Determine if integral domain or field

[Shackleford]

I found the unity to be 1. I'm not sure how to determine 3) for each.

Integral Domain:

1) D has to be commutative ring (under multiplication)
2) D has to have a unity not equal to zero (also under multiplication)
3) D has to have no zero divisors (ab = 0)

Field:

1) F has to be commutative ring (under multiplication)
2) F has to have a unity not equal to zero (also under multiplication)
3) Each nonzero element in F has to have a multiplicative inverse

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110805_224110.jpg?t=1312602377

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110805_224012.jpg?t=1312602392

 

[Dick]

The real numbers have no zero divisors, so it's a pretty safe bet that both are at least integral domains. To determine if there is an inverse, don't start from scratch. Multiply (a+sqrt(2)*b)*(a-sqrt(2)*b) and try to figure out where to go from there.

 

[Shackleford]

The real numbers have no zero divisors, so it's a pretty safe bet that both are at least integral domains. To determine if there is an inverse, don't start from scratch. Multiply (a+sqrt(2)*b)*(a-sqrt(2)*b) and try to figure out where to go from there.

Ah! Duh. I was focusing too much on the rationals a,b and forgetting that the term represents a real number.

So, there's no way to determine if there's an inverse if you start from scratch?

Why did you choose this? Just a good guess?

(a+sqrt(2)*b)*(a-sqrt(2)*b)

 

[Dick]

Ah! Duh. I was focusing too much on the rationals a,b and forgetting that the term represents a real number.

So, there's no way to determine if there's an inverse if you start from scratch?

Why did you choose this? Just a good guess?

(a+sqrt(2)*b)*(a-sqrt(2)*b)

Yes, good guess. It gets rid of the sqrt(2). Which you have to do if you are finally going to get just 1. Now go from there.

 

[Shackleford]

Yes, good guess. It gets rid of the sqrt(2). Which you have to do if you are finally going to get just 1. Now go from there.

(a+sqrt(2)*b)*(a-sqrt(2)*b) = 1

a² - 2b² = 1

a² = 1 + 2b²

If b = 1, then a² = 3

Then, a = sqrt(3), which is not a rational number. This is not a Field.

 

[Dick]

(a+sqrt(2)*b)*(a-sqrt(2)*b) = 1

a² - 2b² = 1

a² = 1 + 2b²

If b = 1, then a² = sqrt(3).

Then, a is not a rational number. This is not a Field.

Ok, my guess was really just a hint. a^2-2*b^2 is unlikely to be 1. Now think about (a+b*sqrt(2))*k*(a-b*sqrt(2))=1 where k is some constant in your system. Can you solve for k if you are working in the rationals?

 

[Shackleford]

Ok, my guess was really just a hint. a^2-2*b^2 is unlikely to be 1. Now think about (a+b*sqrt(2))*k*(a-b*sqrt(2))=1 where k is some constant in your system. Can you solve for k if you are working in the rationals?

Is my counterexample correct? Did I show it's not a Field?

k = [a² - 2b²]^-1

 

[Dick]

Is my counterexample correct? Did I show it's not a Field?

k = [a² - 2b²]^-1

Well, no. Using that value of k, I'd say the inverse of a+sqrt(2) is (a-sqrt(2))/(a^2-2), isn't it? Your main worry over the rationals is whether a^2-2*b^2 could be zero.

 

[Shackleford]

Well, no. Using that value of k, I'd say the inverse of a+sqrt(2) is (a-sqrt(2))/(a^2-2), isn't it? Your main worry over the rationals is whether a^2-2*b^2 could be zero.

Okay, you're right. There is a valid inverse if b = 1.

a² - 2b² = 0

a = sqrt(2)b

 

[HallsofIvy]

$\frac{1}{a+ b\sqrt{2}}$

Rationalise the denominator.

 

[Shackleford]

$\frac{1}{a+ b\sqrt{2}}$

Rationalise the denominator.

You can't?

 

[SammyS]

You can't?

I'm sure he can !

But the point is that you do it.

 

[Shackleford]

I'm sure he can !

But the point is that you do it.

No. I'm saying you can't rationalize the denominator because there will always be root two down there. Is that right?

 

[Dick]

I thought we were getting towards being able to agree that the inverse of (a+b*sqrt(2)) is (a-b*sqrt(2))/(a^2-2*b^2)? In so far as that is defined. Is it defined over the rationals? Let's not go backwards here.

 

[Shackleford]

I thought we were getting towards being able to agree that the inverse of (a+b*sqrt(2)) is (a-b*sqrt(2))/(a^2-2*b^2)? In so far as that is defined. Is it defined over the rationals? Let's not go backwards here.

Oh, sorry. I guess that's next logical conclusion.

Is the inverse defined over the rationals? Not if sqrt(2) is in there.

 

[HallsofIvy]

You are still missing the whole point! $(a+ b)(a- b)= a^2- b^2$. I'm sure you learned that long ago. So what do you get if you multiply both numerator and denominator of
$$\frac{1}{a+ b\sqrt{2}}$$
by
$$a- b\sqrt{2}$$

 

[Hurkyl]

I don't see why people in this thread have an aversion to "starting from scratch". Linear algebra is a very useful technique in abstract algebra, and both of the questions

• What, when multiplied by (a + b$\sqrt{2}$), gives 1?
• What, when multiplied by (a + b$\sqrt{2}$), gives 0?

are linear systems of equations in two unknowns. (treating a and b as "knowns")

Rationalizing denominators is a good trick, but far from required for this problem.

 

[Dick]

Oh, sorry. I guess that's next logical conclusion.

Is the inverse defined over the rationals? Not if sqrt(2) is in there.

What do you mean 'if sqrt(2) is in there'? Do you mean sqrt(2) is in the rationals? Can you elaborate a bit?

 

[Shackleford]

What do you mean 'if sqrt(2) is in there'? Do you mean sqrt(2) is in the rationals? Can you elaborate a bit?

Rationalizing is when you get rid of the square root in the denominator? I was wrong. You can rationalize the denominator with the difference of two squares trick.

[1/sqrt(2)] = sqrt(2)/2

 

[Shackleford]

I don't see why people in this thread have an aversion to "starting from scratch". Linear algebra is a very useful technique in abstract algebra, and both of the questions

• What, when multiplied by (a + b$\sqrt{2}$), gives 1?
• What, when multiplied by (a + b$\sqrt{2}$), gives 0?

are linear systems of equations in two unknowns. (treating a and b as "knowns")

Rationalizing denominators is a good trick, but far from required for this problem.

As established earlier, nothing non-zero multiplied by a real number yields zero. Thus, that system is not possible. The inverse is [a - b*sqrt(2)] / [a² - 2b²].

 

[HallsofIvy]

How can you be certain that $a- 2b^2$ is not 0 for some a and b?

 

[Shackleford]

How can you be certain that $a- 2b^2$ is not 0 for some a and b?

I corrected the typo. It should be a².Then a = sqrt(2)*b.