# Another Euclidean ring

1. Nov 21, 2006

### gonzo

While I'm on the topic, here is another ring I need to show Euclidean. I'll show more of the work this time too. The ring is $Z[{\sqrt 2 \over 2}(1+i)]$

So, using the standard norm difference approach, we pick any element alpha in the field and try and show we can always find an element beta in the ring of integers so that the norm of the difference is less than one. We can start like this:

$$\alpha=a_1 + a_2 \sqrt 2 + a_3 i + a_4 i \sqrt 2$$
$$\beta=b_1 + b_2 \sqrt 2 + b_3 i + b_4 {\sqrt 2 \over 2}(1+i)$$
$$\gamma=\alpha-\beta$$
$$\gamma=(a_1 - b_1) + (a_2 - b_2 - {1 \over 2}b_4) \sqrt 2 + (a_3 - b_3) i + (a_4 - {1 \over 2}b_4) i \sqrt 2$$
$$\gamma=c_1 + c_2 \sqrt 2 + c_3 i + c_4 i \sqrt 2$$

After some pain and agony we get the norm of gamma:

$$N(\gamma)=(c_1^2 - 2c_2^2 - c_3^2 + 2c_4^2)^2 + (2c_1c_3-4c_2c_4)^2$$

The best we can do is force c1, c2 and c3 to be less than 1/2 and c4 to be less than 1/4 (we can switch this for c2 and c4, but the norm is symmetric in c2 and c4, so this doesn't matter).

However, these values don't let you force the norm less than 1. I tried exapanding and recombining in different ways and still nothing.

Any clever ideas?

Last edited: Nov 21, 2006
2. Nov 21, 2006

### Hurkyl

Staff Emeritus
Well, writing things in terms of the basis

$$\left\{1, \frac{1 + i}{\sqrt{2}}, i, \frac{-1 + i}{\sqrt{2}} \right\}$$

got a better-behaved expression:

$$\left( a^2 - c^2 + 2bd \right)^2 + \left( 2ac - b^2 + d^2 \right)^2$$

which I think, with a little calculus, you can show is always less than 1 if a, b, c, and d are all of magnitude less than 1/2.

3. Nov 22, 2006

### gonzo

Really, how? Maybe there are some techniques I don't know that would be helpful here.

One thing I was playing with was using the arithmetic-geometric mean inequality by grouping the monomorphisms into two groups and then adding them.

The problem is this inequality only works if the two terms are positive rational numbers, and I haven't been able to make that limitation work out.

Last edited: Nov 22, 2006
4. Nov 22, 2006

### Hurkyl

Staff Emeritus
The basic idea was that, for example, the first term takes on its largest value of 9/16 only when a = 0 or c = 0. But when that happens, the other term can be 1/16, at most.

Oh, maybe no calculus is needed. Suppose that that expression exceeds 1.

WLOG, we may assume the first term exceeds 1/2, |a| > |c| and bd > 0. The second term must be at least as large as 7/16.

Under these conditions, we must have |c| < 1/4. (we can do better) And thus, the second term cannot exceed 1/4. This is a contradiction.

5. Nov 22, 2006

### Hurkyl

Staff Emeritus
Oh, here's a less clever method. Calculus does come to the rescue!

The maximum of the norm occurs either on the boundary of the cube
[-1/2, 1/2]^4
or somewhere in its interior where its four partial derivatives are all zero.

We can clearly see that the norm is less than one on the boundary, so we have to look at the interior critical points:

if I set

L = a² - c² + 2bd
R = 2ac - b² + d²

then critical points occur when:

aL + cR = 0
cL - aR = 0
bL + dR = 0
dL - bR = 0

which imply, for example:

(a² + c²)R = 0

But you can easily check that the norm cannot exceed 1 of R = 0, and similarly if a² = c² = 0.

6. Nov 22, 2006

### gonzo

Thanks, I follow that. I had the idea about 9/16 limits, but I didn't know how to show rigoursly that maxed out the sum, but your proof by assuming one of them greater than 1 made sense to me.

I figured out another way to do it as well I think using your basis by my original method. If you use your basis for the integer and the simpler one for the non-integer, you can force both C2 and C4 in my norm to be less than 1/4, which works out with a very simple analysis.

I guess part of the trick is finding the right basis to make it work out nicer.

By the way, do you happen to know if the property of having a Euclidean function is "transitive" over field extensions? For example, if K over Q is Euclidean and L over K is Euclidean, does this imply that L over Q is also Euclidean?

7. Nov 22, 2006

### gonzo

Opps, I was wrong again ... even using your basis I can only force either C2 or C4 to be less than 1/4 not both, looks like your method is still the best.

8. Nov 22, 2006

### Hurkyl

Staff Emeritus
Being Euclidean is a property of a ring; I don't see what field extensions have to do with anything. (Incidentally, all fields are Euclidean domains)

9. Nov 22, 2006

### gonzo

Sorry, that's not what I meant. I meant the rings of integers in those fields.

10. Nov 22, 2006

### Hurkyl

Staff Emeritus
Being Euclidean is a property of a ring -- it is not a property of a relationship between two rings -- so your question doesn't really make sense as stated.