Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

On pp.2-3 of Curtis, Abstract Linear Algebra, he gives a definition of a field which seems to fail to exclude a pathological example. He says a field is a set k with two operations (a+b) and (ab) such that:

  • $k$ is an abelian group with +

  • $k - {0}$ (where 0 is the additive identity of the above group) is an abelian group over multiplication

  • for all a,b,c, $a(b + c) = ab + ac$.

It is easy to prove, using distributivity, that for any $a$, $a0 = 0$. But proving that $0a = 0$ seems to require knowing that $(b + c)a = ba + ca$. However, proving that this is the case seems to require knowing that $0a = 0$.

To be clear, if for nonzero $a, b$, $0a = b$, we get something quite pathological. $0b = 0(0a) = (00)a = 0a = b$, so $01 = 0(bb^{-1}) = (0b)b^{-1} = bb^{-1} = 1$. Now for any nonzero a, $a = a1 = a(01) = (a0)1 = 01 = 1$, so the only possible nonzero element is 1. But there's no contradiction here that I see.

Define:

0+0 = 0 = 1+1
0+1 = 1 = 1+0

00 = 0 = 10
11 = 1 = 01

Addition clearly gets you an abelian group. k - {0} is the trivial group, and distributivity is satisfied:

0(0+0) = 00 = 0 = 0 + 0 = 00 + 00
0(0+1) = 01 = 0 + 01 = 00 + 01
0(1+1) = 00 = 0 = 1 + 1 = 01 + 01

1(0+0) = 10 = 0 = 0 + 0 = 10 + 10
1(0+1) = 11 = 1 = 0 + 1 = 10 + 11
1(1+1) = 10 = 0 = 1 + 1 = 11 + 11

I see the same definition for field on p. 34 of Dummit and Foote, 3ed. So have I made a mistake? Or is this definition lacking an additional axiom to exclude out the above case:

  • for any a, b, c, $(a + b)c = ac + bc$

EDIT, Will Jagy: Corrected in Errata for the third edition of Dummit and Foote, see ERRATA PDF

share|improve this question
5  
he says multiplication is also commutative. –  Will Jagy Nov 4 '12 at 3:36
5  
The elements of F - {0} are specified to be commutative. He says nothing about multiplication by zero. –  nham Nov 4 '12 at 3:39
    
I realize it's a totally trivial point. Fields are supposed to be commutative rings, so 0a != 0 makes that not work out. But I just wanted to make sure that I'm not insane and that there actually is a gap in the field axioms presented here. –  nham Nov 4 '12 at 3:50
    
I edited the errata from the Dummit and Foote website into your question, just a link to their pdf. –  Will Jagy Nov 4 '12 at 4:37
    
I read earlier on the same page in Herstein, he gives two distributive laws, exactly as you suggest. See jpegs in my answer. –  Will Jagy Nov 4 '12 at 18:48

1 Answer 1

up vote 7 down vote accepted

EDIT Sunday Morning: I did not give Herstein, or frexids, enough credit. Herstein simply defines a (associative) ring as having both left and right distributive laws. So the later comment about a field being a ring with $1$ such that the nonzero elements form an abelian group is entirely appropriate.

=-=-=-=-=-=-=-=-=-=-=

enter image description here

=-=-=-=-=-=-=-=-=-=-=

I do still own some books that define a field. I'm a bit surprised. All three, essentially, including the linear algebra one, essentially define a commutative ring first, then say that a field is a commutative ring with identity and multiplicative inverses. So it appears that you have found a loophole. Actually Nering simply gives all the axioms in one page.

So, my summary would be that, if a context had not been created before the field definition, there is a problem with the wording in your book. It seems possible that the language is a lazy copy from Herstein, who was a genuine expert. However, example 3.1.3 in the ancient edition I have reads:

$R$ is the set of rational numbers under the usual addition and multiplication of rational numbers. $R$ is a commutative ring with unit element. But even more than that, note that the elements of $R$ different from $0$ form an abelian group under multiplication. A ring with this latter property is called a field.

As you have pointed out, this last line is lazy. Maybe people copied from this, the book is certainly influential. Then, in chapter 5, he correctly defines a field as a commutative ring with unit element and multiplicative inverses for nonzero elements.

In sum, I would not add a second distributive law, I would just demand that multiplication be commutative for all pairs of elements.

I had never heard of Dummit and Foote, so I searched online. I found a page of errata, here is the first page, which corrects page 34 in the third edition:

=-=-=-=-=-=-=

enter image description here

=-=-=-=-=-=-=

share|improve this answer
    
It's interesting that Dummit and Foote also makes the same mistake on page 34. –  Rankeya Nov 4 '12 at 4:11
    
@Rankeya, see the first page of their errata, which catches page 34. I'm guessing this is about fields. –  Will Jagy Nov 4 '12 at 4:27
    
Thank you for pointing this. –  Rankeya Nov 4 '12 at 6:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.