# Subrings and Subfields

1. Jan 21, 2009

### lilcoley23@ho

I'm taking an independent study class over Groups. Rings, and Fields. It's been really confusing. On one page I understand everything completly and the next page I'm completly lost. I'm looking at a problem where I has to show that {x+y(cuberoot of 3) + z(cuberoot of 9) | x,y,z is in Q} is a subring. Now I think I got that...but then I want to prove that it's a subfield. Since I prove that it's a non-empty subset and closed under addition and multiplication by showing that it's a subring, then all I further have to show is that it's a field. (Because to show something is a subfield you just have to show that it's a subset and a field.)

Then to show that the ring is a field you just have to prove it's closed under addition, subtraction, multiplication, division, and it must be cummutative....Right??? So in this case I just have to prove that it's closed under division and cummutative??

I know that my denominator will look like: x^3 + 3y^3 +9z^3 - 9xyz but I really need help!!

Thanks!!

Last edited: Jan 21, 2009
2. Jan 21, 2009

### NoMoreExams

To show something is a field you need to show field axioms. Properties like division for example don't always make sense when talking about abstract fields.

Have you looked at showing these properties hold:

http://mathworld.wolfram.com/FieldAxioms.html

3. Jan 21, 2009

### lilcoley23@ho

So if I can show that all 5 of those properties hold for x+y(cuberoot of 3) + z(cuberoot of 9) than I can prove that x+y(cuberoot of 3) + z(cuberoot of 9) | x,y,z is in Q} is a subfield?

If that's the case can you maybe help me with the first one? Like associativity?

I would really appreciate it. I think I'm making it harder than it has to be...and where does the denominator looking like: x^3 + 3y^3 +9z^3 - 9xyz come into play?

4. Jan 21, 2009

### NoMoreExams

Well you want associativity for both operations. You said you have addition and multiplication so for addition, you want to show

$$(a + b \sqrt[3]{3} + c \sqrt[3]{9} + d + e \sqrt[3]{3} + f \sqrt[3]{9}) + g + h \sqrt[3]{3} + i \sqrt[3]{9} = a + b \sqrt[3]{3} + c \sqrt[3]{9} + (d + e \sqrt[3]{3} + f \sqrt[3]{9} + g + h \sqrt[3]{3} + i \sqrt[3]{9})$$

Can you show that just because rationals are associative under addition?

5. Jan 21, 2009

### lilcoley23@ho

I'm not sure? I know that it's proven that addition and multiplication of rational numbers are associative and cummutative, but I don't know if thats all you have to show to prove it?

Also...are you telling me that if I prove associativity, commutativity, distributity, identity, and inverses for both addition and multiplication for this equation that I have proved that it's a subfield? I thought I had to prove so much more!!

6. Jan 21, 2009

### lilcoley23@ho

I just don't see that...to prove something is a subring it says you have to show that

1. S doesn't equal 0
2. S is closed under subtraction
3. S is closed under multiplication

More precisely:
1. S doesn't equal 0
2. For all a,b in S we have a-b in S and a,b in S

Then for subfield you must show that:
1. S - {0} doesn't equal 0
2. S is closed under subtraction
3. S is closed under multiplication
4. x is in S-{0} ==> x^(-1) is in S

7. Jan 21, 2009

### wsalem

"To show something is a field you need to show field axioms" is not the only way, in fact, it is usually much easier to show that something is a subfield of something.
lilcoley23@ho, your last post contains all you need. Now that you proved that A is a subring of B. You need to do two more things, show that 1!=0, and that the set A\{0} has multiplicative inverses!

p.s What text are you using?

8. Jan 21, 2009

### wsalem

There's a type in your last post though, in case you aren't aware of (but you probably are)
It is For all a,b in S we have a-b in S and a.b in S

9. Jan 21, 2009

### lilcoley23@ho

So to show S was a subring of (R , + , .) I used the more precise version I talked about in my last post:

So, = 0 is in S and so, S is a non empty subset of R

Therefore, S is a subring of R.

Now condition 2:

Let say, x1+y1(cuberoot 3) + z1(cuberoot 9), x2+y2(cuberoot 3) + z2(cuberoot 9) are in S

Then, x1, x2, y1, y2, z1, z2 are in Q

So, (x1-x2), (y1-y2), (z1-z2) are in Q and (x1-x2)+(y1-y2)(cuberoot 3) + (z1-z2)(cuberoot 9) are in S

Therefore, (x1+y1(cuberoot 3) + z1(cuberoot 9)) - (x2+y2(cuberoot 3) + z2(cuberoot 9)) are in S

AND

So: (x1+y1(cuberoot 3) + z1(cuberoot 9) * (x2+y2(cuberoot 3) + z2(cuberoot 9))

==> {x1x2 + 3y1z2 + 3y2z1} + {x1y2 + 3z1z2 + 3x2y1}(cuberoot 3) + {y1y2 + x1z2 + x2z1} (cuberoot 9)

Which shomws that if x1+y1(cuberoot 3) + z1(cuberoot 9), x2+y2(cuberoot 3) + z2(cuberoot 9) are in S

Therefore: (x1+y1(cuberoot 3) + z1(cuberoot 9)) * (x2+y2(cuberoot 3) + z2(cuberoot 9)) are in S

So is that right? And if it is can you please tell me what else I have to show to show it's a subfield...

I've never seen anything with "show that 1!=0, and that the set A\{0} has multiplicative inverses"

And about my book...I'm not sure...I use mostly on-line. Really none of the problems I'm trying to solve are like any in the book!!

Lilcoley23

10. Jan 21, 2009

### wsalem

Regarding "1!=0" and " the set A\{0} has multiplicative inverses", they'are just anyway of rephrasing the subfield test, so nevermind about them.

To show it's a subfield, assume that an element exists and it doesn't equal to zero, now show that its inverse also exists.

11. Jan 21, 2009

### wsalem

I think I haven't been quite explicit enough, sorry,

The last question about the existence of a multiplicative inverse is: is it possible to write $$\frac{1}{ x+y\sqrt[3]{3} + z\sqrt[3]{9} }$$ in the form $$a+b\sqrt[3]{3} + c\sqrt[3]{9}$$. What values of $$a$$, $$b$$, and $$c$$ (for each x, y, and z) do we need then.
Do they belong to $$\mathbf{Q}$$.

12. Jan 28, 2009

### LorenzoMath

$$f$$

13. Jan 28, 2009

### LorenzoMath

well, so you need to show Z[cuberoot of 3] is a field.

Proof 1.
Think of the map Q[X] to Z[cuberoot of 3] that maps X to "cuberoot of 3". This is obviously a surjective ring-homomorphism. The kernel is the ideal generated by an irreducible polynomial X^3-3. Thus we have an isomorphism Q[X]/(X^3-3) onto Z[cuberoot of 3]. Since (X^3-3) is a maximal ideal, Q[X]/(X^3-3) is a field, so Z[cuberoot of 3] is also a field. ////

14. Jan 28, 2009

### lilcoley23@ho

I'm sorry,

but I dont' understand why you're just looking at Z(cuberoot 3) instead of the whole equation? Is the x+y(cuberoot of 3) part of the equation irrelevant?

15. Jan 29, 2009

### LorenzoMath

I denoted a ring generated by "cuberoot of 3" over Z by Z[cuberoot of 3].

Let us denote the cube root of 3 by r for the sake of notation.

Then the set in question is the same as Z[a] because a^3=3.

By the way, in general, if K is a subfield of F and a is an element of F algebraic over K, F[a]=F(a). Here, F(a) denotes the smallest subfield of F that contains K and a.

HERE IS ANOTHER PROOF (basically the same as the one I gave yesterday).

Cuberoot of 3 is a root of an irredicuble polynomial X^3-3. Any nonzero element p of the ring in question is of a form f(cuberoot of 3), where f (=:f(X)) is a polynomial over Z. Since p is not zero, X^3-3 doesn't divide f(X). (If it does, f(p)=0.) As X^3-3 is irredicible, X^3-3 and f(X) are relatively prime. Hence by the Chinese remainder theorem there are nonzero polynomials h(X) and k(X). such that h(X)(X^3-3)+k(X)f(X)=1. Substitute "cuberoot of 3" for X. Then we obtain k(cuberoot of 3)p=1. (f(cuberoot of3)=p by def.) So p has an inverse. ////