MHB Any two fields with 4 elements are isomorphic

[Math Amateur]

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

I need help with Problem 2.34 on page 101.

Problem 2.34 reads as follows:

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Prove that any two fields having exactly four elements are isomorphic.

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Can someone please help me get started on this problem.

Peter

 

[Deveno]

There is actually a more general version of this theorem for any two finite fields of the same cardinality. But we'll stick to this baby version.

We can denote any such field as $F = \{0,1,a,b\}$.

A small detour:

The mapping $\sigma:F \to F$ given by:

$0 \mapsto 0$
$1 \mapsto 1$
$a \mapsto b$
$b \mapsto a$

is a field automorphism.

Since this is clearly bijective, we only need to show it is a ring homomorphism.

To do this, we need to know what the addition and multiplication tables for $F$ actually are. Sums involving 0 are clear, so we need not investigate those.

A small detour to our detour:

Claim: 1 + 1 = 0.

Since $(F,+)$ is an abelian group, we must have either:

1 + 1 = 0, or 1 + 1 + 1 + 1 = 0. Suppose it is the latter. Then $1 + 1 = a$ (or $b$, but since we haven't specified anything about $a$ or $b$, we'll just say it's $a$). It is then clear that $1 + 1 + 1 = a + 1 = b$.

Now by the distributive law (since we're in a field):

$a\cdot a = (1 + 1)(1 + 1) = (1)(1 + 1) + (1)(1 + 1) = (1 + 1) + (1 + 1) = 1 + 1 + 1 + 1 = 0$.

Thus $a$ (which is distinct from 0) is a zero divisor, which cannot happen in a field. This proves our claim.

Claim 2: $x + x = 0$, for all $x \in F$.

Proof: $x + x = x(1) + x(1) = x(1 + 1) = x(0) = 0$ (the proof that $x(0) = 0$ is easy. Do it).

So the additive group $(F,+)$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, under addition modulo 2. This completely determines the addition: for example: $a+1$ cannot be 0 ($a$ is its own additive inverse), it cannot be 1 (since that implies $a = 0$), and it cannot be $a$ (since that implies $0 = 1$ which cannot happen in a field), so $a + 1 = b$.

Since this is "one less letter", we will henceforth write $a+1$ instead of $b$.

Which leads us to multiplication. The products involving 0 or 1 are trivial to define, and I will not bother. That leaves (because of commutativity of the product), just 3 products to investigate:

$a\cdot a$
$a \cdot (a+1)$
$(a+1)\cdot(a+1)$

By the distributive law, once we know $a \cdot a$, we'll know the rest. Now since $(F-\{0\},\cdot)$ forms an abelian group (because we have a field), and this group has 3 elements, it is clear that it is a cyclic group of order 3 (since all groups of order 3 are, since 3 is a prime), so any non-identity element is a generator. This means that:

$a\cdot a \neq 1$, because $a$ must be of (multiplicative) order 3, not order 2. If:

$a \cdot a = a$, then multiplying by $a^{-1}$ yields $a = 1$, which we know isn't true. Thus $a \cdot a = a + 1$.

Now it should be clear that the map $\sigma:F \to F$ above yields a GROUP automorphism of $(F,+)$ (we're just "switching two of the generators of the Klein 4-group"). So we have a FIELD automorphism if and only if $\sigma$ is a group automorphism of the multiplicative group of $F$. But on $F-\{0\}$, the map $\sigma$ is just the inversion map:

$x \mapsto x^{-1}$ (can you see this?)

which is a homomorphism because the multiplicative group is abelian.

Ok, so if we have a second field $F' = \{0',1',a',a'+1\}$, since any ring-homomorphism $\phi$ must take:

$0 \to 0'$
$1 \to 1'$ we either have:

$\phi(a) = a'$
$\phi(a+1) = a'+1$

which is clearly a ring (and thus field) isomorphism, or:

$\phi(a) = a'+1$
$\phi(a+1) = a'$

in which case $\phi \circ \sigma$ is clearly a field isomorphism and thus so is:

$\phi = (\phi \circ \sigma) \circ \sigma^{-1}$

(the composition of two isomorphisms is an isomorphism).

 

[Math Amateur]

There is actually a more general version of this theorem for any two finite fields of the same cardinality. But we'll stick to this baby version.

We can denote any such field as $F = \{0,1,a,b\}$.

A small detour:

The mapping $\sigma:F \to F$ given by:

$0 \mapsto 0$
$1 \mapsto 1$
$a \mapsto b$
$b \mapsto a$

is a field automorphism.

Since this is clearly bijective, we only need to show it is a ring homomorphism.

To do this, we need to know what the addition and multiplication tables for $F$ actually are. Sums involving 0 are clear, so we need not investigate those.

A small detour to our detour:

Claim: 1 + 1 = 0.

Since $(F,+)$ is an abelian group, we must have either:

1 + 1 = 0, or 1 + 1 + 1 + 1 = 0. Suppose it is the latter. Then $1 + 1 = a$ (or $b$, but since we haven't specified anything about $a$ or $b$, we'll just say it's $a$). It is then clear that $1 + 1 + 1 = a + 1 = b$.

Now by the distributive law (since we're in a field):

$a\cdot a = (1 + 1)(1 + 1) = (1)(1 + 1) + (1)(1 + 1) = (1 + 1) + (1 + 1) = 1 + 1 + 1 + 1 = 0$.

Thus $a$ (which is distinct from 0) is a zero divisor, which cannot happen in a field. This proves our claim.

Claim 2: $x + x = 0$, for all $x \in F$.

Proof: $x + x = x(1) + x(1) = x(1 + 1) = x(0) = 0$ (the proof that $x(0) = 0$ is easy. Do it).

So the additive group $(F,+)$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, under addition modulo 2. This completely determines the addition: for example: $a+1$ cannot be 0 ($a$ is its own additive inverse), it cannot be 1 (since that implies $a = 0$), and it cannot be $a$ (since that implies $0 = 1$ which cannot happen in a field), so $a + 1 = b$.

Since this is "one less letter", we will henceforth write $a+1$ instead of $b$.

Which leads us to multiplication. The products involving 0 or 1 are trivial to define, and I will not bother. That leaves (because of commutativity of the product), just 3 products to investigate:

$a\cdot a$
$a \cdot (a+1)$
$(a+1)\cdot(a+1)$

By the distributive law, once we know $a \cdot a$, we'll know the rest. Now since $(F-\{0\},\cdot)$ forms an abelian group (because we have a field), and this group has 3 elements, it is clear that it is a cyclic group of order 3 (since all groups of order 3 are, since 3 is a prime), so any non-identity element is a generator. This means that:

$a\cdot a \neq 1$, because $a$ must be of (multiplicative) order 3, not order 2. If:

$a \cdot a = a$, then multiplying by $a^{-1}$ yields $a = 1$, which we know isn't true. Thus $a \cdot a = a + 1$.

Now it should be clear that the map $\sigma:F \to F$ above yields a GROUP automorphism of $(F,+)$ (we're just "switching two of the generators of the Klein 4-group"). So we have a FIELD automorphism if and only if $\sigma$ is a group automorphism of the multiplicative group of $F$. But on $F-\{0\}$, the map $\sigma$ is just the inversion map:

$x \mapsto x^{-1}$ (can you see this?)

which is a homomorphism because the multiplicative group is abelian.

Ok, so if we have a second field $F' = \{0',1',a',a'+1\}$, since any ring-homomorphism $\phi$ must take:

$0 \to 0'$
$1 \to 1'$ we either have:

$\phi(a) = a'$
$\phi(a+1) = a'+1$

which is clearly a ring (and thus field) isomorphism, or:

$\phi(a) = a'+1$
$\phi(a+1) = a'$

in which case $\phi \circ \sigma$ is clearly a field isomorphism and thus so is:

$\phi = (\phi \circ \sigma) \circ \sigma^{-1}$

(the composition of two isomorphisms is an isomorphism).

Thanks for the help Deveno ...

Now working through your post in detail ...

Thanks again,

Peter

Just a simple preliminary questions:

You write:

"Since $(F,+)$ is an abelian group, we must have either:

1 + 1 = 0, or 1 + 1 + 1 + 1 = 0."

Why/how does this follow?

Peter

 

[Deveno]

One thing I omitted, because it was not part of the problem, was proving a field of four elements actually EXISTS.

This is not as obvious as it sounds: for example, no field with 6 elements exists.

However, this is easily remedied:

$F = \Bbb Z_2[x]/\langle x^2 + x + 1\rangle$

is easily seen to be a field, as $x^2 + x + 1$ is irreducible over $\Bbb Z_2$ (having no roots in $\Bbb Z_2$, and thus no linear factors).

It is also easily seen that we have precisely 4 distinct cosets:

$0 + \langle x^2 + x + 1\rangle$
$1 + \langle x^2 + x + 1\rangle$
$x + \langle x^2 + x + 1\rangle$
$x+1 + \langle x^2 + x + 1\rangle$

and using the usual definitions of coset addition and multiplication it is easily seen that this gives a field isomorphic to those in my prior post.

Alternatively, one could verify that the field axioms are satisfied for $F = \{0,1,a,a+1\}$ used in my first post, but this is way too much tedium for me to stand.

 

[Math Amateur]

Thanks for the help Deveno ...

Now working through your post in detail ...

Thanks again,

Peter

Just a simple preliminary questions:

You write:

"Since $(F,+)$ is an abelian group, we must have either:

1 + 1 = 0, or 1 + 1 + 1 + 1 = 0."

Why/how does this follow?

Peter

Hi Deveno,

I am still working on the question above, namely:

You write:

"Since $(F,+)$ is an abelian group, we must have either:

1 + 1 = 0, or 1 + 1 + 1 + 1 = 0."

Why/how does this follow?

I have been reading on finite fields and abelian groups - just skimming the topics for usable results - some near misses but nothing directly on target ...

However I did find the following text from a discussion of a field of 4 elements on math.stackexchange:

"The additive group of F is an abelian group with 4 elements. The order of 1 in this group divides 4, so it is either 2 or 4."

So if the order of 1 is 2 then 1 + 1 = 0 and if the order of 1 is 4 then 1 + 1 + 1 + 1 = 0 as you said ... ...

But then the question becomes: why can we claim that "The order of 1 in this group divides 4" - why does this follow from (F, +) being an abelian group with 4 elements?

Can someone please help?

Peter

PS I think I need to thoroughly revise the theory of finite abelian groups!***EDIT***

A little further reading has made me aware that if g is an element of a finite group G then (as a consequence of Lagrange's Theorem) the order of g divides the order of G - so then the order of 1 divides the order of F which is the result I wanted ... ...

So the preliminary step that was worrying me did not depend on the group involved being abelian ... ... oh well, all's well that ends well ... ...

Can someone please confirm that my thinking/analysis above is correct?

Peter

 

[Deveno]

Yes, an abelian group is a group, so Lagrange's theorem applies.

Since the (additive) order of 1, which is equal to the order of the (additive) subgroup generated by 1 is a divisor of 4, and 1 is a non-identity element (1 $\neq$ 0), it must have order 2 or 4.

********

Alternatively, by the structure theorem on finitely generated abelian groups, we have:

$(F,+) \cong \Bbb Z_4$ or $(F,+) \cong \Bbb Z_2 \times \Bbb Z_2$, since 2 is the only prime dividing 4 and the only multiplicative decompositions are either 2*2 or 4 (we would have some more choices in an abelian group of order 8: 2*2*2,2*4, or 8).