TITLE: Help understanding fields.
QUESTION [9 upvotes]: Hi guys I have a test this tuesday and I am given practice questions to do , and I have trouble understanding fields. Like I know by definition what they are, but applying them is kind of confusing. this is the question I have.
1) Which of the following algebraic structures (with the usual operations) are fields? Explain briefly.
a) Integers 
b) Rational numbers 
c) Real numbers 
d) $2 \times 2$ matrix 
e) Integers Modulo 2 
f) Integer modulo 4 
g) Real numbers modulo $2\pi$
2) Is $\{ a+b2^{\frac{1}{4}} | a,b \in \mathbb{Q} \}$ a subfield of real numbers? Explain.
My attempt:
1) What I know for sure is that {rational numbers, real numbers, complex numbers} are field and {$2 \times 2$ matrix} is not a field.
I am not sure if a, f, g, h are fields or not.
I know for it to be a field all the operations have to be valid. I want to say that a and g are field but I am not sure please help out on that one . I know {$2 \times 2$ matrix} is not a field because the division does not apply. 
If someone could explain that to me it would be really helpful. 
and for 2)
To show this I did addition and subtraction first so let $a,b,c,d \in \mathbb{Q}$ so let $x=a+b2^{\frac{1}{4}}$ and $y=c+d2^{\frac{1}{4}}$ so $x + y = a+b2^{\frac{1}{4}} + c+d2^{\frac{1}{4}}$ and that implies that $(a+c) + (b+d)2^{\frac{1}{4}}$ which is good. and same goes for subtraction that $x-y = (a-c) +(b-d)2^{\frac{1}{4}}$
and now for multipication and division this is where I got confused. I have a feeling its not a subfield but its not working out for multipication I am getting an extra variable what I did for multipication was if
$x \times y = (a+b2^{\frac{1}{4}}) \cdot (c+d2^{\frac{1}{4}})$ that implies when you simplify $ac + (ad+bc)2^{\frac{1}{4}} + bd\sqrt{2}$ and thats not in our property. so i dont know please help out.
Any help or hints would be greatly appreciated.
Thanks a lot guyss

REPLY [3 votes]: For these questions, I might advise going down this list.

What are the two operations? Is the ring closed under both operations?
Are both operations commutative?
Does each operation have an identity?
Does each element have an inverse under each operation?

Here, it seems every case has a 'clear' pair of operations with identity. But not every pair is commutative (matrices). Not every pair is closed (hint). And not operation has an inverse (hint).
In the comments, you have seen that the integers are not a field because the element $3$ does not have a multiplicative inverse in the ring, for example.
In a sense, I've espoused the 'every nonzero element is a unit' idea of a field. But you could also show that something is not a field by finding zero-divisors, i.e. a pair of nonzero numbers whose product is zero. It's not always good to do so, but sometimes it's clear that such a thing exists. (Recall that zero-divisors can't be units)