1

I'm trying to figure out which of the axioms (binary operation [so associativity and distributivity], additive/multiplicative inverses, commutativity, and the existence of an additive/multiplicative identity) fail to be satisfied for $\mathbb{R}/2\mathbb{Z}$ to not be a field. I realize that $2\mathbb{Z}$ is not an ideal of $\mathbb{R}$ and that furthermore my group is not even an integral domain (e.g., $\sqrt{2} \cdot \sqrt{2} = 0$), so it cannot possibly be a field. I just can't seem to determine what property of fields (in terms of the axioms) that it does not satisfy. To be clear, I am using the notation $\mathbb{R}/2\mathbb{Z}$ to refer to the set of cosets of reals such that two reals are in the same coset if their difference is an even integer. Where have I gone wrong with this?

rigged
  • 131
  • 2
    A field needs an operation of multiplication. – Angina Seng Apr 22 '20 at 18:52
  • Is there something wrong with the usual multiplication and then using the coset that the product belongs to? – rigged Apr 22 '20 at 18:53
  • To form a quotient ring, you need to mod out by an ideal.$;$The set $2\mathbb{Z}$ is not an ideal of $\mathbb{R}$. – quasi Apr 22 '20 at 18:53
  • 3
    $\Bbb R/2\Bbb Z$ is certainly a group, but since $2\Bbb Z$ is not an ideal in $\Bbb R$ the multiplication in $\Bbb R$ does not define a multiplication in $\Bbb R/2\Bbb Z$ which, thus, is not a ring. – Andrea Mori Apr 22 '20 at 18:54
  • For example, $0 + 2 \mathbb Z = 2 + 2 \mathbb Z$, but we have $0 \cdot \frac 1 2 + 2 \mathbb Z = 2 \mathbb Z \neq 2 \cdot \frac 1 2 + 2 \mathbb Z = 1 + 2 \mathbb Z$. – cxx Apr 22 '20 at 18:56
  • Thank you, that clarifies the situation sufficiently. – rigged Apr 22 '20 at 18:58
  • See also Quotients of rings not necessarily by ideals – Bill Dubuque Apr 22 '20 at 19:01
  • 2
    If you visit the link above, be aware that the currently accepted answer is apparently incorrect, claiming that the denominator of the quotient doesn't even have to be an additive group to make a quotient ring. – rschwieb Apr 22 '20 at 19:14
  • Note that your actual question, in the post, is not the question you raise in the title. You agree it cannot be a field because it has zero divisors, your question lies elsewhere... – Arturo Magidin Apr 22 '20 at 20:17
  • @rschwieb: The accepted answer there is being disingenuous; it is defining the equivalence relation by partitioning using translates of the coset $1+{0}$ instead of the ideal ${0}$, and then claiming that he is taking a quotient modulo something that is not an ideal. It would be like saying you can mod out a group by something that is not a subgroup if you take a normal subgroup $N$, an element $g\notin N$, and then claim you are “moding out by $gN$” by taking the partition induced by translates of $gN$. – Arturo Magidin Apr 22 '20 at 20:39
  • @ArturoMagidin Yup. ${}{}{}{}{}{}$ – rschwieb Apr 22 '20 at 20:42

2 Answers2

1

You already have a problem showing $\mathbb R/2\mathbb Z$ is even a ring.

The operation used on cosets for quotient rings ($(x+I)(y+I):=xy+I$) does not produce a well-defined multiplication on your cosets.

You need the thing in the bottom of the quotient to be an ideal of $\mathbb R$ for the operation to be well-defined (and there are only two ideals, $\{0\}$ and $\mathbb R$ itself.)

rschwieb
  • 153,510
  • As I stated in the question, I'm aware of the fact that $2\mathbb{Z}$ is not an ideal, since $\mathbb{R}$ is a field so has only trivial ideals. Where does the operation fail to be well-defined? (Edit: Another commenter provided an explicit counterexample so I will accept this answer). – rigged Apr 22 '20 at 18:57
  • What about a similar construction (which is also not a field) of multiplication on $[0, 2)$: $a \circ b = a \cdot b - \left \lfloor{\frac{a \cdot b}{2}}\right \rfloor$, where $\circ$ denotes multiplication in this new sense and $\cdot$ denotes multiplication in $\mathbb{R}$, so $1.5 \circ 1.5 = 0.25$. Obviously this still contains zero divisors, but why is multiplication not well-defined here? – rigged Apr 22 '20 at 20:24
  • @rigged Is addition supposed to still be regular addition in $\mathbb R$? Or what? – rschwieb Apr 22 '20 at 20:26
  • Ah, forgot to clarify but it should be defined an analogous way so the set remains closed: $a \bigoplus b = a + b - 2\left(\left \lfloor{\frac{a + b}{2}}\right \rfloor\right)$. Also edited the first one since I made a mistake. The subtracted part in the multiplication sense should also be multiplied by 2, so they represent the intuitive sense of "mod 2". – rigged Apr 22 '20 at 20:27
  • 1
    @rigged: It’s not a ring. You have $\frac{1}{2}(1+1) = \frac{1}{2}(0) = 0$, but $\frac{1}{2}(1) + \frac{1}{2}(1) = \frac{1}{2}+\frac{1}{2} = 1\neq 0$. So your multiplication does not distribute over your addition. (The addition is fine, it’s just the addition of the group $\mathbb{R}/2\mathbb{Z}$). – Arturo Magidin Apr 22 '20 at 20:54
  • @rigged: (There is no “well-definedness” issue in this multiplication: each element has a unique name, and the [corrected] definition sends pairs of elements of the domain into the domain; it defines an operation, just one that is not compatible with the addition to make a ring) – Arturo Magidin Apr 22 '20 at 22:56
1

Multiplication of cosets via representatives is not well-defined.

For instance, note that $0+2\mathbb{Z} =2+2\mathbb{Z}$. But if you try to define $$(a+2\mathbb{Z})(b+2\mathbb{Z}) = ab+2\mathbb{Z}$$ then the answer depends on which representative you use: $$\begin{align*} \left(\frac{1}{2}+2\mathbb{Z}\right)(0+2\mathbb{Z}) &= 0+2\mathbb{Z}\\ \left(\frac{1}{2} + 2 \mathbb{Z}\right)(2+2\mathbb{Z}) &= 1+2\mathbb{Z} \end{align*}$$ but $0+2\mathbb{Z} \neq1+2\mathbb{Z}$.

So the multiplication operation on cosets via representatives is not well defined; you don’t have a multiplication on cosets (at least, not an obvious one, and not one inherited from multiplication in $\mathbb{R}$.

Just as in the case of groups where we can define an operation on cosets via representatives if and only if the subgroup is normal, in any ring $R$, if $T$ is a subring of $R$, then we can define multiplication of cosets in $R/T$ using representatives if and only if $T$ is an ideal.

Theorem. Let $R$ be a ring, and let $T$ be a subring. The operation on cosets $R/T$ given by $$(r+T)(s+T) = rs+T$$ is well defined if and only if $T$ is a two-sided ideal of $R$.

Proof. The standard proof shows that if $T$ is a two-sided ideal, then the multiplication is well defined.

Conversely, assume the multiplication is well defined, and let $a\in T$, $r\in R$. Since $a+T=0+T$, we have that, because multiplication is well defined, $$\begin{align*} 0+T = (r+T)(0+T) &= (r+T)(a+T) = ra+T\\ 0+T = (0+T)(r+T) &= (a+T)(r+T) = ar+T\\ \end{align*}$$ This means $ra,ar\in T$.

Thus, for $a\in T$ and $r\in R$, $ar,ra\in T$. This proves $T$ is a two-sided ideal. $\Box$

Arturo Magidin
  • 398,050
  • Thank you, I should have recalled exactly why the group to quotient by must be a two-sided ideal (it was a homework problem for me to prove this a while back, even), and I see the parallel to the requirement that the subgroup to quotient by must be normal in the case of quotient groups. – rigged Apr 22 '20 at 20:50