MHB Group of units - Rotman - page 36 - Proposition 1.52

[Math Amateur]

I am reading Joseph Rotman's book Advanced Modern Algebra.

I need help in fully understanding the proof of Proposition 1.52 on page 36.

Proposition 1.52 and its proof reads as follows:

View attachment 2676
The part of the proof on which I need help/clarification is Rotman's argument where he establishes that each $$ r \in U( \mathbb{I}_m ) $$ has an inverse in $$ U( \mathbb{I}_m ) $$.

As can be seen in the text above Rotman's argument (which I must say confuses me) reads as follows:

"If $$ (a,m) = 1 $$ then $$ [a][x] = 1 $$ can be solved for x in $$ \mathbb{I}_m $$. Now (x,m) = 1 for rx + sm = 1 for some integer s, and so (x,m) = 1. Hence $$ [x] \in U( \mathbb{I}_m ) $$, and so each $$ r \in U( \mathbb{I}_m ) $$ has an inverse in $$ U( \mathbb{I}_m ) $$."

I confess I cannot follow Rotman's argument above! (maybe MHB members will find it clearer than I do?)

Can someone please provide a clear and rigorous restatement of Rotman's argument regarding inverses in $$ U( \mathbb{I}_m ) $$.

Particular points of confusion are as follows:

1. Rotman writes: "If $$ (a,m) = 1 $$ then $$ [a][x] = 1 $$ can be solved for x in $$ \mathbb{I}_m $$." - I cannot see exactly why this follows:

2. Rotman writes: "Now (x,m) = 1 for rx + sm = 1 for some integer s, and so (x,m) = 1." - I do not follow this statement at all: indeed I think it may be badly expressed whatever he means ... ...

3. I do not follow the rest of his statements - probably because I do not follow 1 and 2 above.

Help and clarification would be appreciated

Peter

Note that Rotman uses the symbol $$\mathbb{I}_m$$ for $$ \mathbb{Z}/ m \mathbb{Z} $$.

 

[Deveno]

Ok, what Rotman is saying is:

$[k] \in \Bbb Z_m$ is a unit if and only if $\text{gcd}(k,m) = 1$.

$(\impliedby)$: Suppose $\text{gcd}(k,m) = 1$. Using the (extended) Euclidean algorithm for finding the gcd, we can find integers $a,b$ such that:

$ak + bm = 1$ (Bezout's Identity).

If we reduce this equation mod $m$, we get:

$[a][k] + [0] = [a][k] + [0] = [a][k] = [1]$.

Evidently, then, $[a]$ is an inverse for $[k]$ in $\Bbb Z_m$.

$(\implies)$: On the other hand, suppose $[k]$ is a unit. This means that for some $ \in \Bbb Z_m$ we have:

$[k] = [1]$.

This means that:

$(u + am)(k + bm) = 1 + cm$, for some integers $a,b,c$. It then follows that:

$uk + (ak + bu + abm - c)m = 1$, which shows that $\text{gcd}(k,m) = 1$.

************

I will show you how we actually DO this, for a specific $k$ and $m$. Suppose $m = 24$ and $k = 7$. We want to show $[7]$ is a unit. To do so, we need to find its inverse. Note that $\text{gcd}(7,24) = 1$, so we ought to be able to do this.

What we want to do, is find $a,b$ so that $7a + 24b = 1$. So this is what we do:

24 = 3*7 + 3
7 = 3*2 + 1

therefore:

1 = 7 - 6 = 7 - 3*2 = 7 - (24 - 3*7)*2 = 7 - 2*24 + 6*7 = 7*7 + (-2)*24

so $a = 7$ and $b = -2$. We don't need $b$, we just want $a$, and since $0 \leq 7< 24$, we don't even need to reduce.

And indeed: $[7]*[7] = [49] = [48] + [1] = [2*24] + [1] = [2][24] + [1] = [2][0] + [1] = [0] + [1] = [1]$.

Now Rotman shows closure of $U(\Bbb Z_m)$ by showing if $\text{gcd}(k,m) = 1$ and $\text{gcd}(k',m) = 1$, then:

$\text{gcd}(kk',m) = 1$.

but I find it easier to note that if:

$[a][k] = [1]$ and $[a'][k'] = [1]$, then: $[aa'][k][k'] = [aa'kk'] = [(ak)(a'k')] = [ak][a'k'] = ([a][k])([a'][k']) = [1][1] = [1]$

so that evidently $[aa']$ is an inverse for $[k][k']$, so the product of two units is again a unit.

Again, let's use $m = 24$ as an example. Clearly $[5]$ is also a unit. Note that $[5]$ is its own inverse:

$[5][5] = [25] = [1] + [24] = [1] + [0] = [1]$.

What we have done above says that $[5][7] = [35] = [11]$ is also a unit, with inverse:

$[5*7] = [35] = [11]$.

We verify:

$[11][11] = [121] = [1] + [120] = [1] + [5][24] = [1] + [5][0] = [1] + [0] = [1]$.

 

[Math Amateur]

Ok, what Rotman is saying is:

$[k] \in \Bbb Z_m$ is a unit if and only if $\text{gcd}(k,m) = 1$.

$(\impliedby)$: Suppose $\text{gcd}(k,m) = 1$. Using the (extended) Euclidean algorithm for finding the gcd, we can find integers $a,b$ such that:

$ak + bm = 1$ (Bezout's Identity).

If we reduce this equation mod $m$, we get:

$[a][k] + [0] = [a][k] + [0] = [a][k] = [1]$.

Evidently, then, $[a]$ is an inverse for $[k]$ in $\Bbb Z_m$.

$(\implies)$: On the other hand, suppose $[k]$ is a unit. This means that for some $ \in \Bbb Z_m$ we have:

$[k] = [1]$.

This means that:

$(u + am)(k + bm) = 1 + cm$, for some integers $a,b,c$. It then follows that:

$uk + (ak + bu + abm - c)m = 1$, which shows that $\text{gcd}(k,m) = 1$.

************

I will show you how we actually DO this, for a specific $k$ and $m$. Suppose $m = 24$ and $k = 7$. We want to show $[7]$ is a unit. To do so, we need to find its inverse. Note that $\text{gcd}(7,24) = 1$, so we ought to be able to do this.

What we want to do, is find $a,b$ so that $7a + 24b = 1$. So this is what we do:

24 = 3*7 + 3
7 = 3*2 + 1

therefore:

1 = 7 - 6 = 7 - 3*2 = 7 - (24 - 3*7)*2 = 7 - 2*24 + 6*7 = 7*7 + (-2)*24

so $a = 7$ and $b = -2$. We don't need $b$, we just want $a$, and since $0 \leq 7< 24$, we don't even need to reduce.

And indeed: $[7]*[7] = [49] = [48] + [1] = [2*24] + [1] = [2][24] + [1] = [2][0] + [1] = [0] + [1] = [1]$.

Now Rotman shows closure of $U(\Bbb Z_m)$ by showing if $\text{gcd}(k,m) = 1$ and $\text{gcd}(k',m) = 1$, then:

$\text{gcd}(kk',m) = 1$.

but I find it easier to note that if:

$[a][k] = [1]$ and $[a'][k'] = [1]$, then: $[aa'][k][k'] = [aa'kk'] = [(ak)(a'k')] = [ak][a'k'] = ([a][k])([a'][k']) = [1][1] = [1]$

so that evidently $[aa']$ is an inverse for $[k][k']$, so the product of two units is again a unit.

Again, let's use $m = 24$ as an example. Clearly $[5]$ is also a unit. Note that $[5]$ is its own inverse:

$[5][5] = [25] = [1] + [24] = [1] + [0] = [1]$.

What we have done above says that $[5][7] = [35] = [11]$ is also a unit, with inverse:

$[5*7] = [35] = [11]$.

We verify:

$[11][11] = [121] = [1] + [120] = [1] + [5][24] = [1] + [5][0] = [1] + [0] = [1]$.

Thanks Deveno ... just skimmed through your post and am now working through it in detail to ensure I fully understand what you have said ...

From my brief read, your post is extremely clear and helpful (especially since it includes a concrete example!) ... ... so if you ever write a book, then please let me know ... it would likely be so much more easy to follow than the extant texts ...

Thanks again,

Peter