# Thread:

1. ## Ideals, Euclidean Domains

$J$ is an Euclidean Domain with $v:J-[0]\rightarrow\mathbb{Z}\cup[0]$. Let $1$ be the multiplicative identity.Let
$I=[a\in{J}|v(a)>v(1)]$
Is $I$ an ideal?

The only thing confusing me about this definition is whether the elements $0$ and $1$ are in $I$, as if they aren't then $I$ isn't a subring, so it's not an ideal.
$0$ isn't in the domain of $v$, so I'm not sure if it's in $I$ and $v(1)=v(1)$, so $1$ should not be in $I$.

2. ## Re: Ideals, Euclidean Domains

I think you've a little confusion about ideals.
You're correct that $1 \notin I$, so that I isn't a ring. But not containing 1 doesn't mean that I isn't an ideal. The "usual" situation is for an ideal not to contain 1. The only ideal in ring that contains 1 is the entire ring itself.

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$\text{Proof: Let }I \text{ be an ideal in a ring } R \text{ such that }1 \in I.$

$\text{Then for all }r \in R, \text{ since } 1 \in I, \text{ have that } r1 = r \in I.$

$\text{Thus if }r \in R, \text{ then } r \in I, \text{ and so } R \subset I.$

$\text{But also }I \subset R, \text{ and therefore } I = R.$

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Being an ideal means being a subgroup of the addition group of the ring, and being closed under multiplication by *any* element of the ring.

Think of the ideal E of even numbers in the integers. E = (2) in (Z, +, *, 0, 1). 0 is even, and addition and subtraction of even numbers is again an even number, so E is a subgroup of the additive group of the integers. Also, an even times *any* integer (even or odd) is again even. Thus E is an ideal. But it's *not* a ring, since it doesn't contain 1.

Note that there's nothing special about 2 for that. The ideal I = (n) in (Z, +, *, 0, 1) behaves the exact same way: sums and differences of integers that are both multiples of n are again multiples of n, 0 is a multiple of n (0 = n0), so (n) is a subgroup of (Z, +, 0). If m is *any* integer, then m times a multiple of n is again a multiple of n, so m times anything in (n) is also in (n). Thus (n) is an ideal.

So ideals differ from rings in that they don't necessarily contain a 1, and that the defined multiplication in the ideal is BIGGER than the ideal itself. We define multiplication in the ideal of even integers as mutliplication of an even integer times ANY integer, even or not.

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As to your problem, I always get confused about things like whether 0 is in I. in fact, I'd have to look it up, and write it down carefully since, as you note, 0 is not in the domain of v. Thus whether 0 is in I, or not, is a matter of the truth of the vacuous proposition and the technical definition of that set. This is something that I won't say one way or another off the top of my head (my "guess" is that 0 is in I, but don't quote me.) You're right that 0 be in I *is* necessary for I to be an ideal (since as a subgroup of the additive group of J, I must contain the identity of the additive group, 0.)

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However, you can see that that's not an ideal w/o answering whether 0 in I or not. Counter examples are easy.
Consider J = the usual Euclidean Domain of the integers. J = (Z, +, *, 0, 1), where v(x) = |x|. Then I is the set of integers whose absolute value is greater than the absolute value of 1 (i.e. I is the set of integers whose absolute value is greater than or equal to 2.) Is 5 in I? Is -4 in I? Is (5) + (-4) in I? Is I a subgroup of the additive group of the ring (i.e. of (Z, +, 0))? Is I an ideal of the Euclidean domain that's the ring of integers?
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