Is Every Non-Zero Element in a Finite Ring Invertible?

[mathmari]

Hey!

Let $R$ be a finite non-trivial ring.
We suppose that for each $r,s\in R$ with $rs=0$ then either $r=0$ or $s=0$.
I want to show that $R$ is a division ring.

Could you give me a hint how we could show that each element $x\in R\setminus \{0\}$ has an inverse? (Wondering)

 

[Deveno]

Consider the mapping, for $a \in R -\{0_R\}$:

$L_a: R -\{0_R\} \to R$, given by $L_a(r) = ar$.

Note that we actually have $\text{im }L_a \subseteq R -\{0_R\}$, since if $ar = 0_R$, by our definition of $R$, we have $r = 0_R$, a contradiction.

So, is $L_a$ injective? Let's investigate:

Suppose $ar = as$, for $r,s \neq 0_R$

Then $ar - as = 0_R$, so that: $a(r - s) = 0_R$. Since $a \neq 0_R$, we have $r - s = 0_R$, that is, $r = s$. So $L_a$ is injective.

Now consider the map $R_a$ which sends $r \mapsto ra$. Use the injectivity of these two maps (and the finiteness of $R$!), to show that $R - \{0_R\}$ is a group (this is a standard exercise of group theory). If you have trouble ,let me know.

(hint #1: the finiteness of $R$ implies both maps are also surjective. That means there is some $r \neq 0_R$ in $R$ such that $ar = a$. Try to find a clever way to show what $rb$ might be, for any non-zero $b$).

 

[mathmari]

(hint #1: the finiteness of $R$ implies both maps are also surjective. That means there is some $r \neq 0_R$ in $R$ such that $ar = a$. Try to find a clever way to show what $rb$ might be, for any non-zero $b$).

Do we show in that way that the ring contains the identity element? (Wondering)

 

[Deveno]

Yes-we will need that to show $R$ has units, right?

EDIT: you will need to show that $R$ has a TWO-SIDED identity, since we cannot assume $R$ is commutative. You will then need to show each non-zero element of $R$ has a TWO-SIDED inverse. You will need to invoke associativity of multiplication several times.

 

[mathmari]

(hint #1: the finiteness of $R$ implies both maps are also surjective.

How does the surjectivity follow from the finiteness of $R$ ? (Wondering)

 

[Deveno]

Do you know the Pigeonhole Principle?

Suppose $S$ is a finite set, with $f:S \to S$ injective. If $s \in S$ has no pre-image in $S$ under $f$, then we have $|S|$ domain elements, with at most $|S| - 1$ images to send them to. So some element must have more than one pre-image, contradiction.

 

[mathmari]

Do you know the Pigeonhole Principle?

Suppose $S$ is a finite set, with $f:S \to S$ injective. If $s \in S$ has no pre-image in $S$ under $f$, then we have $|S|$ domain elements, with at most $|S| - 1$ images to send them to. So some element must have more than one pre-image, contradiction.

So, we consider the mappings $L_a:R\setminus \{0\}\rightarrow R\setminus\{0\}$ and $R_a:R\setminus \{0\}\rightarrow R\setminus\{0\}$ and not the mappings $L_a:R\setminus \{0\}\rightarrow R$ and $R_a:R\setminus \{0\}\rightarrow R$, right? (Wondering)

 

[Deveno]

So, we consider the mappings $L_a:R\setminus \{0\}\rightarrow R\setminus\{0\}$ and $R_a:R\setminus \{0\}\rightarrow R\setminus\{0\}$ and not the mappings $L_a:R\setminus \{0\}\rightarrow R$ and $R_a:R\setminus \{0\}\rightarrow R$, right? (Wondering)

Yes, the surjectivity is something we need.

 

[mathmari]

So, since $L_a$ and $R_a$ are surjective, we have that for each $a\in R\setminus\{0\}$ there is a $r\in R\setminus\{0\}$ such that $L_a(R)=a$ and $R_a(r)=a$, i.e., $ar=a$ and $ra=a$.
Therefore, $R$ has a unity, right? (Wondering)

 

[Deveno]

Your argument is much like the story:

"Once upon a time, they lived happily ever after."

It starts out right, it ends right, but something seems to be missing in the middle.

We have no reason to expect that for EACH $a \in R\setminus\{0\}$ it is "the same $r$".

But we know that for a SPECIFIC $a$, there is some $r$ (let's call it $r_a$ for now, to indicate it might depend on $a$) such that:

$ar_a = a$.

Let us investigate $r_ab$. Here is what we know:

$ab = (ar_a)b = a(r_ab)$.

This means $a(b - r_ab) = 0$.

By our conditions on $R$, since $a \neq 0$, we have $b - r_ab = 0$, so that $b = r_ab$.

This is true for *every* $b \in R\setminus\{0\}$.

So $r_a$ is a left-identity for $R\setminus\{0\}$.

A similar argument (using $R_a$) shows there is some non-zero element of $R$ such that:

$s_aa = a$. Taking any non-zero $b$, we have:

$bs_aa = ba$, and thus $(b_sa - b)a = 0$, and so $bs_a = b$.

So now we have a left-identity ($r_a$) and a right-identity ($s_a$).

Thus:

$r_a = r_as_a$ (since $s_a$ is a right-identity), and:

$r_as_a = s_a$ (since $r_a$ is a left-identity).

So $r_a = s_a$, and this is an identity for all of $R\setminus\{0\}$. Let's call it $u_a$, now.

In the above, $a$ is FIXED, and $b$ is arbitrary. The question remains: if we do this for "some other fixed element" (say $x$) of $R\setminus\{0\}$, do we have $u_x = u_a$? Yes, and this is easy to prove:

$u_a = u_au_x = u_x$.

So, yes, $R\setminus\{0\}$ has a multiplicative identity.

Now, on to inverses...

 

[mathmari]

So $r_a = s_a$, and this is an identity for all of $R\setminus\{0\}$. Let's call it $u_a$, now.

In the above, $a$ is FIXED, and $b$ is arbitrary. The question remains: if we do this for "some other fixed element" (say $x$) of $R\setminus\{0\}$, do we have $u_x = u_a$? Yes, and this is easy to prove:

$u_a = u_au_x = u_x$.

So, yes, $R\setminus\{0\}$ has a multiplicative identity.

So, $u_a$ (resp. $u_x$) is the identity element for every element but with fixed $a$ (resp. $x$), right? (Wondering)

Now, on to inverses...

Let $a\in R\setminus\{0\}$.
We have that $L_a$ is surjectiv, so there is a $r\in R\setminus\{0\}$ such that $L_a(r)=u_a \Rightarrow ar=u_a$.
We also have that $R_a$ is surjectiv, so there is a $\tilde{r}\in R\setminus\{0\}$ such that $R_a(\tilde{r})=u_a \Rightarrow \tilde{r}a=u_a$.
So, we have that $ar=u_a=\tilde{r}a$.
It is left to show that $r=\tilde{r}$, right? (Wondering)

We have that $ar=u_a=\tilde{r}a \Rightarrow u_ar=\tilde{r}ar \Rightarrow r=\tilde{r}u_a \Rightarrow r=\tilde{r}$.
Is this correct? (Wondering)

 

[Deveno]

So, $u_a$ (resp. $u_x$) is the identity element for every element but with fixed $a$ (resp. $x$), right? (Wondering)

The last part of my prior post show the identity is independent of $a$, so we can just call it $u$, or $1$.

Let $a\in R\setminus\{0\}$.
We have that $L_a$ is surjectiv, so there is a $r\in R\setminus\{0\}$ such that $L_a(r)=u_a \Rightarrow ar=u_a$.
We also have that $R_a$ is surjectiv, so there is a $\tilde{r}\in R\setminus\{0\}$ such that $R_a(\tilde{r})=u_a \Rightarrow \tilde{r}a=u_a$.
So, we have that $ar=u_a=\tilde{r}a$.
It is left to show that $r=\tilde{r}$, right? (Wondering)

Yes, note that for inverses, $r$ and $\tilde{r}$ *will* depend on $a$.

We have that $ar=u_a=\tilde{r}a \Rightarrow u_ar=\tilde{r}ar \Rightarrow r=\tilde{r}u_a \Rightarrow r=\tilde{r}$.
Is this correct? (Wondering)

Yes, the trick is to put "$a$ in the middle":

$r = ur = (\tilde{r}a)r = \tilde{r}(ar) = \tilde{r}u = \tilde{r}$

Note how I wrote it to emphasize where associativity is used.

 

[mathmari]

The last part of my prior post show the identity is independent of $a$, so we can just call it $u$, or $1$.

Yes, note that for inverses, $r$ and $\tilde{r}$ *will* depend on $a$.
Yes, the trick is to put "$a$ in the middle":

$r = ur = (\tilde{r}a)r = \tilde{r}(ar) = \tilde{r}u = \tilde{r}$

Note how I wrote it to emphasize where associativity is used.

Ah ok...

So, we have that $ar=u=ra$.

Therefore, each non-zero element has an inverse, i.e., $R$ is a division ring, right? (Wondering)

 

[Deveno]

Yep. So what can you say about finite integral domains?

 

[mathmari]

Yep. So what can you say about finite integral domains?

We can say that each finite integral domain is a field, right? (Wondering)

 

[Deveno]

Exactly!

 

[mathmari]

I want to show also that each integral domain with finitely many ideals is a field.

Could you give me a hint how we could show that? (Wondering)

 

[Deveno]

Here is a hint:

If for every $a \neq 0 \in R$ we have $(a) = R$, then every such $a$ is a unit, and thus $R$ is a commutative division ring-that is, a field.

So, we may assume without loss of generality that there is some non-zero $a$ with $(a) \neq R$.

Consider the ideals:

$(a^k) : k \in \Bbb Z^+$.

There are infinitely many $k$, but only finitely many ideals of $R$, by assumption.

Another approach:

show that for $a \neq 0$ that $I \mapsto aI$ is an injective map on the finite set of all ideals of $R$.

 

[mathmari]

If for every $a \neq 0 \in R$ we have $(a) = R$, then every such $a$ is a unit, and thus $R$ is a commutative division ring-that is, a field.

Is $(a)$ a principal ideal? (Wondering)

So, we may assume without loss of generality that there is some non-zero $a$ with $(a) \neq R$.

Consider the ideals:

$(a^k) : k \in \Bbb Z^+$.

There are infinitely many $k$, but only finitely many ideals of $R$, by assumption.

So, it must hold that $(a^i)=(a^j)$, for $i\neq j$, or not? (Wondering)

Then we have the following:
$a^i\in (a^i)=(a^j)\Rightarrow a^i=(a^j)^nb\Rightarrow a^i=a^{nj}b$, for $n\in \mathbb{N}$ and for some $b$.
So, we have that $a^{nj}b-a^i=0 \Rightarrow a^i(a^{nj-i}b-1)=0$.
Since $R$ is an integral domain, there are no zero divisors.
So, it must be $a^{nj-i}b-1=0 \Rightarrow a^{nj-i}b=1$.
Does this mean that $a$ is invertible? (Wondering)

Another approach:

show that for $a \neq 0$ that $I \mapsto aI$ is an injective map on the finite set of all ideals of $R$.

If $I, J$ are ideals of $R$ with $aI=aJ$, then there are $i\in I$ and $j\in J$ such that $ai=aj \Rightarrow ai-aj=0 \Rightarrow a(i-j)=0$.
Since $R$ is an integral domain, there are no zero divisors.
So, it must be $i=j\in J$.
Therefore, we have that $I\subseteq J$.
We also have that $j=i\in I$, implies that $J\subseteq I$.
Therefore, $I=J$.

So, $I \mapsto aI$ is an injective map.

Is this correct? (Wondering)

Do we have to show that this map is also surjective? (Wondering)

 

[Deveno]

Is $(a)$ a principal ideal? (Wondering)

Yes, the principal ideal generated by $a$.

So, it must hold that $(a^i)=(a^j)$, for $i\neq j$, or not? (Wondering)

Then we have the following:
$a^i\in (a^i)=(a^j)\Rightarrow a^i=(a^j)^nb\Rightarrow a^i=a^{nj}b$, for $n\in \mathbb{N}$ and for some $b$.
So, we have that $a^{nj}b-a^i=0 \Rightarrow a^i(a^{nj-i}b-1)=0$.
Since $R$ is an integral domain, there are no zero divisors.
So, it must be $a^{nj-i}b-1=0 \Rightarrow a^{nj-i}b=1$.
Does this mean that $a$ is invertible? (Wondering)

Yes, with inverse $a^{nj - i - 1}b$.

If $I, J$ are ideals of $R$ with $aI=aJ$, then there are $i\in I$ and $j\in J$ such that $ai=aj \Rightarrow ai-aj=0 \Rightarrow a(i-j)=0$.
Since $R$ is an integral domain, there are no zero divisors.
So, it must be $i=j\in J$.
Therefore, we have that $I\subseteq J$.
We also have that $j=i\in I$, implies that $J\subseteq I$.
Therefore, $I=J$.

So, $I \mapsto aI$ is an injective map.

Is this correct? (Wondering)

Do we have to show that this map is also surjective? (Wondering)

No, because it is an injective map from a finite set to itself.

 

[mathmari]

Yes, the principal ideal generated by $a$.

So, we take cases if there is an element that generates the ring or not, so if the ring is principal or not, right? (Wondering)

No, because it is an injective map from a finite set to itself.

So, since it is an injective map from a finite set to itself, we have that the map is also surjective.
So, there is a $i\in I$ such that $ai=1$. Therefore, $a$ has an inverse.

Is this correct? (Wondering)

 

[Deveno]

So, we take cases if there is an element that generates the ring or not, so if the ring is principal or not, right? (Wondering)

If for *every* $a\neq 0 \in R$ we have $(a) = R$ then every non-zero $a$ is a unit, and $R$ is a field. We don't need the ring to be *cyclic* (generated by one element), but if it is, it's a field (there are cyclic rings that aren't fields, but none of them are integral domains).

But, of course, this usually isn't the case-we might have a non-cyclic integral domain (these do occur). We don't care if $R$ is principal or not, we just want to find *some* $a \neq 0$ such that $(a) \neq R$. We then consider the chain:

$(a) \supseteq (a^2) \supseteq (a^3) \supseteq (a^4) \cdots$

This descending chain must stabilize, since we only have finitely many ideals to choose from.

The descending chain condition listed above is called the Artinian property. So, as a bonus, you will have proven: any Artinian integral domain is a field.

So, since it is an injective map from a finite set to itself, we have that the map is also surjective.
So, there is a $i\in I$ such that $ai=1$. Therefore, $a$ has an inverse.

Is this correct? (Wondering)

Since $R$ is an ideal of $R$, it follows that there is some ideal $I$ such that $aI = R$. And then what you wrote follows, since $1 \in R$.

 

[mathmari]

Since $R$ is an ideal of $R$, it follows that there is some ideal $I$ such that $aI = R$. And then what you wrote follows, since $1 \in R$.

We have that $R$ is an integral domain. Why is $R$ is an ideal of $R$ ? (Wondering)

Is it because an integral domain is commutative, and then for all $a,r\in R$ we have that $ar\in R$ and $ra\in R$ ? (Wondering)

 

[Deveno]

We have that $R$ is an integral domain. Why is $R$ is an ideal of $R$ ? (Wondering)

Is it because an integral domain is commutative, and then for all $a,r\in R$ we have that $ar\in R$ and $ra\in R$ ? (Wondering)

For ANY ring $R$, we have $R$ is an ideal of $R$ (it is the kernel of the 0-morphism that sends everything to 0).