0

I just don't understand this concept at all and I can't find anywhere an explanation thay's easy enough to understand!

For example, the quotient ring $$\mathbb{Z}/6\mathbb{Z}$$

What are the elements in this?

Could someone maybe explain in VERY simple terms as I'm not great at the subject (at all). So sorry if this is an offensively stupid question.

Lindsay
  • 13
  • Do you know what cosets are? BTW, your question is neither stupid nor offensive. – saulspatz May 14 '20 at 17:56
  • the elements are $6\mathbb Z$, $1+6\mathbb Z$, $2+6\mathbb Z$, $ 3+6\mathbb Z, 4+6\mathbb Z, $ and $5+6\mathbb Z$ – J. W. Tanner May 14 '20 at 17:56
  • A description of the elements is given in the definition of the quotient ring. They are the cosets $a+6\Bbb{Z}$. Here $a$ can be any integer, but $a+6\Bbb{Z}=b+6\Bbb{Z}$ whenever $a-b$ is a multiple of six. Write down a few such cosets to see it, if you are not familiar with the idea: $$4+6\Bbb{Z}={4,10,16,\ldots,-2,-8,\ldots}=-2+6\Bbb{Z}.$$ – Jyrki Lahtonen May 14 '20 at 18:00
  • In this old answer I try to describe the philosophy/thinking that leads to quotient groups. It's more or less the same with quotient rings: to make the algebraic structure to match the phenomenon we want to describe, we are lead to make identification of elements of some other structure. Here we identify two integers, if their difference is a multiple of six. Some problems call for such identifications. – Jyrki Lahtonen May 14 '20 at 18:06
  • @saulspatz yes I do! Thank you! – Lindsay May 14 '20 at 18:06
  • @J.W.Tanner thank you for being so helpful! – Lindsay May 14 '20 at 18:07
  • @JyrkiLahtonen thank you so much for being so helpful, I'm very grateful – Lindsay May 14 '20 at 18:07

3 Answers3

3

Elements of quotient are classes of equivalence (some subsets) of original ring.

If we take $\mathbb Z / 6\mathbb Z$, elements are equal if they differ by exactly $6$ (there difference is in $6\mathbb Z$). So elements of quotient rings are sets:

0) $\{0, 6, -6, 12, -12, 18, -18, \ldots\} = 0 + 6\mathbb Z$

1) $\{1, 7, -5, 13, -11, 19, -17, \ldots\} = 1 + 6\mathbb Z$

2) $\{2, 8, -4, 14, -10, 20, -16, \ldots\} = 2 + 6\mathbb Z$

and so on.

To take, for example, sum of such elements, we need to sum elements of it member-wise: $$(1 + 6\mathbb Z) + (2 + 6\mathbb Z) = \{1, 7, -5, \ldots\} + \{2, 8, -4, \ldots\} = \{3, 9, -3, 9, 15, 3, -3, 3, -9, \ldots\} = 3 + \mathbb Z$$

($6\mathbb Z$ been ideal guarantees that member-wise sum or product of two equivalence classes is also an equivalence class)

mihaild
  • 15,368
  • Thank you so much! Is there any way I could ask just a quick follow up? How would I then find the zero divisors in the quotient ring? – Lindsay May 14 '20 at 18:08
  • It deserves it's own question. There is a notion of prime ideal, and quotient ring has zero divisiors iff ideal we took quotient by is not prime. In case of $\mathbb Z$, prime ideals are exactly ideals generated by prime number. More generally, we need if $ab$ is in ideal, then at least of $a$ and $b$ is in ideal. – mihaild May 14 '20 at 18:11
2

The elements are cosets of the form $a + 6\mathbb{Z}$. For example, $0 + 6\mathbb{Z}$ is an element, and so is $5 + 6 \mathbb{Z}$.

What do we mean by such a coset? The element $5 + 6 \mathbb{Z}$ is $$ 5 + 6 \mathbb{Z} = \{ 5 + 6n : n \in \mathbb{Z} \}.$$ Thus a first answer to your question is that all elements in $\mathbb{Z}/6\mathbb{Z}$ are of the form $a + 6 \mathbb{Z}$.

But there is more to the story. For example, $5 + 6\mathbb{Z}$ is the same coset as $11 + 6\mathbb{Z}$. Why is this true? This is because the sets $$ 5 + 6\mathbb{Z} = \{ 5 + 6n : n \in \mathbb{Z}\}$$ and $$ 11 + 6\mathbb{Z} = \{ 11 + 6m : m \in \mathbb{Z}\}$$ are the same set. In this situation, we sometimes say that $5$ and $11$ (as elements of $\mathbb{Z}$) represent the same coset in $\mathbb{Z}/6\mathbb{Z}$. Similarly $3$ and $9$ represent the same coset.

A complete set of coset representatives (i.e. a set of numbers in $\mathbb{Z}$ that represent every coset in $\mathbb{Z}/6\mathbb{Z}$) are the numbers $0, 1, 2, 3, 4, 5$. Stated differently, the elements of $\mathbb{Z}/6\mathbb{Z}$ are exactly $$ 0 + 6\mathbb{Z}, 1 + 6\mathbb{Z}, 2+ 6\mathbb{Z}, 3+ 6\mathbb{Z}, 4+ 6\mathbb{Z}, 5+ 6\mathbb{Z}.$$ Since writing "$+ 6\mathbb{Z}$" is annoying, some people denote the coset $a + 6\mathbb{Z}$ by $\overline{a}$, so you might find that some will say that the elements of $\mathbb{Z}/6\mathbb{Z}$ are exactly the cosets $$ \overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5}.$$ These are the same statement, just with different notation.

It is a good exercise to make sure that you understand what it means to add and multiply these cosets --- consider going through the proof that a quotient ring is a ring explicitly for this example.

davidlowryduda
  • 91,687
  • Thank you so much! Is there any way I could ask just a quick follow up? How would I then find the zero divisors in the quotient ring? – Lindsay May 14 '20 at 18:09
1

Consider in the set $Z$ of integers the following relation: you say that an integer $x$ is related to an integer $y$ if and only if $6$ divides their difference. You can check that this relation is reflevive, symmetric and transitive, hence an equivalence relation. Call $\mathbb{Z}/6\mathbb{Z}$ the quotient set of $\mathbb{Z}$ with respect to that equivalence. Hence, an element of the quotient is by definition an equivalence class of elements of $\mathbb{Z}$. For instance, the class of $0$ is the set of all integers $x$ such that $x-0$ is divisible by $6$, hence the class of all multiples of $6$; the class of $1$ will be the set of all integers such that $x-1$ is divisible by $6$, hence $1, 7, 13, 19, \ldots$. And so on. You can also check that the classes of $0, 1, 2, 3, 4$ and $5$ are all the distinct classes you can have, hence $\mathbb{Z}/6\mathbb{Z}$ has cardinality $6$,

Another, but equivalent way of thinking, is to consider the additive structure of $\mathbb{Z}$ as a group under addition. Then $6\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$ (every subgroup of an abelian group is obviously normal), hence you can form the quotient group whose elements are left (or right) cosets of $6\mathbb{Z}$. Of course, each coset of $6\mathbb{Z}$ corresponds to each distinct equivalence class of $\mathbb{Z}$ modulo the congruence equivalence relation we have considered above.

Maryam
  • 1,889