# Proof of closure under addition and multiplication in a field

1. Feb 24, 2008

### karnten07

1. The problem statement, all variables and given/known data

Does anyone know a generic way of showing that a field is closed under multiplication and addition? Please, thanks
2. Relevant equations

3. The attempt at a solution

Just need to prove that a+b and ab are in the field that each element a and b are from. Any ideas??

2. Feb 24, 2008

### Mystic998

There's not really a generic way to do it. It all depends on what set you're claiming is a field. Incidentally, calling it a field before you show it's closed under the operations is bad form.

3. Feb 24, 2008

### karnten07

The field im trying to show is a field is of the form Q$$\sqrt{}$$d where Q is the set of rational numbers and d is in the set of complex numbers, C.

4. Feb 24, 2008

### Hurkyl

Staff Emeritus
Yes -- the definition of "field" mandates that it is closed under multiplication and addition.

I suspect you meant to ask something else -- judging by your wording, did you mean to ask about showing whether a subset of a field (with the induced arithmetic operations) is a subfield?

5. Feb 25, 2008

### Mystic998

Okay, so what you want to do is take arbitrary elements of the field, namely elements of the form $a + b\sqrt{d}$ and $c + e\sqrt{d}$, with $a, b, c, e \in \mathbb{Q}$, and show that you get something of that form back when you multiply or add them.

6. Feb 25, 2008

### HallsofIvy

Staff Emeritus
As Mystic988 said in the first post, how you show a set if closed under operations (and so is a field) depends on how the set and operations are defined. How you would show that "a field" is of the form Q($\sqrt{d}$) where d is a complex number depends upon exactly how "a field" is defined!

Exactly how is your field defined? Mystic988's suggestion is assuming you already know d and want to show that the set of numbers of the form a+ b$\sqrt{d}$ is a field. That is quite correct if d is an integer but if, for example, d is a trancendental number, it is not at all correct.

Again, exactly what is the problem you are working on?

7. Feb 25, 2008

### karnten07

It's okay, i did the assignment now, i apologise for my poor wording of the question. But i understand how to do the question now, thanks guys

8. Feb 25, 2008

### Mystic998

Oh, yeah, I was assuming we were talking about quadratic rationals (I think that's the right name. My brain is not functioning fully at this very moment), not more exotic field extensions. Sorry about that.

9. Feb 25, 2008

### HallsofIvy

Staff Emeritus
I started to answer the question in exactly the same way!