prime ideal $J$.

So it appears that in some cases it is easy to choose an ideal $J$ whose properties are conveyed to a quotient ring $A/J$.