A field has exactly two ideals: the zero ideal, which contains only the additive identity and the whole field itself. These are known as trivial ideals. Further if a commutative ring, with distinct additive and multiplicative identities, has no ideals other than the trivial ones, then it must be a field. These two facts are elegant in their symmetry and simplicity. In this article, we will explore why these facts are true. Familiarity with algebraic structures such as groups, rings and fields is assumed.

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## Definition of Ideals

A left ideal of a ring \( R \) is a subset \( I \subseteq R \) such that \( I \) is an additive subgroup of \( R \) and for all \( a \in I \) and \( r \in R \), we have \( r \cdot a \in I \). We say that a left ideal absorbs multiplication from the left by any ring element; or equivalently, that it is closed under left multiplication by any ring element.

Similarly, a right ideal of a ring \( R \) is a subset \( I \subseteq R \) such that \( I \) is an additive subgroup of \( R \) and for all \( a \in I \) and \( r \in R \), we have \( a \cdot r \in I \). We say that a right ideal absorbs multiplication from the right by any ring element; or equivalently, that it is closed under right multiplication by any ring element.

In a commutative ring \( R \), every left ideal is also a right ideal and vice versa. This is because for all \( a \in I \) and \( r \in R \), we have \( r \cdot a = a \cdot r \). Therefore, when working with commutative rings, we do not need to distinguish between left and right ideals and we simply refer to them as ideals. In this case, the ideal is said to absorb multiplication by any ring element; or equivalently, it is said to be closed under multiplication by any ring element.

## Examples of Ideals

Consider the set of even integers \[ \langle 2 \rangle = \{ 2n : n \in \mathbb{Z} \}. \] This is an ideal of \( \mathbb{Z} \). Indeed, if we multiply any even integer by any integer, the result is an even integer. In other words, the set of even integers absorbs multiplication by any integer. Equivalently, the set of even integers is closed under multiplication by any integer.

Let us see another example. Consider the ring of polynomials in the indeterminate \( t \) with integer coefficients, denoted \( \mathbb{Z}[t] \). The set \[ \langle 2, t \rangle = \{ 2f + tg : f, g \in \mathbb{Z}[t] \} \] is an ideal of \( \mathbb{Z}[t] \). Every element of this ideal is a linear combination of \( 2 \) and \( t \) with polynomial coefficients. If we take any element \( 2f + tg \in \langle 2, t \rangle \) where \( f, g \in \mathbb{Z}[t] \) and multiply it by any polynomial \( h \in \mathbb{Z}[t] \), we obtain \( 2fh + tgh \), which is again an element of \( \langle 2, t \rangle \). Hence \( \langle 2, t \rangle \) absorbs multiplication by any element of \( \mathbb{Z}[t] \), i.e. it is closed under multiplication by elements of \( \mathbb{Z}[t] \).

## Known Results

For the sake of brevity, we assume the following standard results.

Proposition 1. Let \( R \) be a ring. Then, for all \( a \in R \), we have \[ a \cdot 0 = 0 \cdot a = 0. \]

Proposition 2. Let \( R \) be a ring and let \( a \in R \). Then \begin{align*} I_L &= \{ r \cdot a : r \in R \}, \\ I_R &= \{ a \cdot r : r \in R \} \end{align*} are respectively a left ideal and a right ideal of \( R \). If \( R \) is commutative, then \( I_L = I_R \) and we write \[ \langle a \rangle = \{ a \cdot r : r \in R \} \] and say that \( \langle a \rangle \) is an ideal of \( R \) generated by \( a \).

## Ideals of Fields

In this section, we show that a field \( K \) has only two ideals: \( \{ 0 \} \) and \( K \) itself.

Clearly \( \{ 0 \} \) is an ideal of \( K \), as it satisfies the definition of an ideal. It is the trivial additive subgroup of \( K \) and by Proposition 1, for all \( r \in K \), we have \( r \cdot 0 = 0 \in \{ 0 \} \).

Now \( K \) is also an ideal of itself. Since \( K \) is an additive group by the definition of a field, it is an additive subgroup of itself. Moreover, as a field, \( K \) is closed under multiplication, so for all \( a, r \in K \) we have \( a \cdot r \in K \).

We will now show that \( \{ 0 \} \) and \( K \) are the only ideals of \( K \). Let \( I \) be an ideal of \( K \). There are two cases to consider: \( I = \{ 0 \} \) and \( I \ne \{ 0 \} \). Suppose \( I \ne \{ 0 \} \). Then there exists a non-zero element \( b \in I \). Since \( b \ne 0 \) and \( K \) is a field, \( b \) has a multiplicative inverse \( b^{-1} \in K \). Since \( b \in I \), \( b^{-1} \in K \) and \( I \) is closed under multiplication by any element of \( K \), we have \[ 1 = b \cdot b^{-1} \in I. \] Now, let \( c \in K \). Since \( 1 \in I \), \( c \in K \) and \( I \) is an ideal of \( K \), we get \[ c = 1 \cdot c \in I. \] Thus \( K \subseteq I \) and since \( I \subseteq K \) by definition, we conclude \( I = K \). Therefore the only ideals of \( K \) are \( \{ 0 \} \) and \( K \) itself.

## Rings With Trivial Ideals

We now show that if \( R \) is a commutative ring with \( 1 \ne 0 \) and the only ideals of \( R \) are \( \{ 0 \} \) and \( R \) itself, then \( R \) must be a field. To do this, we first show that every non-zero element of \( R \) has a multiplicative inverse in \( R \). Let \( a \in R \) with \( a \ne 0 \). We now show that there exists a multiplicative inverse \( a^{-1} \in R \). By Proposition 2, the set \[ \langle a \rangle = \{ a \cdot r : r \in R \}. \] is an ideal of \( R \). Since \( a = a \cdot 1 \in \langle a \rangle \), we have \( \langle a \rangle \ne \{ 0 \} \). By assumption, the only ideals of \( R \) are \( \{ 0 \} \) and \( R \), so it must be that \( \langle a \rangle = R \). Therefore \( 1 \in \langle a \rangle \) and \[ 1 = a \cdot s \] for some \( s \in R \). Thus \( a \) has a multiplicative inverse \( s \in R \) and this holds for every non-zero \( a \in R \).

The remaining properties of fields, namely, associativity and commutativity of addition and multiplication, the existence of distinct additive and multiplicative identities, the existence of additive inverses and the distributivity of multiplication over addition, are inherited from the ring \( R \). Therefore \( R \) is a field.

## Conclusion

To summarise, any commutative ring with distinct additive and multiplicative identities that has only trivial ideals is a field and every field has only trivial ideals.

Note that every field is also a commutative ring with distinct additive and multiplicative identities. Therefore, we can say that every field is a commutative ring with distinct additive and multiplicative identities and only trivial ideals and vice versa. It is neat how the two facts align so nicely.