A homomorphism between rings is a function that preserves the operations of addition and multiplication in the ring. More specifically, if \(R\) and \(S\) are rings, then a ring homomorphism is a map \(\phi : R \rightarrow S\) satisfying
\begin{align*}
\phi( a + b ) & = \phi( a ) + \phi(b)\\
\phi( a b ) & = \phi( a ) \phi(b)
\end{align*}
for all \(a, b \in R\text{.}\) If \(\phi : R \rightarrow S\) is a one-to-one and onto homomorphism, then \(\phi\) is called an isomorphism of rings.
The set of elements that a ring homomorphism maps to \(0\) plays a fundamental role in the theory of rings. For any ring homomorphism \(\phi : R \rightarrow S\text{,}\) we define the kernel of a ring homomorphism to be the set
\begin{equation*}
\ker \phi = \{ r \in R : \phi( r ) = 0 \}\text{.}
\end{equation*}
For any integer \(n\) we can define a ring homomorphism \(\phi : {\mathbb Z} \rightarrow {\mathbb Z}_n\) by \(a \mapsto a \pmod{n}\text{.}\) This is indeed a ring homomorphism, since
\begin{align*}
\phi( a + b ) & = (a + b) \pmod{n}\\
& = a \pmod{n} + b \pmod{n}\\
& = \phi( a ) + \phi(b)
\end{align*}
and
\begin{align*}
\phi( a b ) & = ab \pmod{n}\\
& = a \pmod{n}\cdot b \pmod{n}\\
& = \phi( a ) \phi(b)\text{.}
\end{align*}
The kernel of the homomorphism \(\phi\) is \(n {\mathbb Z}\text{.}\)
Let \(C[a, b]\) be the ring of continuous real-valued functions on an interval \([a,b]\) as in Example 1.1.5. For a fixed \(\alpha \in [a, b]\text{,}\) we can define a ring homomorphism \(\phi_{\alpha} : C[a, b] \rightarrow {\mathbb R}\) by \(\phi_{\alpha} (f ) = f( \alpha)\text{.}\) This is a ring homomorphism since
\begin{gather*}
\phi_{\alpha}( f + g ) = (f + g)( \alpha) = f(\alpha) + g(\alpha) = \phi_{\alpha}( f ) + \phi_{\alpha}(g )\\
\phi_{\alpha}( f g ) = (f g)( \alpha) = f(\alpha) g(\alpha) = \phi_{\alpha}( f ) \phi_{\alpha}(g )\text{.}
\end{gather*}
Ring homomorphisms of the type \(\phi_{\alpha}\) are called evaluation homomorphisms.
An ideal in a ring \(R\) is a subring \(I\) of \(R\) such that if \(a\) is in \(I\) and \(r\) is in \(R\text{,}\) then both \(ar\) and \(ra\) are in \(I\text{;}\) that is, \(rI \subset I\) and \(Ir \subset I\) for all \(r \in R\text{.}\)
Let \(R\) be a ring with identity and suppose that \(I\) is an ideal in \(R\) such that \(1\) is in \(I\text{.}\) Since for any \(r \in R\text{,}\)\(r1 = r \in I\) by the definition of an ideal, \(I = R\text{.}\)
If \(a\) is any element in a commutative ring \(R\) with identity, then the set
\begin{equation*}
\langle a \rangle = \{ ar : r \in R \}
\end{equation*}
is an ideal in \(R\text{.}\) Certainly, \(\langle a \rangle\) is nonempty since both \(0 = a0\) and \(a = a1\) are in \(\langle a \rangle\text{.}\) The sum of two elements in \(\langle a \rangle\) is again in \(\langle a \rangle\) since \(ar + ar' = a(r + r')\text{.}\) The inverse of \(ar\) is \(-ar = a (-r) \in \langle a \rangle\text{.}\) Finally, if we multiply an element \(ar \in \langle a \rangle\) by an arbitrary element \(s \in R\text{,}\) we have \(s(ar) = a(sr)\text{.}\) Therefore, \(\langle a \rangle\) satisfies the definition of an ideal.
The zero ideal \(\{ 0 \}\) is a principal ideal since \(\langle 0 \rangle = \{ 0 \}\text{.}\) If \(I\) is any nonzero ideal in \({\mathbb Z}\text{,}\) then \(I\) must contain some positive integer \(m\text{.}\) There exists a least positive integer \(n\) in \(I\) by the Principle of Well-Ordering. Now let \(a\) be any element in \(I\text{.}\) Using the division algorithm, we know that there exist integers \(q\) and \(r\) such that
\begin{equation*}
a = nq + r
\end{equation*}
where \(0 \leq r \lt n\text{.}\) This equation tells us that \(r = a - nq \in I\text{,}\) but \(r\) must be \(0\) since \(n\) is the least positive element in \(I\text{.}\) Therefore, \(a = nq\) and \(I = \langle n \rangle\text{.}\)
The set \(n {\mathbb Z}\) is ideal in the ring of integers. If \(na\) is in \(n{\mathbb Z}\) and \(b\) is in \({\mathbb Z}\text{,}\) then \(nab\) is in \(n {\mathbb Z}\) as required. In fact, by Theorem 1.3.6, these are the only ideals of \({\mathbb Z}\text{.}\)
Let \(a,b\in \ker\phi\) be arbitrary. The \(\phi(a+b)=\phi(a)+\phi(b)=0+0=0\) by the definitions of a homomorphism and kernel, thus \(a+b\in \ker\phi\text{.}\) Now let \(r\in R\) be arbitrary. We must show that \(ar\) and \(ra\) are in \(\ker \phi\text{.}\) Indeed,
In our definition of an ideal we have required that \(rI \subset I\) and \(Ir \subset I\) for all \(r \in R\text{.}\) Such ideals are sometimes referred to as two-sided ideals. We can also consider one-sided ideals; that is, we may require only that either \(rI \subset I\) or \(Ir \subset I\) for \(r \in R\) hold but not both. Such ideals are called left ideals and right ideals, respectively. Of course, in a commutative ring any ideal must be two-sided. In this text we will concentrate on two-sided ideals.
Let \(r+I\) and \(s +I\) be in \(R/I\text{.}\) We must show that the product \((r + I)(s + I) = rs + I\) is independent of the choice of coset; that is, if \(r' \in r+I\) and \(s' \in s+I\text{,}\) then \(r's'\) must be in \(rs+I\text{.}\) Since \(r' \in r+I\text{,}\) there exists an element \(a\) in \(I\) such that \(r' = r + a\text{.}\) Similarly, there exists a \(b \in I\) such that \(s' = s + b\text{.}\) Notice that
\begin{equation*}
r' s' = (r+a)(s+b) = rs + as + rb + ab
\end{equation*}
and \(as + rb + ab \in I\) since \(I\) is an ideal; consequently, \(r' s' \in rs + I\text{.}\)
Addition can similarly be shown to be well-defined. We will leave as an exercise the verification of the associative law for multiplication and the distributive laws. These will follow from the definitions of a ring.
Let \(I\) be an ideal of \(R\text{.}\) The map \(\phi : R \rightarrow R/I\) defined by \(\phi( r ) = r + I\) is a ring homomorphism of \(R\) onto \(R/I\) with kernel \(I\text{.}\)
Certainly \(\phi : R \rightarrow R/I\) is a surjective function. It remains to show that \(\phi\) works correctly under ring multiplication and addition. Let \(r\) and \(s\) be in \(R\text{.}\) Then
The map \(\phi : R \rightarrow R/I\) is often called the natural or canonical homomorphism. In ring theory we have isomorphism theorems relating ideals and ring homomorphisms. We now will prove only the First Isomorphism Theorem for rings in this chapter and leave the proofs of the other two theorems as exercises. The first isomorphism is one of the fundamental theorems of modern algebra.
Let \(\psi : R \rightarrow S\) be a ring homomorphism. Then \(\ker \psi\) is an ideal of \(R\text{.}\) If \(\phi : R \rightarrow R/\ker \psi\) is the canonical homomorphism, then there exists a unique isomorphism \(\eta: R/\ker \psi \rightarrow \psi(R)\) such that \(\psi = \eta \phi\text{.}\)
Although we have given a procedure to produce the image \(\eta(r+K)\text{,}\) it is not clear that this procedure gives a well-defined function. The issue is the first step: since \(\phi\) may not be injective, there may be multiple choices of pre-images of \(\eta(r+K)\text{,}\) which could in turn give different results for a fixed input. To see this does not happen, let \(r',r''\in R\) be two arbitrary \(\phi\)-preimages of \(r+K\text{.}\) Then there exists \(k\in K\) such that \(r'=r''+k\text{,}\) which gives
by definition of a homomorphism and the kernel. It follows that \(\eta\) defined this way is well-defined. Since every element of \(R\) is contained in a \(K\)-coset, it follows that every element of \(\psi(R)\) is the image of some \(rK\in R/K\text{.}\)
We finally show \(\eta\) is injective. Recall that the zero element of \(R/K\) is the coset \(K = 0+K\text{.}\) Every non-trivial element of the quotient is of the form \(r+K\) with \(r\not\in K\text{.}\) Suppose towards a contradiction that \(\eta\) is not injective. Then there is some non-zero element \(r+K\) such that \(\eta(r+K)=\psi(r)=0\text{,}\) but this contradicts \(r\not\in K = \ker \psi\text{.}\) It follows that \(\eta\) is injective.
Let \(I\) be an ideal of a ring \(R\text{.}\) Then \(S \mapsto S/I\) is a one-to-one correspondence between the set of subrings \(S\) containing \(I\) and the set of subrings of \(R/I\text{.}\) Furthermore, the ideals of \(R\) containing \(I\) correspond to ideals of \(R/I\text{.}\)
