# Prove $+$ and $\times$ are well-defined on quotient rings

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.

• @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. – 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:

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

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

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

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

Hint:

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