Show 2, 3, and 4 +/- sqrt(10) are prime in Z(sqrt(10))."

[miss rohde]

Hi,

I have this problem that I don't understand;

"Consider the set Z(sqrt10) = {m + n*sqrt10 | integer m, n}. The number (m
+ n*sqrt10) is called a unit if m^2 -10n^2 = ± 1, since it has an inverse
(that is, since (m + n*sqrt10) ×± (m - n*sqrt10) = 1). For example,
3+sqrt10 is a unit, and so is 19-6*sqrt10. Pairs of cancelling units can be
inserted into any factorisation, so we ignore them. Nonunit numbers of
Z(sqrt10) are called prime if they cannot be written as a product of two
nonunits.

Show that 2, 3, and 4 +/- sqrt(10) are prime in Z(sqrt(10))."


Hint: if 2 = (a + b*sqrt10 ) (c + d*sqrt10) then
4 = (a^2 -10b^2 )(c^2 -10d^2 ).

Furthermore the square of any integer mod 10 is 0, 1, 4, 5, 6, or 9."

Well that tells me that I have some options

4 = 1 * 4
4 = 6*9 (mod10)
4 = 4*6 (mod 10

but now how do I continue?? please help me!

Best regards
Bettina

[William Elliot]

Post by miss rohde
I have this problem that I don't understand;

I have problem reading you post as it includes weird graphic characters.
What where I have marked ? are you saying?

Post by miss rohde
"Consider the set Z(sqrt10) = {m + n*sqrt10 | integer m, n}. The number
(m + n*sqrt10) is called a unit if m^2 -10n^2 = ? 1, since it has an
inverse (that is, since (m + n*sqrt10) ?? (m - n*sqrt10) = 1). For
example, 3+sqrt10 is a unit, and so is 19-6*sqrt10. Pairs of cancelling
units can be inserted into any factorisation, so we ignore them. Nonunit
numbers of Z(sqrt10) are called prime if they cannot be written as a
product of two nonunits.
Show that 2, 3, and 4 +/- sqrt(10) are prime in Z(sqrt(10))."

Prime or irreduciable? In these rings, sometimes there a difference
between prime and irreduciable.

Post by miss rohde
Hint: if 2 = (a + b*sqrt10 ) (c + d*sqrt10) then
4 = (a^2 -10b^2 )(c^2 -10d^2 ).

Furthermore the square of any integer mod 10 is 0, 1, 4, 5, 6, or 9."

Well that tells me that I have some options
4 = 1 * 4
4 = 6*9 (mod10)
4 = 4*6 (mod 10
but now how do I continue?? please help me!

Standard way is to define norm n(a + b.sqr k) for square free k
as a^2 - k.b^2 (which includes the norm of complex numbers)

Then show n(rs) = n(r) n(s)
r|s ==> n(r) | n(s)

r unit iff n(r) = +-1
by definition r is a unit when some s with rs = 1.
n(r) prime ==> r irreducible

[miss rohde]

I don't quite understand what you mean. I guess the whole problem has to do
with the number 6 having two representations:

2 * 3 = 6 = (4+sqrt10)*(4-sqrt10)

If we consider the set of all numbers of the form m + n sqrt 10 when m and n
integers, then the product of any two such numbers is again of the same
form, and we can call such a number "prime" if it can't be factored in a
non-trivial way. (this is what my math.book says). - and then I'm supposed
to show that

2, 3, and 4+/- sqrt10 are all "prime" in this system.


"Consider the set Z(sqrt10) = {m + n*sqrt10 | integer m, n}. The number
(m + n*sqrt10) is called a unit if m^2 -10n^2 = 1, since it has an
inverse (that is, since (m + n*sqrt10) (m - n*sqrt10) = 1). For example,
3+sqrt10 is a unit, and so is 19-6*sqrt10. Pairs of cancelling units can be
inserted into any factorisation, so we ignore them. Nonunit numbers of
Z(sqrt10) are called prime if they cannot be written as a product of two
nonunits.

Show that 2, 3, and 4 +/- sqrt(10) are prime in Z(sqrt(10))."

Prime or irreduciable? In these rings, sometimes there a difference between
prime and irreduciable.

I guess "irreduciable" (4+/-sqrt 10 can never be prime - I think!)

Hint: if 2 = (a + b*sqrt10 ) (c + d*sqrt10) then
4 = (a^2 -10b^2 )(c^2 -10d^2 ).
Furthermore the square of any integer mod 10 is 0, 1, 4, 5, 6, or 9."

Well that tells me that I have some options

4 = 1 * 4
4 = 6*9 (mod10)
4 = 4*6 (mod 10



Standard way is to define norm n(a + b.sqr k) for square free k as a^2 -
k.b^2 (which includes the norm of complex numbers)

Post by William Elliot
Then show n(rs) = n(r) n(s)
r|s ==> n(r) | n(s)

how do I show this?

Post by William Elliot
r unit iff n(r) = +-1
by definition r is a unit when some s with rs = 1.
n(r) prime ==> r irreducible

I'm sorry but I don't really follow you??

[a***@hotmail.com]

Post by miss rohde
I don't quite understand what you mean. I guess the whole problem has to do
2 * 3 = 6 = (4+sqrt10)*(4-sqrt10)
If we consider the set of all numbers of the form m + n sqrt 10 when m and n
integers, then the product of any two such numbers is again of the same
form, and we can call such a number "prime" if it can't be factored in a
non-trivial way. (this is what my math.book says). - and then I'm supposed
to show that
2, 3, and 4+/- sqrt10 are all "prime" in this system.
"Consider the set Z(sqrt10) = {m + n*sqrt10 | integer m, n}. The number
(m + n*sqrt10) is called a unit if m^2 -10n^2 = 1, since it has an
inverse (that is, since (m + n*sqrt10) (m - n*sqrt10) = 1). For example,
3+sqrt10 is a unit, and so is 19-6*sqrt10. Pairs of cancelling units can be
inserted into any factorisation, so we ignore them. Nonunit numbers of
Z(sqrt10) are called prime if they cannot be written as a product of two
nonunits.
Show that 2, 3, and 4 +/- sqrt(10) are prime in Z(sqrt(10))."
Prime or irreduciable? In these rings, sometimes there a difference between
prime and irreduciable.
I guess "irreduciable" (4+/-sqrt 10 can never be prime - I think!)
Hint: if 2 = (a + b*sqrt10 ) (c + d*sqrt10) then
4 = (a^2 -10b^2 )(c^2 -10d^2 ).
Furthermore the square of any integer mod 10 is 0, 1, 4, 5, 6, or 9."
Well that tells me that I have some options
4 = 1 * 4
4 = 6*9 (mod10)
4 = 4*6 (mod 10
Standard way is to define norm n(a + b.sqr k) for square free k as a^2 -
k.b^2 (which includes the norm of complex numbers)

Post by William Elliot
Then show n(rs) = n(r) n(s)
r|s ==> n(r) | n(s)

how do I show this?

Post by William Elliot
r unit iff n(r) = +-1
by definition r is a unit when some s with rs = 1.
n(r) prime ==> r irreducible

I'm sorry but I don't really follow you??

There is a confusion in what you are saying, because one traditionally
makes a distinction between prime and irreducible. Failure to split
into a product non-units is generally called irreducible, while p being
prime means that p | ab implies p | a or p | b. In non-unique
factorization domains, the two definitions are not identical.

Mr. Elliott described for you the extremely useful norm function N(x)
where in your ring, Z(sqrt(10)) we have N(a + b*sqrt(10)) = a^2 -
10*b^2 and
N(rs) = N(r)N(s). This can be proved by brute forc multiplication,
although there are prettier ways to do it which generalize in important
ways. Now 2 and 3 reducible would mean that there are elements of norm
2 or 3 (since having a norm of 1 is equivalent to being a unit - prove
this!). Then 2 or 3 would be squares mod 10 which is not true. Also
N(4 + sqrt(10)) = 6, so if it were reducible that would also imply the
existence of an element of norm 2 or 3 which we have already seen to be
impossible.


Regards,
Achava

[William Elliot]

Post by a***@hotmail.com

Post by miss rohde
Show that 2, 3, and 4 +/- sqrt(10) are prime in Z(sqrt(10))."
Prime or irreduciable?
Standard way is to define norm n(a + b.sqr k) for square free k as a^2 -
k.b^2 (which includes the norm of complex numbers)

Post by William Elliot
Then show n(rs) = n(r) n(s)
r|s ==> n(r) | n(s)

how do I show this?

Post by William Elliot
r unit iff n(r) = +-1
by definition r is a unit when some s with rs = 1.
n(r) prime ==> r irreducible

I'm sorry but I don't really follow you??

There is a confusion in what you are saying, because one traditionally
makes a distinction between prime and irreducible. Failure to split
into a product non-units is generally called irreducible, while p being
prime means that p | ab implies p | a or p | b. In non-unique
factorization domains, the two definitions are not identical.
Mr. Elliott described for you the extremely useful norm function N(x)
where in your ring, Z(sqrt(10)) we have N(a + b*sqrt(10)) = a^2 -
10*b^2 and
N(rs) = N(r)N(s). This can be proved by brute forc multiplication,
although there are prettier ways to do it which generalize in important
ways. Now 2 and 3 reducible would mean that there are elements of norm
2 or 3 (since having a norm of 1 is equivalent to being a unit - prove
this!). Then 2 or 3 would be squares mod 10 which is not true. Also
N(4 + sqrt(10)) = 6, so if it were reducible that would also imply the
existence of an element of norm 2 or 3 which we have already seen to be
impossible.

The pettier way is to define the conjugate of r = a + b.sqr k as
r* = a - b.sqr k. Then just as for complex numbers n(r) = rr*,
(r + s)* = r* + s*, (rs)* = r* s*, by brute force.

This will greatly ease the rest. Start by showing
r | s iff r* | s*
and do it for any k, even negative square free integers and
not just for k = 10. Then apply the results to k = 10, ie Z[sqr 10].

BTW, usual notation for the complex conguate is r^* instead of my
quick and lazy r*.

[miss rohde]

Achava and William Elliot thank you very much for taking the time to help
me. I'm not sure that I understand everything but I'll definitly be working
on it.

Best regards
Bettina Rohde