0

Context: I am new to this. I started my course yesterday.

I know that the operations $+$ and $\cdot$ are required to satisfy the field axioms. So how can $F=\{0\}$ be a field? Recall the additive and multiplicative identity: there exist 2 different elements $0$ and $1$ in $F$ such that $ a + 0 = a$ and $a \cdot 1 = a$

But there is only one element in $F$? Shouldn't that mean that this rule has not been satisfied and thus $F$ is not a field?

My teacher mentioned something about $0=1$ can someone explain that?

Update my lecture notes:

enter image description here

CountDOOKU
  • 1,071
  • Your teacher is partially correct, although instead of a field I think your teacher implies an Abelian group. An Abelian group with one element $G={e}$, has an element that acts as both the additive identity, (traditionally notated as $0$) and the multiplicative identity ($1$).For all elements, $a\in G$, $e$ satisfies the multiplicative identity $a*e = e$, and the additive identity $a+e = a$. – Graviton Feb 26 '21 at 06:30
  • 2
    Related: https://math.stackexchange.com/questions/427078/is-0-a-field?rq=1 – Xander Henderson Feb 26 '21 at 06:31
  • 6
    If that's what your teacher said, they are incorrect. "Field" universally means $0 \neq 1$. It's much like the fact that 1 is not considered a prime number--we'd constantly be adding silly assumptions if we allowed ${0}$ to be a field. – Joshua P. Swanson Feb 26 '21 at 06:32
  • 1
    Agreed. If you want ${0}$ to be something, it can be a ring. – 311411 Feb 26 '21 at 06:34
  • https://en.wikipedia.org/wiki/Field_with_one_element – Servaes Feb 26 '21 at 06:45
  • 3
    {$0$} is not a field. Every common definition of a field assumes $0\ne 1$. And even if we allow this field it has no interesting properties whatsoever. – Peter Feb 26 '21 at 08:52
  • @Servaes I think just posting that link without comment confuses matters. The relevant point is that the "field with one element" is not actually a field. – xxxxxxxxx Feb 26 '21 at 10:28
  • @JoshuaP.Swanson Hi Mr Swason, thanks for commenting. I updated my lecture notes. He did say that $0≠1$ – CountDOOKU Feb 26 '21 at 10:28

1 Answers1

8

Under the usual axioms $\{0\}$ is not a field. E.g., if $F$ is s field and $0$ its additive neutral then $F\setminus\{0\}$ is a group under multiplication - but groups cannot be empty. Put differently, one often takes $0\ne 1$ as one of the axioms of a field.

Martin Sleziak
  • 56,060
Hagen von Eitzen
  • 382,479