In Representations of Cyclic Groups, we were able to determine what every cyclic group would look like. We will now look at what a cyclic field, that is a field generated by a single element, would look in a specific case. In particular, we will find the field generated by the number \(1\).

**Field Generated by \(1\)**

Find the field, \(F\), generated by \(1\) using the usual addition and multiplication of real numbers.

**Definition**

Before we start giving what this field will look like, we need to recall the definition of a field. Therefore, we note that \(\langle A, +, *\rangle\) is a field for the operations \(+\) and \(*\) if

- \(\langle A, +, *\) is a commutative ring with identity.
- For every non-zero element, \(a\), of \(A\), there exists a \(a^{-1}\) such that \(a*a^{-1}=a^{-1}*a=1\) where \(1\) is the multiplicative identity.

Therefore, in order to find the field, \(F\), generated by \(1\), we will start with the set \(\left\{1\right\}\) and transitively close this set under each of the required operation.

**Group**

We know that in order to have a field, the set \(F\) must be a commutative group under addition. Because we are using the usual addition for real numbers, we know that addition will be associative and commutative. However, we must ensure that our set is closed under addition, additive identities and additive inverses.

Since \(F\) must be closed under addition and we know that \(1 \in F\), it must also contain \(1+1=2\). Furthermore, \(1+1+1=3 \in F\). If we continue this process, we note that \(\mathbb{N} \subset F\). That is, if we close the set \(\left\{1\right\}\) under addition, we find the semigroup generated by \(1\) to be the set of natural numbers.

Next, since the set must be closed for the additive identity, we must have that \(0 \in F\). That is, the monoid generated by \(1\) will be the set of whole number, \(\mathbb{N} \cup \left\{0\right\}\).

Now that we have an identity, we can find additive inverses. That is, for all \(x \in F\) we need an element \(y\) such that \(x+y=0\). Therefore, for all \(x \in F\), we must have \(-x \in F\). Since we know that \(\mathbb{N} \cup \left\{0\right\} \subseteq F\), this implies that all the negative integers will also be in \(F\). Combining these, we realize that \(\mathbb{Z} \subseteq F\) and that the group generated by \(1\) would be the set of integers.

**Field**

Now that we’ve closed the set under addition, the additive identity and additive inverses, we must close the set under multiplication as well. We know that this multiplication will be associative and commutative since it is the usual addition for real numbers, so we can focus on finding the closure for the required properties.

We begin with the set \(\mathbb{Z}\) since we already know that \(\mathbb{Z} \subseteq F\). Since this set is already closed under products and the multiplicative identity, we won’t need to add any more elements to satisfy closure here. That is, the ring with identity generated by \(1\) is precisely the set of integers.

Now, we still need to make this a field by closing the set for multiplicative inverses for all elements other than zero. Therefore, if we let \(n \in \mathbb{Z}\) we must have that there exists an \(y \in F\) such that \(ny=1\). That is, \(\frac{1}{n} \in F\). However, by adding all numbers of the from \(\frac{1}{n}\) for all \(n \in \mathbb{Z}\), we have created a set which is no longer closed under addition. So will start at the beginning and close this new set under the required properties.

If we do this; however, we note that this implies that, for any \(p,q \in \mathbb{Z}\) with \(q \neq 0\), we have that \(\frac{p}{q} \in \mathbb{Z}\). That is, we have that \(\mathbb{Q} \subseteq F\).

Now, if we can show that \(\mathbb{Q}\) is a field, we will have that \(F \subseteq \mathbb{Q}\) since \(F\) is the smallest field containing \(1\). We, therefore, provide a proof of this.

**Rational Numbers are a field**

Let \(\mathbb{Q}\) be the set of rational numbers using the usual addition and multiplication for real numbers. Then, let \(a, b \in \mathbb{Q}\). By definition, we know that \(a=\dfrac{p_{1}}{q_{1}}\) and \(b=\dfrac{p_{2}}{q_{2}}\) for some \(p_{1},p_{2},q_{1},q_{2} \in \mathbb{Z}\) with \(q_{1} \neq 0\) and \(q_{2} \neq 0\).

**Group**

Note that

\begin{align*}

a+b&=\frac{p_{1}}{q_{1}}+\frac{p_{2}}{q_{2}} \\

&=\frac{p_{1}q_{2}+p_{2}q_{1}}{q_{1}q_{2}}.

\end{align*}

Since the integers are closed under sums and products, we have that \(p_{1}q_{2}+p_{2}q_{1}, q_{1}q_{2} \in \mathbb{Z}\). Furthermore, since \(q_{1} \neq 0\) and \(q_{2} \neq 0\) and the are no zero-divisors for the real numbers, we must have that \(q_{1}q_{2} \neq 0\). Hence, \(\mathbb{Q}\) is closed for addition.

Furthermore, since we are using the usual addition for real numbers, we have that this addition must be associative and commutative.

Next, since \(0=\frac{0}{q}\) for any integer \(q\), we have that \(0 \in \mathbb{Q}\). Since \(0\) is the identity for the real numbers, it also must be the identity for the rationals. Hence, \(\mathbb{Q}\) has an additive identity. As a further note, since \(-a\) is the inverse of \(a\) for the real numbers it must also be the inverse for the rationals. Also, because \(-a=\frac{-p_{1}}{q_{1}}\), and \(-p_{1}, q_{1} \in \mathbb{Z}\) with \(q_{1} \neq 0\), we have that \(-a \mathbb{Q}\), so \(\mathbb{Q}\) is closed for additive inverses. Hence, \(\mathbb{Q}\) is an Abelian group under the usual addition for real numbers.

**Field**

We next note that

\begin{align*}

a*b&=\frac{p_{1}}{q_{1}}\frac{p_{2}}{q_{2}} \\

&=\frac{p_{1}p_{2}}{q_{1}q_{2}}.

\end{align*}

Since \(p_{1}p_{2}, q_{1}q_{2} \in \mathbb{Z}\) and \(q_{1}q_{2} \neq 0\), we have that \(\mathbb{Q}\) is closed under multiplication.

Now, note that \(1=\frac{1}{1} \in \mathbb{Q}\) and \(1a=a1=a\) for all \(a \in \mathbb{Q}\), so \(\mathbb{Q}\) has a multiplicative identity.

Next, we note that if \(a \neq 0\) and \(a=\frac{p_{1}}{q_{1}}\), then \(p_{1} \neq 0\). Therefore, \(\frac{q_{1}}{p_{1}} \in \mathbb{Q}\). Also

\begin{aling*}

a\frac{q_{1}}{p_{1}}&=\frac{p_{1}}{q_{1}}\frac{q_{1}}{p_{1}} \\

&=1,

\end{align*}

so \(\mathbb{Q}-\left\{0\right\}\) is closed under multiplicative inverses.

Hence, \(\mathbb{Q}\) is a field for the usual addition and multiplication of real numbers.

**Conclusion**

We have now shown that \(\mathbb{Q} \subseteq F\) and that \(F \subseteq \mathbb{Q}\). Hence \(F=\mathbb{Q}\). That is, the field generated by the number \(1\), for the usual addition and multiplication of real numbers is the set of all rational numbers.

I hope this helps you with your study and understanding of fields and generating elements. If it did, be sure to share this with anyone else that may need the help.