# Does <1,x> = <1> U <x> where U denotes union in Z[x]

1. Apr 20, 2007

### pivoxa15

Does <1,x> = <1> U <x> where U denotes union in Z[x]

It should be yes because what else could it be?

Last edited: Apr 20, 2007
2. Apr 20, 2007

### Hurkyl

Staff Emeritus
Did you notice that <1> = Z[x]?

3. Apr 20, 2007

### pivoxa15

Good point.

Maybe I should ask does <a,b>=<a> U <b> where a,b in R.

4. Apr 20, 2007

### Hurkyl

Staff Emeritus
<a, b> is certainly the smallest ideal of R that contains <a> U <b>. Do you have any reason to think <a> U <b> is actually an ideal?

5. Apr 21, 2007

### pivoxa15

I= <a> U <b> does seem like an ideal to me because it consists of elements generated by a or b. So any r in R with rI will either be in <a> or <b> so rI is a subset of I.

Last edited: Apr 21, 2007
6. Apr 21, 2007

### pivoxa15

My question is what is <a,b>?

7. Apr 22, 2007

### Hurkyl

Staff Emeritus
Being an ideal is more than simply being closed under multiplication by anything in R.

<a, b> is, by definition, the smallest ideal of R that contains a and b.

8. Apr 22, 2007

### pivoxa15

An ideal must also be a subset of R which I=<a>U<b> is so I don't see a problem with I being an ideal.

9. Apr 22, 2007

### Hurkyl

Staff Emeritus
And it must be closed under addition.

10. Apr 22, 2007

### matt grime

Then you''re not thinking very hard. Or not doing some examples. Or making a mistake with what <a>U<b> is. Something is in <a>U<b> if it is <a> OR it is in <b>. That will not in general make an ideal. In fact it is an ideal if and only if <a> is contained in <b> or vice versa.

One thing I've noticed, pixova, is that you don't seem to work out any examples to figure out what is going on. Intuition is good, but you need to develop it, and examples help.

Let's take the simplest example of Z, <2> and <3>. What is <2>U<3>? It is the set of things that are multiples of 2 *or* of 3. That is not an ideal.

Last edited: Apr 22, 2007
11. Apr 22, 2007

### pivoxa15

True I should construct more examples. Let I=<2,3> which is the smallest ideal which contains 2 and 3. So <2,3>={a2+3b |a,b in Z}. 10 is in I but not in <2>U<3> so <2>U<3> is not an ideal.

12. Apr 23, 2007

### matt grime

Is that a typo? 10 is certainly in <2>U<3>, since it is a multiple of 2.

On the example thing, if 10 wasn't a typo, the example is yours to play with, so you can always take the simplest thing - I'm always amazed at students who come up with some weird or wonderful counter example when frequently, say, 0 would do.

Of course here 0 isn't the right thing to look at. But what about 1? Or what about 2+3? It is the integers we're talking about here. The integers are PID. You've known for ages (the euclidean algorithm) that it is a PID, even if you don't know the words: <a,b>=<gcd(a,b)>, so picking any two coprime elements like 2 and 3, <2,3>=Z=/=<2>U<3>.

So, using knowledge of Z, and the good old Euclidean algorithm tells you it is wrong in Z.

Now, by picking the *simplest* possible thing to look at show that <a>U<b> is an ideal in any ring if and only if <a> is a subset of <b> or vice versa.

Last edited: Apr 23, 2007
13. Apr 24, 2007

### pivoxa15

Mybad, an error that went unnoticed by my behalf. 1 would be a counter example. I have a more general question:

When we talk about <a,b> we have to mention whether it is generating a subring or an ideal. If an ideal than <a,b> = {Aa+Bb|all A,B in R} as this set is the smallest ideal containing both a and b.

However what about <a,b> as generating a subring of R? It would have to be the smallest subring containing a and b. However it may be a smaller set than {Aa+Bb|all A,B in R} although in some cases it is exactly that set. i.e <2,x> as an ideal or subring of Z[x] is the set of all polynomials in Z[x] with even constant term, including 0. Is there a formula or algorithm to determine <a,b> as generating a subring of R?

Last edited: Apr 24, 2007
14. Apr 24, 2007

### Hurkyl

Staff Emeritus
That's not a ring: it doesn't contain 1.

You've probably studied subrings -- e.g. I bet you know what the subring $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{C}$ is! (If not, you'll probably talk about it eventually)

15. Apr 25, 2007

### pivoxa15

From what I have studied, 1 dosen't need to be in a ring. But 0 must be.

My lecturer talks about cases of a ring containing 1 which implicitly means there are rings without 1 but we are talking about rings with 1. He would never 'say' a ring with 0 because it would be a tautology as 0 must by definition be in any ring.

Is $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{C}$ just {a+bsqrt(2)|a,b in Q}

Last edited: Apr 25, 2007
16. Apr 25, 2007

### Hurkyl

Staff Emeritus
I always hear rumors that there are people out there who define the word "ring" without requiring that it have 1, but I've never actually met someone or read a text that does so.

I've only ever seen "ring with 1" used as a synonym for "ring", to clarify the situation if the reader/listener wasn't sure what definition the author/speaker is using. I've only ever seen "ring without 1" used as an entire phrase -- i.e. "ring without 1" does not mean "ring" "without 1". In particular a "ring without 1" need not be a ring, and may, in fact, have a multiplicative identity.

17. Apr 26, 2007

### pivoxa15

Here is something new for you, "A more significant difference is that some authors (such as I. N. Herstein) omit the requirement that a ring have a multiplicative identity. These authors call rings which do have multiplicative identities unital rings, unitary rings, or simply rings with unity." in http://en.wikipedia.org/wiki/Ring_(mathematics)

Hungerford dosen't specify 1 must be in a ring although Artin does.

However I do think it is a trivial matter.

Last edited: Apr 26, 2007
18. Apr 26, 2007

### matt grime

Really? Perhaps, because every ring can be extended to a unital ring, and there is a universal such, however, that extension might ruin some important properties. To be honest, the only non-unital ring that I've met was the subspace of Endos of a countable dimensional banach space with finite trace.