Subrings of Integers are Sets of Integer Multiples

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Theorem

Let $\struct {\Z, +, \times}$ be the integral domain of integers.


The subrings of $\struct {\Z, +, \times}$ are the rings of integer multiples:

$\struct {n \Z, +, \times}$

where $n \in \Z: n \ge 0$.


There are no other subrings of $\struct {\Z, +, \times}$ but these.


Proof

From Integer Multiples form Commutative Ring, it is clear that $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.

We also note that when $n = 0$, we have:

$\struct {n \Z, +, \times} = \struct {0, +, \times}$

which is the null ring.

When $n = 1$, we have:

$\struct {n \Z, +, \times} = \struct {\Z, +, \times}$

From Null Ring and Ring Itself Subrings, these extreme cases are also subrings of $\struct {\Z, +, \times}$.


From Subgroups of Additive Group of Integers, the only additive subgroups of $\struct {\Z, +, \times}$ are $\struct {n \Z, +}$.

So there can be no subrings of $\struct {\Z, +, \times}$ which do not have $\struct {n \Z, +}$ as their additive group.

Hence the result.

$\blacksquare$


Examples

Even Integers

Let $2 \Z$ be the set of even integers.

Then $\struct {2 \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$.


Sources

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