Does every infinite field contain the integers as a subring?

Question

I simply ask because if $1+1=2(1)=2$ then this would imply that all positive integers are contained, and as every element in a field has a negative all the negative integers are contained. At the same time every "reciprocal" would be contained as every element has an inverse. Which leads me to believe at least some section of the rational numbers (excluding integers) is also contained in every infinite field.

How close am I to correct, as surely this would also imply that the integers are a subring of every infinite ring as well?

EDIT: Okay I think the qualifier here is if it has characteristic $0$.

Answer 1

Your reasoning (maybe formalised a little bit further) shows that every (infinite) field of characteristic $0$ contains (a isomorphic copy of) the ring of integers, hence also (a isomorphic copy of) the field of rational numbers (which is the field of fractions of the ring of integers). As pointed out in the comments, a field of characteristic $p$ contains a isomorphic copy of $\mathbb{Z}/p\mathbb{Z}$.

Answer 2

No. Any copy of $\mathbb{Z}$ must be the ring generated by $1$, but in any infinite field of prime characteristic $p$, $$\underbrace{1 + \cdots + 1}_p = 0,$$ and so the ring generated by $1$ is not isomorphic to the integers.

Edit Under the condition that the field have characteristic $0$, this is true, essentially because in such a field the above equation does not hold for any (nonzero) $p$.