Integral Ideal is Ideal of Ring

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Theorem

Let $J$ be a non-empty subset of the set of integers $\Z$.


Then:

$J$ is an integral ideal

if and only if:

$J$ is an ideal of the ring of integers $\struct {\Z, +, \times}$.


Proof

Let $J \subseteq \Z$ fulfil the conditions of an integral ideal:

$(1): \quad n, m \in J \implies m + n \in J, m - n \in J$
$(2): \quad n \in J, r \in \Z \implies r n \in J$


First note that $J$ is non-empty by definition.


Then from $(1)$ we have in particular:

$n, m \in J \implies m - n \in J$

Thus by the One-Step Subgroup Test:

$\struct {J, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$.


Then from $(2)$ we have:

$n \in J, r \in \Z \implies r n \in J$

and by Integer Multiplication is Commutative it follows that:

$n \in J, r \in \Z \implies n r \in J$


Thus it is seen that the conditions are fulfilled for $J$ to be an ideal of the ring of integers $\struct {\Z, +, \times}$.

$\Box$


Suppose $J \subseteq \Z$ such that:

$(3): \quad \struct {J, +}$ is a subgroup of $\struct {\Z, +}$
$(4): \quad \forall n \in J: \forall r \in R: n r \in J \land r n \in J$

From $(4)$, $(2)$ immediately follows.

From $(3)$ it follows from the One-Step Subgroup Test that:

$\forall n, m \in J: m - n \in J$

which is a restatement of $(1)$.

Thus the conditions are fulfilled for $J$ to be an integral ideal.

$\blacksquare$

Sources

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