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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.


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

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he says multiplication is also commutative. – Will Jagy Nov 4 '12 at 3:36
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
up vote 8 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


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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

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