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Section Maps of Rings

Worksheet Part 1: Ring Homomorphisms

The goal of these problems is to explore the properties of ring homomorphisms and practice using the First Isomorphism Theorem for rings.

1.

Determine all ring homomorphisms from \(\Z_{20}\) to \(\Z_{30}\text{.}\)
Solution.
As we discussed in the Daily Prep video and notes, every ring homomorphism \(\phi\) out of \(\Z_n\) is determined by \(\phi(1)\text{.}\) For \(\phi: \Z_{20}\to\Z_{30}\text{,}\) we also need \(|\phi(1)|\mid \gcd(20,30)\) and \(\phi(1)^2=\phi(1) \bmod 30\text{.}\) Since \(\gcd(20,30)=10\text{,}\) the first condition implies \(\phi(1)\) is a multiple of 3, i.e \(\phi(1)\in \{0,3,6,9,12,15,18,21,24,27\}\text{.}\) Of these numbers, only \(0, 6, 15\text{,}\) and \(21\) are idempotents. Therefore there are four ring homomorphisms from \(\Z_{20}\) to \(\Z_{30}\text{:}\)
\begin{equation*} \phi_1(x)=0, \quad \phi_2(x)=6x, \quad \phi_3(x)=15x, \quad \phi_4(x)=21x. \end{equation*}

2.

Suppose \(\phi: \Z\oplus \Z\to \Z\oplus \Z\) is a ring homomorphism. What are the possibilities for \(\phi((1,0))\text{?}\)
Solution.
Because \((1,0)^2=(1,0)\text{,}\) \(\phi((1,0))\) must also be an idempotent. \(\Z \oplus \Z\) has four idempotent elements,
\begin{equation*} (0,0), (1,0), (0,1), (1,1), \end{equation*}
and any of these could be \(\phi(1,0)\text{.}\)

3.

Determine all ring homomorphisms \(\Z\oplus \Z \to \Z\oplus \Z\text{.}\)
Solution.
A ring homomorphism out of \(\Z\oplus \Z\) is determined by \(\phi((1,0))\) and \(\phi((0,1)\text{,}\) since
\begin{equation*} \phi((a,b))=\phi(a\cdot(1,0)+b\cdot(0,1))=a\cdot\phi((1,0))+b\cdot\phi((0,1)). \end{equation*}
We know there are only four possibilities for \(\phi((1,0))\text{,}\) so let’s analyze each case. The same four possibilities exist for \(\phi((0,1))\text{,}\) but not all of these combinations work together, since \((1,0)(0,1)=(0,0)\) implies that
\begin{equation*} \phi((1,0))\phi((0,1))=\phi(0,0)=(0,0). \end{equation*}
  1. \(\phi((1,0))=(0,0)\text{.}\) Then \(\phi((0,1))\) can be any of the four idempotents, giving us the maps
    \begin{equation*} (a,b)\mapsto (0,0),\quad (a,b)\mapsto (b,0),\quad (a,b)\mapsto (0,b),\quad (a,b)\mapsto (b,b)\text{.} \end{equation*}
  2. \(\phi((1,0))=(1,0)\text{.}\) Then \(\phi((0,1))\) is either \((0,0)\) or \((0,1)\text{,}\) giving us the maps
    \begin{equation*} (a,b)\mapsto (a,0),\quad (a,b)\mapsto (a,b)\text{.} \end{equation*}
  3. \(\phi((1,0))=(0,1)\text{.}\) Then \(\phi((0,1))\) is either \((0,0)\) or \((1,0)\text{,}\) giving us the maps
    \begin{equation*} (a,b)\mapsto (0,a), \quad (a,b)\mapsto (b,a)\text{.} \end{equation*}
  4. \(\phi((1,0))=(1,1)\text{.}\) Then \(\phi((0,1))=(0,0)\) and we get the map
    \begin{equation*} (a,b)\mapsto (a,a) \text{.} \end{equation*}

4.

Let \(R\) be a commutative ring of prime characteristic \(p\text{.}\) Show that the Frobenius map, \(x\mapsto x^p\text{,}\) is a ring homomorphism from \(R\) to \(R\text{.}\) (This map is extremely important to the theory and applications of finite fields.)
Solution.
Let \(a,b \in R\) and \(f\) be the Frobenius map. In a commutative ring the Binomial Theorem holds (where the integer binomial coefficient multiplying the ring element \(a^kb{p-k}\) means repeated addition that many times). Then
\begin{align*} f(a+b)\amp=(a+b)^p\\ \amp=\sum_{k=0}^p \binom{p}{k} a^kb^{p-k}\\ \amp=a^p +b^p+\sum_{k=1}^{p-1} \dfrac{p!}{k!(p-k)!}a^kb^{p-k}\text{.} \end{align*}
All of the factors in the denominator of binomial coefficient fraction of the remaining terms of the sum are less than \(p\) and therefore do not divide \(p\text{,}\) so each term in the sum is added a multiple of \(p\) times, and thus is 0 since \(\char R=p\text{.}\) So we have \(f(a+b)=a^p+b^p=f(a)+f(b)\text{.}\) Now \(f(ab)=(ab)^p=a^pb^p=f(a)f(b),\) since \(R\) is commutative. Therefore the Frobenius map is a ring homomorphism.

Worksheet Part 2: Maps of Rings with Identity

In this part, you will keep working with ring homomorphisms, with a focus on maps out of \(\Z\) to other rings with identity.

1.

Show that \((\Z \oplus \Z)/(\langle a\rangle \oplus \langle b\rangle)\) is ring-isomorphic to \(\Z_a\oplus \Z_b\text{.}\)
Solution.
We will apply the First Isomorphism Theorem for rings. The natural map from \(\Z \oplus \Z\) to \(\Z_a\oplus \Z_b\) is the map
\begin{equation*} \phi((x,y))=(x\bmod a, y\bmod b) \text{.} \end{equation*}
This is a homomorphism because the map \(x\mapsto x\bmod n\) is a homomorphism for any \(n\) and the operations of a direct sum of rings are componentwise. The map is surjective because for any \(0\leq x \leq a-1\) and \(0\leq y \leq b-1\) we have that the element \((x,y)\in \Z\oplus \Z\) maps to \((x,y)\) in \(\Z_a\oplus \Z_b\text{.}\) Finally, the kernel of the map is the set of pairs \((x,y)\) so that \(x\bmod a = 0\) and \(y\bmod b =0\text{,}\) i.e. \(x\) is a multiple of \(a\) and \(y\) is a multiple of \(b\text{,}\) i.e. \((x,y) \in \langle a\rangle \oplus \langle b\rangle\text{.}\)

2. Test for Divisibility by 11.

Let \(n\) be an integer with decimal representation
\begin{equation*} a_ka_{k-1}\dots a_1a_0=\sum_{i=0}^{k} a_i 10^i\text{.} \end{equation*}
Prove that \(n\) is divisible by \(11\) if and only if
\begin{equation*} a_0-a_1+a_2-\dots(-1)^ka_k=\sum_{i=0}^k (-1)^i a_i \end{equation*}
is divisible by \(11\text{.}\)
Solution.
Apply the natural map \(\phi:\Z\to\Z_{11}\text{,}\) which produces a ring isomorphism \(\Z/(11\Z)\isom \Z_{11}\text{.}\) We have \(n\) is divisible by \(11\) if and only if \(\phi(n)=0\text{.}\) But
\begin{align*} \phi(n)\amp=\sum_{i=0}^{k} a_i \phi(10)^i\\ \amp =\sum_{i=0}^{k} a_i (-1)^i\text{,} \end{align*}
so the result follows.

3.

Prove or disprove that the field of real numbers is ring-isomorphic to the field of complex numbers.
Solution.
This is false. If \(\phi:\C \to \R\) is a ring isomorphism, then \(\phi(1)=1\) and \(\phi(-1)=-1\) by the properties of ring isomorphisms. Then
\begin{equation*} \phi(i)^2=\phi(i^2)=\phi(-1)=-1\text{,} \end{equation*}
so \(\phi(i)\) is a square root of \(-1\) in \(\R\text{.}\) But no such real number exists, so \(\C\) and \(\R\) are not ring-isomorphic.

4.

Prove Every Field Contains \(\Z_p\) or \(\Q\) by showing that if \(S=\im(\phi)\) for the natural map \(\phi:\Z\to F\text{,}\) then either \(S\isom\Z_p\) or \(T=\{ab^{-1}\mid a,b\in S, b\neq 0\}\isom \Q\text{.}\)
Solution.
If \(\char F = p\text{,}\) then by A Ring with 1 Contains \(\Z\) or \(\Z_n\) we have \(S\isom \Z_p\text{.}\) If \(\char F = 0\text{,}\) then \(S\isom \Z\text{.}\) Let \(f:\Q \to T\) be the map
\begin{equation*} f(a/b)=(a\cdot 1)(b\cdot)^{-1}\text{.} \end{equation*}
This map is well-defined because if \(a/b=c/d\) in \(\Q\) we have \(ad=bc\) and \(b,d\neq 0\) so
\begin{equation*} (a\cdot 1)(d\cdot 1)=(b\cdot 1)(c\cdot 1) \end{equation*}
and this implies
\begin{equation*} f(a/b)=(a\cdot 1)(b\cdot)^{-1}=(c\cdot 1)(d\cdot 1)^{-1}=f(c/d)\text{.} \end{equation*}
Then we have
\begin{align*} f(a/b+c/d) \amp = f((ad+bc)/bd)\\ \amp = ((ad+bc)\cdot 1)/(bd \cdot 1)\\ \amp = (ad \cdot 1)/(bd\cdot 1)+(bc \cdot 1)/(bd\cdot 1)\\ \amp = \frac{(a\cdot 1)(d\cdot 1)}{(b\cdot 1)(d\cdot 1)}+\frac{(b\cdot 1)(c\cdot 1)}{(b\cdot 1)(d\cdot 1)}\\ \amp = \frac{a\cdot 1}{b\cdot 1}+\frac{c\cdot 1}{d\cdot 1}\\ \amp = f(a/b)+f(c/d)\text{.} \end{align*}
Similarly,
\begin{align*} f((a/b)(c/d)) \amp = f(ac/bd)\\ \amp = (ac \cdot 1)/(bd \cdot 1)\\ \amp = \left(\frac{a\cdot1}{b\cdot 1}\right)\left(\frac{c\cdot1}{d\cdot 1}\right)\\ \amp = f(a/b)f(c/d)\text{,} \end{align*}
so \(f\) is a ring homomorphism. It is surjective by construction. If \(f(a/b)=0\) then \((a\cdot 1)(b\cdot 1)^{-1}=0\) so \(a\cdot 1=0\) which forces \(a=0\) since \(\char(F)=0\text{,}\) so \(a/b=0\text{.}\) Therefore \(f\) is injective, and we have \(T\isom \Q\text{.}\)
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