2.2. Ring homomorphisms.
After one understand the fundamentals of group theory, they go on to construct maps between different groups. This is the same strategy we'll follow here, since we can definitely define ring homomorphisms between rings.
The ring homomorphisms are also useful since they can help us deduce when two rings \(R\) and \(S\) are the "same," a concept which evolves into the concept of isomorphisms.
Let \(R\) and \(S\) be rings, and \(f:R \to S\). We define \(f\) to be a ring homomorphism if it preserves addition and multiplication; that is, if
for all \(a, b \in R\). If \(f\) is a bijection, then we say that \(f\) is a ring isomorphism.
\textcolor{NavyBlue}{Note that a ring homomorphism is simply a group homomorphism, with the extra condition of which preserves multiplication of the ring elements.} Therefore the following proposition, which hold for group homomorphisms, holds for ring homomorphisms too.
Let \(R\) and \(S\) be rings and \(f: R \to S\) a ring homomorphism. Then
- 1. if \(0_R \in R\) and \(0_S \in S\) are zero elements, then \(f(0_R) = 0_S\).
- 2. if \(f(-a) = -f(a)\) for all \(a \in R\)
- 3. \(f(a_1a_2\cdots a_n) = f(a_1)f(a_2)\cdots f(a_n)\) for all \(a_1, a_2, \dots a_n \in R\)
- 4. \(f(a_1 + a_2 + \cdots + a_n) = f(a_1) + f(a_2) + \cdots + f(a_n)\) for all \(a_1, a_2, \dots a_n \in R\).
Observe that
Since \((R, +)\) is a group, subtract \(\phi(r)\).
\textcolor{red}{Note that it is not necessarily true that \(f(1_R) = 1_S\).} In group theory, it was always guaranteed that we could map the identity element from one group to another. In our case, that's still true: \(f(0_R) = 0_S\). Group identity of \((R,+)\) is still mapped to the identity of \((S, +)\). But this is mapping the additive identity of \(R\) to the additive identity of \(S\).
What we're saying is that *multiplicative identities may not always be mapped to each other. *
Now since we can't always guarantee that
\(f(1_R) = 1_S\), \textcolor{red}{we also can't guarantee that \(f(a^{-1}) =
f(a)^{-1}\) for some invertible \(a \in R\).}
However, there is a clear cut case for when these things do
happen.
\
Let \(R\) and \(S\) be rings and \(\phi:R \to S\) a nonzero ring homomorphism. Denote \(1_R \in R\) and \(1_S \in S\) to be the respective multiplicative identities. Then
- 1. If \(\phi(1_R) \ne 1_S\) then \(\phi(1_R)\) is a zero divisors of \(S\).
- 2. If \(S\) is an integral domain then \(\phi(1_R) = 1_S\).
- 3. If \(\phi(1_R) = 1_S\) and \(u \in R\) is a unit then \(\phi(u)\) is a unit in \(S\). In other words, \(\phi(R^*) \subset S^*\)
- 4. If \(\phi(1_R) = 1_S\) and if \(u \in R\) has an inverse \(u^{-1} \in R\) then \(\phi(u^{-1}) = \phi(u)^{-1}\).
\textcolor{MidnightBlue}{An immediately corollary is this: \(\phi: R \to S\) is a not nonzero ring homomorphism if and only if \(\phi(1_R) \ne 0_S\). Furthermore, If \(S\) is an integral domain then \(\phi(R^*) \subset S^*\) for any homomorphism \(\phi: R \to S\). }
- 1. Suppose \(\phi(1_R) \ne 1_S\). Since \(1_R1_R = 1_R\), we know that
Since \(\phi(1_R) \ne 1_S\), either \(\phi(1_S) = 0\) or it is a zero divisor of \(S\).
Suppose \(\phi(1_R) = 0_S\) and let \(a \in R\). Then
Thus we see that \(\phi\) send every element of \(R\) to \(0_S\). However, this cannot be the case since we supposed that \(\phi\) is a nonzero homomorphism. Therefore \(\phi(1_R)\ne 0\), leaving us with no choice but to conclude that \(\phi(1_R)\) is a zero divisor in \(S\) as desired. * 2. Suppose \(S\) is an integral domain, and that \(\phi(1_R) \ne 1_S\) for the sake of contradiction. Then observe for any \(a \in R\)
Since \(\phi(1_R) \ne 1_S\), and \(\phi\) is a nonzero homomorphism, this implies that \(\phi(a)\) and \(\phi(1_R) - 1_S\) are zero divisors in \(S\) for at least one \(a \in R\). However, this is a contradiction since \(S\) is an integral domain and hence has no zero divisors. Thus by contradiction \(\phi(1_R) = 1_S\). * 3. Suppose \(\phi(1_R) = 1_S\) and let \(u\) be a unit in \(R\). Then \(uv = 1_R\) for some \(v \in R\). So
Therefore, \(\phi(u)\) is a unit in \(S\). Next, since \(uu^{-1} = 1_R\),
as desired. * 4. Suppose \(\phi(1_R) = 1_S\) and that \(u \in R\) has some inverse \(u^{-1} \in R\). Since \(uu^{-1} = 1_R\),
as desired.
\noindentExamples.\ Let \(n \in \ZZ\), and define the function \(f: \ZZ \to \ZZ\) as
Then this is a homomorphism if and only if \(n = 0\) or 1. Suppose otherwise. Then observe that the second condition of the definition of a ring homomorphism specifies that
This is only true if \(n = 0\) or 1, which is our contradiction.
Instead, we can construct the following function to form a homomorphism between \(\ZZ\) and \(\ZZ/n\ZZ\), where \(n\) is a positive integer. Let \(f: \ZZ \to \ZZ/n\ZZ\) such that
where \([m] = \{k \in \ZZ \mid k = m \mbox{ mod } n\}\). \
Suppose we construct a homomorphism \(\phi: \mathbb{R}[x] \to S\). (Recall that \(\RR[x]\) is the set of finite polynomials with coefficients in \(\RR\)). Define \(\phi\) as
First, observe that this is surjective, since for any \(a + bi \in \mathbb{C}\) we can send \(a + bx \in \RR\) to this element via \(\phi\). Therefore \(\im(\phi) = \mathbb{C}\).
Let us now describe \(\ker(\phi)\). First suppose that \(p(i) = 0\) for some \(p(x) \in \RR[x]\). At this point, we know that \(p(x)\) must be at least a second degree or greater polynomial. Therefore we can express \(p(x)\) as
for some \(q(x) \in \RR[x]\). Then
but this implies that \(a + bi = 0 \implies a = b = 0\). Therefore, \(p(i) = 0\) if and only if \(p(x) = q(x)(x^2 + 1)\) some \(q(x) \in \RR[x]\). In other words,
\ \ As in group theory, we have the following theorem regarding isomorphisms. We won't prove this again.
Let \(R\) and \(S\) be rings. A ring homomorphism \(f: R \to S\) is an isomorphism if and only if there exists a h omomorphism \(g: S \to R\) such that \(f \circ g\) is the identity map on \(R\) and \(g \circ f\) is the identity map on \(S\).
With the ring homomorphism defined, we again have \(\ker(f)\) and \(\im(f)\) as valid and important concepts.
Let \(R\) and \(S\) be rings and \(f: R \to S\) a ring homomorphism. Then we define
and
Suppose \(f: R \to S\) is a ring homomorphism. Then
- 1. The kernal \(\ker(f)\) is a subring of \(R\).
- 2. The image \(\im(f)\) is a subring of \(S\).
Caveat: Recall that "subrings" are rings that might not possibly contain \(1\), the multiplcative identity.
- 1. We can show this using the Subring Criterion. As we stated before, \(f(0) = 0\). Hence \(0 \in \ker(f)\) so that \(\ker(f)\) is nonempty.
To prove this, observe that
Since \((R, +)\) is a group, we can subtract \(f(0)\) from both sides to get \(f(0) = 0\).
Next, we want to show that \(r_1, r_2 \in \ker(f) \implies r_1 - r_2 \in \ker(f)\). Since we showed that \(f(-r) = -f(r)\) for all \(r \in R\), we know that
Hence, we see that \(r_1 - r_2 \in \ker(f)\).
Now again suppose \(r_1r_2 \in \ker(f)\). Then
so that \(r_1r_2 \in \ker(f)\). By the subring test, we see that \(\ker(f)\) is a subring of \(R\). * 2. We can similarly prove this via the Subring Test. First, observe that \(f(0) = 0\), so that \(0 \in \im(f)\). Hence, \(\im(f)\) is nonempty.
Next, suppose \(s_1, s_2 \in \im(f)\). Then we want to show that \(s_1 - s_2 \in \im(f)\). Now
This shows that \(s_1 - s_2 \in \im(f)\). Finally, we'll show that \(s_1s_2 \in \im(f)\). Observe that
Hence we see that \(s_1\times s_2 \in \im(f)\). Thus \(\im(f)\) is a subring of \(R\).
Finally, we end this section by noting that two important and useful mathematical identites continue to hold in the context of rings. We won't offer their proofs though since they are a bit tedious.
Let \(R\) be a ring and let \(a_1, a_2, \dots, a_m\) and \(b_1, b_2, \dots, b_n\) be elements of \(R\). Then
[Binomial Theorem] Let \(R\) be a ring (with identity) and let \(a, b \in R\) with \(ab = ba\). Then for any \(n \in \mathbb{N}\)