3.2 Fields

Definition 3.2.1. A commutative ring in which the set of non - zero elements form a group with respect to multiplication is called a field.

Alternatively, a field is an integral domain where every non zero element has a multiplicative unity.

Or An integral domain where every non - zero element is a unit.

A relationship between classes of rings is a follows \[\text {Fields} \subset \text {Integral domain} \subset \text {Commutative Rings} \subset \text {Rings}\]

Example 3.2.2.

1.

\(\mathbb {Z}_p\, ,\, p\) is prime \(\, \{0\, , 1\, , 2\, \cdots \cdots \, p-1\}\)

2.

\(\mathbb {Z}_3[i] = \big \{a + bi\, \big |\, a\, , \, b \in \in \mathbb {Z}_3\big \} = \{ 0,\, 1, \, 2, \, i, \, 1 +i,\, 2 + i, \, 1 + 2i,\, 2 + 2i\}\)

Check for multiplicative inverses! These include \(\, i\cdot 2i = 1\, i\cdot 2i = (1\cdot 2)i^2 = -2 =1\).

\(\implies \, i\) and \(2i\) are multiplicative inverses for each other. \(1\cdot 1 = 1\, , \, (1 + i)(2 + i) = 2 - 1 + 3i = 1 + 3i = 1\,\)

also \(\, 2\cdot 2 = 4 = 1\) mod\(_3\) and \((1 + 2i)(2 + 2i) = -2 + 6i = 1 + 6i = 1\)

3.

\(\mathbb {Q}\big [\sqrt {2}\big ] = \big \{a + b\sqrt {2}\, \big |\, a\, ,\, b\in \mathbb {Q}\big \}\). Clearly \(\mathbb {Q}\big (\sqrt {2}\big )\) is a ring. We show that every non zero element has an inverse \(\, \big (a + b\sqrt {2}\big )\cdot \big (c + d\sqrt {2}\big ) = 1\) \begin {align*} c + d\sqrt {2} & = \dfrac {1}{a + b\sqrt {2}} = \dfrac {a - b\sqrt {2}}{a^2 + 2b^2}\\\\ & = \dfrac {a}{a^2 + 2b^2} - \dfrac {b\sqrt {2}}{a^2 + 2b^2}\, \in \mathbb {Q} \sqrt {2}\hspace {0.5cm}\text {which is a field}\\ \end {align*}

\(\bullet \) By the same argument, one can check that \(\mathbb {Z}\sqrt {2}\,\) is not a field.

Since \(\, \dfrac {a}{a^2 + 2b^2}\,\) and \(\, - \dfrac {b\sqrt {2}}{a^2 + 2b^2}\,\) may be fractions and not integers.

Theorem 3.2.3. Every finite integral domain is a field.

Proof. Let \(D\) be a finite integral domain.
We need to show that each \(0\in a\in D\) has a multiplicative inverse. i.e, \(\exists \, b \in D \ni ab = 1\). We need to show that \(1\) is among the elts \(\{ ax\, :\, x\in D\}\). To do this, we consider the mapping \(\lambda _a :\, D\longrightarrow D.\, \hspace {0.2cm} x \longrightarrow ax\), for each \(x\in D\).
If \(\, \lambda _a\) is onto, then \(\lambda _a(x) = 1\) for some \(x\in D\). Thus we just need to show that \(\lambda _a\) is onto. Since \(\lambda _a\) is a mapping of a finite set onto itself, if suffices to show that \(\lambda _a\) is one to one for this imply onto.
To show one\(-\)one, assume \(\lambda _a(x_1) = \lambda _a(x_2)\). This implies \(ax_1 = ax_2\) by \(D\) is an integral domain, so \(ax_1 - ax_2 = 0\implies x_1 - x_2 = 0\) since \(a\neq 0\).\(\, \implies x_1 = x_2\). Thus \(\lambda _a\,\) is onto as required.

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Corollary 3.2.4. \(\mathbb {Z}_n\) is a field if and only if \(n\) is a prime.

Proof. We know that \(\mathbb {Z}_n\) is an integral domain if and only if \(n\) is prime. Now, by the previous theorem, \(n\) is prime implies \(\mathbb {Z}_n\) is finite and thus a field.

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Example 3.2.5.

1.: \(\mathbb {Z}_p\, ,\, p\,\) prime
2.: \(\mathbb {Z}_3[i]\)
3.: \(\mathbb {Q}\big [\sqrt {2}\big ]\)

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