“up to associates” in the Euclidean domain
“up to associates” in the Euclidean domain
We know that the Euclidean Domain has the property of Unique Factorization.
More precisely, every nonzero element in a Euclidean ring $R$ can be uniquely written (up to associates) as a product of prime elements or is a unit in $R$.
The word "up to associates" confusing me a bit.
P.S. Let's consider the example in the euclidean domain $mathbbZ[i]$ and consider the following prime factorizations such as: $$(2+i)(1+i) quadtextand quad (-1+2i)(1-i)$$
Note that $2+isim -1+2i$ and $1+isim 1-i$.
Can anyone explain me the meaning of the phrase "up to associates" in the above example, please?
If $ab$ is one prime factorization, then for any unit $u$, we can get another prime factorization $(au)(u^-1b)$, So the factorization is only unique except for ("up to") the replacement of $a$ by the associate $au$ and $b$ by the corresponding associate $u^-1b$, where $u$ is any unit. In your example, $a=2+i$, $b = 1+i$, and $u=i$.
– Bungo
Aug 27 at 16:19
3 Answers
3
Uniqueness of factorization up to associates means that if $rin R$, nonzero and not a unit, is written as
$$
r = p_1p_2dots p_m = q_1q_2dots q_n
$$
with $p_i$ and $q_j$ irreducible, then
Two elements $a$ and $b$ are associate if there is a unit $u$ with $b=ua$.
This happens also in the integers: for instance, $6=2cdot3=(-3)(-2)$.
In your case, $2+i$ is associate to $-1+2i$ and $1+i$ is associate to $1-i$.
Let's consider another example: $(2+i)(1+i)$ and $(-1+2i)(1+i)$. Are these prime factorizations equal? Since $2+isim -1+2i$ and 1+isim 1+i$.
– RFZ
Aug 27 at 18:02
@RFZ The products are different. They're factorizations of associate elements.
– egreg
Aug 27 at 19:36
Hmm Thanks! I was thinking that uniqueness of factorization up to associates is something like that: for example, we have two primes $pi_1, pi_2$ then $pi_1pi_2$ is the same as $upi_1pi_2$ where $u$ - some unit of $R$.
– RFZ
Aug 27 at 19:50
In other words, if we have one factorization into primes and we multiply it by some unit $u$ then these two factorizations are the same!
– RFZ
Aug 27 at 19:51
@RFZ Not really; say $r=pq$, with $p$ and $q$ irreducible; if $u$ is a unit, then $r=(pu)(qu^-1)$ is another factorization, but the factors are associate to the ones in the other factorization.
– egreg
Aug 27 at 20:10
This means the prime factors are determined only up to a unit factor. Indeed in the Gaussian integers, the group of units is $;1,-1,i,-i$ and indeed
$$2+i=(-i)(-1+2i),qquad 1+i=i(1-i).$$
We have the same situation in $mathbf Z$, where $mathbf Z^times=1,-1$, and, for instance
$$6=2cdot 3=(-2)cdot(-3)$$
In the ordinary integers the factorizations
$$
6 = 2times 3 = (-2) times (-3)
$$
are equivalent up to associates because $2$ and $-2$ are associates because
$$
-2 = (-1) times 2
$$
and $(-1)$ is a unit - it has a multiplicative inverse.
In the Gaussian integers $2+i$ and $-1 + 2i$ are associates because
$$
-1 + 2i = i times (2 +i)
$$
and $(-i)$ is a unit because it has a multiplicative inverse, namely $i$.
In the integers you don't usually have to fuss with "up to associates" since there is a natural way to specify that primes should be positive. In the Gaussian integers there is no good way to distinguish among the associates
$$
2+i, -2 -i, -1 + 2i, 1 - 2i .
$$
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Related : math.stackexchange.com/questions/1933073/…
– Arnaud D.
Aug 27 at 16:16