Unit (ring theory)

An el­e­ment \(x\) of a non-triv­ial ring­noteThat is, a ring in which \(0 \not = 1\); equiv­a­lently, a ring with more than one el­e­ment. is known as a unit if it has a mul­ti­plica­tive in­verse: that is, if there is \(y\) such that \(xy = 1\). (We speci­fied that the ring be non-triv­ial. If the ring is triv­ial then \(0=1\) and so the re­quire­ment is the same as \(xy = 0\); this means \(0\) is ac­tu­ally in­vert­ible in this ring, since its in­verse is \(0\): we have \(0 \times 0 = 0 = 1\).)

\(0\) is never a unit, be­cause \(0 \times y = 0\) is never equal to \(1\) for any \(y\) (since we speci­fied that the ring be non-triv­ial).

If ev­ery nonzero el­e­ment of a ring is a unit, then we say the ring is a field.

Note that if \(x\) is a unit, then it has a unique in­verse; the proof is an ex­er­cise.

If \(xy = xz = 1\), then \(zxy = z\) (by mul­ti­ply­ing both sides of \(xy=1\) by \(z\)) and so \(y = z\) (by us­ing \(zx = 1\)).

Examples

  • In \(\mathbb{Z}\), \(1\) and \(-1\) are both units, since \(1 \times 1 = 1\) and \(-1 \times -1 = 1\). How­ever, \(2\) is not a unit, since there is no in­te­ger \(x\) such that \(2x=1\). In fact, the only units are \(\pm 1\).

  • \(\mathbb{Q}\) is a field, so ev­ery ra­tio­nal ex­cept \(0\) is a unit.

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