Question from self-studying Halmos' Finite Dimensional Vector Spaces For section 1 on Fields, there is a question 2c:
2.
a) Is the set of all positive integers a field?
b) What about the set of all integers?
c) Can the answers to both these question be changed by re-defining addition or multiplication (or both)?
My answer initially to 2c was that for a), as long as there no longer was the addition axiom dealing with (-a) + (a) = 0 and the multiplicative axiom dealing with reciprocals, it would be a field. For b) it would be a field if the reciprocal axiom did not exist. 
However, I looked at a suggested answer here (https://drexel28.wordpress.com/2010/09/21/halmos-chaper-one-section-1-fields/)
and completely did not understand the lemma:
I don't understand why the things are in parentheses, or what the things represent. I don't understand what card F = n means, and why in the sentence, plus and multiplication signs are in circles. I don't understand what is meant by F x F --> F. I don't understand cardinalities, or what legitimate binary operations are. I don't understand the use of inverses either. 
In the section on Vector Space Examples, Halmos writes that, "Let P be set of all polynomials with complex coefficients, in a variable t. To make P into a complex vector space, we interpret vector addition  and scalar multiplication as the ordinary addition of two polynomials and the multiplication of a polynomial by a complex number; the origin in P is the polynomial identically zero.


*

*What does it mean for something to be identically zero? 


What else should I read along with Halmos to get a good understanding of Linear Algebra and Abstract Algebra? I'm also reading Spence Freidberg Insel, Artin, Fraleigh.
 A: You point out correctly the field axioms that $\Bbb{N}$ and $\Bbb{Z}$ do not satisfy. Just a (maybe nitpicky) note on wording: it's better to say that they are not fields because they do not satisfy the field axioms than to say they would be fields if the definition of fields were different.
Now onto the link with the answer: 
The first thing in parenthesis is $(\mathfrak{F}, \cdot, +)$; this is a common way of specifying which operations, $\cdot$ and $+,$ make your set $\mathfrak{F}$ a field. This distinction is important if you're studying different operations on the same set.
The cardinality of $F$, card($F$), is the number of elements of $F$. The link says card($F)=n$, but note it does not assume it is finite. We say two sets have the same cardinality if there exists a bijection (a function that is both one-to-one and onto) between them. The sets $\Bbb{Z}, \Bbb{N},$ and $\Bbb{Q}$ are all infinite, and even though $\Bbb{N} \subsetneq \Bbb{Z} \subsetneq \Bbb{Q}$, the three have the same cardinality: they are all countable sets. This is why the lemma proved applies to them. 
I think the plus and times signs are in circles to distinguish them from the initial operations on $\mathfrak{F}$, but this may be a convention that I'm not aware of. To understand the proof, you don't need to worry about that; all you need to work with is the definition of the operations. 
The polynomial $f(x)$ that is identically zero is just the zero polynomial: $f(x)=0$ for any $x$.
About the books: I learned abstract algebra from Artin and linear algebra from Friedberg and Axler. 
EDIT: 
For the definition of binary operation, see here: http://mathworld.wolfram.com/BinaryOperation.html
What the whole solution does is use the operations given in $(\mathfrak{F}, \cdot, +)$ to define new operations $\oplus$ and $\otimes$ on a set $F$ that has the same cardinality as $\mathfrak{F}$. Since they have the same cardinality, there is a bijection $\theta$ from $F$ to $\mathfrak{F}$, and this bijection has an inverse $\theta^{-1}$ from $\mathfrak{F}$ to $F$. 
A: I just want to answer one of your questions. If we have a map, say g, such that $g:\mathbb{F}\times \mathbb{F}\rightarrow \mathbb{F}$ this means that g is a function with domain in $\mathbb{F} \times \mathbb{F}$ with its codomain in $\mathbb{F}$. Here, $\mathbb{F}$ means the field.
In other words, if $\mathbb{F} = \mathbb{R}$, a function that maps $\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is of the form:$$z=f(x,y)$$ where a point $(x,y) \in \mathbb{R} \times \mathbb{R}$ and the image point $z \in \mathbb{R}$.
