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Given that K is an arbitrary field, is it true that 1 ∈ K, and hence 0 ∈ K?

K. Claesson
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  • By definition of a field, there is a $0$ and a $1$ element, typically called the additive and multiplicative identity. – B. Mehta Nov 22 '17 at 16:32

1 Answers1

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A field $K$ will always have an additive identity - called "$0_K$" - and a multiplicative identity - called "$1_K$." (And it's worth noting that one of the field axioms asserts that these are distinct.) So in this sense the answer to your question is "yes."

However, note the subscripts in my previous paragraph! There is not a single thing "$0$" which is an element of every field - one field's additive identity need not be in a different field. Each field has its own additive and multiplicative identity, but asking "Is $0\in K$?" falls prey to a subtle confusion.

We do often write "$0, 1$" instead of "$0_K, 1_K$" when there's only one field being talked about, or when all the fields being talked about share the same additive and multiplicative identities (e.g. the fields $\mathbb{Q},\mathbb{R},\mathbb{C}$ all "overlap"$^1$). But this is technically an abuse of notation.

$^1$Well, as usually understood. But this is a case where it's valuable to ignore the set theorists (of which I am one).

Noah Schweber
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  • This is a bit of a tangent to the original question, but based on your answer, does one need to prove that 1 ∈ K in besides proving that x + y ∈ K, x - y ∈ K, x * y ∈ K, and x/y ∈ K, where x,y ∈ K, in order to show that the set K is a field. – K. Claesson Nov 22 '17 at 18:35