I am asked to prove that the $+$ and $\times$ operations are well-defined on quotient rings. Having little experience with similar questions I looked up the proofs online and found that most of them feature a step I don't entirely understand.

Normally, they would begin with something like this:

Let $r+I=r'+I$ and $s+I=s'+I$, then $r-r',s-s' \in I$.

I have run through several examples in my head and realise that this is correct, but I don't really understand where the general conclusion comes from, especially since $r-r'$ or $s-s'$ isn't necessarily $0$. I'm probably missing something extremely obvious, but being new to quotient rings I just don't see why $r-r' \in I$, and a similar argument applies to the $\times$ proof.


Let's do an example. Consider the ring $\mathbb{Z}[x]$ (polynomials in $x$ with integer coefficients) and the ideal $I=(x)$ (polynomials with $0$ constant term).

Let's take $2+(x)$ and $3 + (x)$ in the quotient ring $\mathbb{Z}[x]/(x)$. Now, their product is defined as $$(2+(x))(3+(x)) = 6 + (x).$$ But notice, $2+(x) = 2+x + (x)$, so we could've also multiplied as $$(2+x+(x))(3+(x)) = 6 + 3x + (x).$$ Now, notice that $6+3x+(x)$ and $6+(x)$ are equal, so BOTH multiplications give the SAME answer. This is what well-defined means.

Thus, to check that the multiplication is well-defined in general, you have to show that picking different representatives of an ideal $r+I$ gives the same multiplication/addition.

  • 2
    $\begingroup$ @Geneten48 Ideals are subgroups under addition. So $a, a' \in I$ implies $a-a' \in I,$ which implies your $r-r' \in I$. That's right. $\endgroup$ – Dzoooks Apr 24 '19 at 22:45

The key is that, for an ideal $I$ and element $r$ we simply have $$r+I=I\ \iff\ r\in I$$ Now apply this to $r=r_2-r_1$.


Suppose I have a ring $R$, it has $0, +, -, \times$ as required.

Let's additionally endow $R$ with an equivalence relation, $\simeq$ that is transitive, reflexive, and symmetric as required, but a priori unrelated to the structure of $R$ .

$(R, \simeq)$ is now a ring with an equivalence relation. Let's say that $(\simeq)$ respects the ring structure of $R$ if all of the following hold for all $a, b, c$ in $R$ .

$$ a \simeq b \implies (-a) \simeq (-b) \\ a \simeq b \implies c + a \simeq c + b \\ a \simeq b \implies a\times c \simeq b \times c \\ a \simeq b \implies c \times a \simeq c \times b $$

In other words, once you have two ring elements $a, b$ that are equivalent, if you can construct a probe out of constant ring elements and ring functions that send $a$ and $b$ to nonequivalent ring elements, then $\simeq$ does not respect the ring structure of $R$.

Suppose $\simeq$ respects the ring structure of $R$ .

$(R, \simeq)$ is not our quotient ring, but we can construct a new ring $R' = R/\simeq$ whose elements are the equivalence classes that $\simeq$ partitions $R$ into. We adorn each equivalence class a designated canonical element, denoted $x_c$ where $x$ is in $R'$ .

Now we need to define $0, +, -, \times$ in $R'$

$0$ is whatever class in $R'$ contains $0$.

$x + y = z \stackrel{\text{def}}{\iff} x_c + y_c \in z $

$-x = z \stackrel{\text{def}}{\iff} -(x_c) \in z $

$x \times y = z \stackrel{\text{def}}{\iff} x_c \times y_c \in z $ .

Because $\simeq$ respects the ring structure of $R$, it does not matter which elements of the equivalence classes we picked as canonical.

Ring Ideals give us an equivalence relation that can be used to construct a $\simeq$ that respects the ring structure.

In particular, let $I$ be a ring ideal and let $\simeq_I$ be the equivalence relation associated with the ideal $I$.

$$ x \simeq_I y \stackrel{\text{def}}{\iff} x - y \in I $$

A ring ideal is a subset of a ring $R$ satisfying two properties:

  • Strong multiplicative closure. $\forall r \in R \mathop. r \times I \subset I$ and $\forall r \in R \mathop. I \times r \subset I $ . In words, $I$ is closed under multiplication by any ring element $R$. (I have never heard anyone else call it strong multiplicative closure, but it's a decent mnemonic).
  • Additive closure $\forall w \in I \mathop. w + I \subset I $. $I$ is closed under addition. Equivalently, $(I,+)$ is an Abelian group.

So, if we check that $(R, \simeq_I)$ respects the ring structure of $R$, we know that $R/I$ is a bona fide ring.

Checking negation:

$$ a \simeq b \\ a - b \in I \\ -(a - b) \in I \\ -a - (-b) \in I \\ -a \simeq -b $$

Checking addition

$$ a \simeq b \\ a - b \in I \\ a + (c - c) - b \in I \\ (a + c) - (c + b) \in I \\ (a + c) - (b + c) \in I \\ a + c \simeq b + c $$

Checking left multiplication

$$ a \simeq b \\ a - b \in I \\ c \times (a - b) \in I \\ c \times a - c \times b \in I \\ c \times a \simeq c \times b $$

Checking right multiplication

$$ a \simeq b \\ a - b \in I \\ (a - b) \times c \in I \\ a \times c - b \times c \in I \\ a \times c \simeq b \times c $$



$r+I=r'+I$ means $r-r'\in I$. Similarly, $s-s'\in I$.

Now you have to show that, if $r-r'$ and $s-s'\in I$, then

  • $(r+s)-(r'+s')\in I$. Observe that $(r+s)-(r'+s')=(r-r')+(s-s')$.
  • $rs-r's'\in I$. Rewrite it as $\;rs-r's'=(r-r')s+r's-r's'=\dots$
  • $\begingroup$ I think you meant $r+I$ where you wrote $r=I$ at the beginning $\endgroup$ – J. W. Tanner Apr 24 '19 at 23:08
  • 1
    $\begingroup$ Oh! yes. It's fixed. Thanks for pointing it! $\endgroup$ – Bernard Apr 24 '19 at 23:38

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