# Show the following sets are rings

• October 3rd 2011, 02:46 PM
ehpoc
Show the following sets are rings
I am asked to show that the following sets are rings. And determine if they commutative.

A set that is a ring...... if you didn't know is (I am sure you did though)...

Quote:

1. Closure under addition. For all a, b in R, the result of the operation a + b is also in R.c[›] 2. Associativity of addition. For all a, b, c in R, the equation (a + b) + c = a + (b + c) holds. 3. Existence of additive identity. There exists an element 0 in R, such that for all elements a in R, the equation 0 + a = a + 0 = a holds. 4. Existence of additive inverse. For each a in R, there exists an element b in R such that a + b = b + a = 0 5. Commutativity of addition. For all a, b in R, the equation a + b = b + a holds.
So I am asked to "show" the following sets of rings. Furthermore my prof said ...
You have to show the rules hold, for *all* numbers in
the set.

So I have to show that the rule holds for all values in the set for...

1. Z[x], the set of all polynomials in x with integer coeeficients.

So i know already that all these rules hold for all integer values. And multiplying and dividing
any polynomial with integer coefficients together will render another polynomial with integer coefficients. So how do I show this obvious fact for all value in x?

I also have to show for...

2.The set of all matrices with rational entries

And this last one

3.Q(√2) = { a + b√2 | a, b are rationals }

Does this previous one mean the set of all rational numbers a,b such that (a+b√2)

Also for number 3 I must show that it forms a field so i must show that there is a multiplicative inverse for every non zero entry. I understand what this means, but really have no clue how to show it. :S

Any help in any amount would be so appreciated!!!!

EDIT: Another related question

Quote:

6. Make a Venn diagram illustrating the case Ac∩Bc∩C = Ac∩B∩Cc = ∅, B ≠ C, and B ∩ C ≠ ∅.
What do these small C's mean?
• October 4th 2011, 02:42 AM
Deveno
Re: Show the following sets are rings
1. the first 5 conditons you have listed amount to showing (R,+,0) is an abelian group. there are some important rules you did not list:

(R,*) is an associative semi-group:

a*b (often writtten just ab) must be in R (closure of multiplication)

(a*b)*c = a*(b*c) (associativity of multplication)
*******
other important rules a ring must satisfy:

a*(b+c) = a*b + a*c (distributivity of left-multiplication)

(a+b)*c = a*c + b*c (distributivity of right multiplication). we have to check these separately because it may not be the case that a*b = b*a (such rings are commutative).

if a ring IS commutative, checking one distributive axiom will suffice.

a ring may also possess a multiplicative element, called unity, and wriiten 1, as a multiplicative identity (a*1 = 1*a for all a in R).

if a commutative ring with unity possesses multiplicative inverses for all non-zero a in R, it is a field

(this can be said more succintly by stating R is a ring with (R-{0},*,1) an abelian group).

now for Z[x], pick 3 "arbitrary polynomials":

$p(x) = a_0 + a_1x +...+a_mx^m, q(x) = b_0 + b_1x +...+b_nx^n$ and
$r(x) = c_0 + c_1x +...+c_sx^s$

and show the axioms hold. you may find it helpful to make all 3 polynomials the same degree by setting t = max({deg(p),deg(q),deg(r)}), and using coefficients of 0, for powers of x between m and t, n and t, and s and t.

a caveat: since you are dealing with integer coefficients, "dividing" is something you should stay away from.

3. $\mathbb{Q}(\sqrt{2})$ is the set of all real numbers of the form a+b√2, where a,b are rational. addition will be easy....multiplication is messy, but not too hard. to find an inverse, look at what (a + b√2)(a - b√2) is. if this is rational (and non-zero), say it's q, then what would (1/q)(a + b√2)(a - b√2) be?

be sure to state your final answer only in terms of a and b.

for 6: the small c's mean: "complement". if your "universal set" is U, then Ac = U - A = {everything in U, except the stuff in A}. for example, if U = {1,2,3,4,5}, and A = {2,4}, then Ac = {1,3,5}.

in a venn diagram, if A is drawn as a circle (colored in), then Ac would be everything colored in outside the circle, and the circle A left blank (clear).
• October 4th 2011, 11:49 AM
ehpoc
Re: Show the following sets are rings
Quote:

and show the axioms hold.
This is where my problem is. How do I show that all the rules ?

Do I have to go through each and every rule and apply that rule to my three polynomials?

That is going to be a lot of wasted paper LOL.

Can I use a polynomials of length 2?

EDIT:

Using the polynomials you listed to prove closuer under multiplication I did the following

p(x).q(x)=(a0*b0)+(a0*b1)x+(a0*b2)x^2+(a1*b0)x+(a1 *b1)x^2+(a1*b2)x^3+(a2*b0)x^2+(a2*b1)x^3+(a2*b2)x^ 4

Clearly since is in the set since all polynomials with integer coeffcients multiply to create only other polynomials with integer coefficients.

Is this a correct proof for showing closure of multiplication of polynomials with integer coefficients?
• October 4th 2011, 11:09 PM
Deveno
Re: Show the following sets are rings
no, because not all polynomials are of degree 2. you have to prove it for polynomials in Z[x] of any degree. here is a hint that will save some paper:

$p(x) = \sum_{i = 0}^m a_ix^i$

so, for example, you can write:

$p(x)q(x) = \sum_{k = 0}^{m+n}(\sum_{i+j = k} a_ib_j)x^k$

which is fairly compact notation.

proofs that "such and such" is a ring can be tedious to write out. even with the best of notation, you will burn several pages. for Z[x], where possible, try to reduce a property to that same property for Z. same thing for Q(√2), try to reduce an axiom to that same axiom for Q.
• October 5th 2011, 06:31 AM
ehpoc
Re: Show the following sets are rings
Ughh I am so crappy with summations I can't remember the last time I used them. Definitely not in Calculus or linear Algebra.

As for your summation, I am thinking about it as nested for loops from computer programming :P

So k=0 and the loop runs until k=m+n? What is m and n?

And for every time the first loop runs the inner loop runs but I do not see an upper bound.....

What do so when k =1 i+j=k? I am lost once I get to the second summation.

Can you give me an example of how you would prove the same for a 2x2 matrices? I don't want you to do my work for me, but I just really want to see it so I can break it down and learn it.

Also how do you prove associativity of addition?

For commutativity I just added p(x) and q(x) together and stated that because all of their coefficients are integers and all integers value are commutative, then all polynomials with integer coefficients are commutative.

Thanks for your help so far!
• October 5th 2011, 11:43 AM
Deveno
Re: Show the following sets are rings
with the product of p(x), q(x), m is the degree of p (the highest power with a non-zero coefficient) and n is the degree of q. k runs from 0 to m+n.

for each value of k, the second sum runs over all combinations of i,j that sum to k. for example, 0 can only be written as 0+0, so the constant term of p(x)q(x)

is $a_0b_0$, 1 can be summed in two ways: 1+0 and 0+1, so the coefficient of x in p(x)q(x) is $a_0b_1 + a_1b_0$.

2 can be summed in 3 ways: 0+2, 1+1, and 2+0, so the coefficient of $x^2$ in p(x)q(x) is $a_0b_2 + a_1b_1 + a_2b_0$.

your "proof" that polynomials commute (additively) is too vague. here is a better one: suppose t = max{deg(p), deg(q)}.

for all $x^i$ in q(x) where deg(q) < i ≤ t (this set may be empty), define $b_j = 0$ and similarly for $x^i,\ a_i$ in p(x). then:

$p(x) + q(x) = \sum_i^t a_ix^i + \sum_i^t b_ix^i = \sum_i^t(a_i + b_i)x^i$

$=\sum_i^t(b_i+a_i)x^i$ <---commutativity of addition for integers used here

$= \sum_i^t b_ix^i + \sum_i^t a_ix^i = q(x) + p(x)$

for 2x2 matrices, use "typical matrices":

$A = \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}, B = \begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix}, C = \begin{bmatrix}c_{11}&c_{12}\\c_{21}&c_{22} \end{bmatrix}$

here is the left-distributive law:

$A(B + C) = \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}( \begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix} + \begin{bmatrix}c_{11}&c_{12}\\c_{21}&c_{22} \end{bmatrix})$

$= \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}\begin{bmatrix}b_{11}+c_{11}&b_{12}+c _{12}\\b_{21}+c_{21}&b_{22}+c_{22} \end{bmatrix}$

$= \begin{bmatrix}a_{11}(b_{11}+c_{11})+a_{12}(b_{21} +c_{21})&a_{11}(b_{12}+c_{12})+a_{12}(b_{22}+c_{22 })\\a_{21}(b_{11}+c_{11})+a_{22}(b_{21}+c_{21})&a_ {21}(b_{12}+c_{12})+a_{22}(b_{22}+c_{22}) \end{bmatrix}$

$= \begin{bmatrix}a_{11}b_{11}+a_{11}c_{11}+a_{12}b_{ 21}+a_{12}c_{21}&a_{11}b_{12}+a_{11}c_{12}+a_{12}b _{22}+a_{12}c_{22}\\a_{21}b_{11}+a_{21}c_{11}+a_{2 2}b_{21}+a_{22}c_{21}&a_{21}b_{12}+a_{21}c_{12}+a_ {22}b_{22}+a_{22}c_{22} \end{bmatrix}$

(here we have used the distributive law for the rationals)

$= \begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}+a_{11}c_{ 11}+a_{12}c_{21}&a_{11}b_{12}+a_{12}b_{22}+a_{11}c _{12}+a_{12}c_{22}\\a_{21}b_{11}+a_{22}b_{21}+a_{2 1}c_{11}+a_{22}c_{21}&a_{21}b_{12}+a_{22}b_{22}+a_ {21}c_{12}+a_{22}c_{22} \end{bmatrix}$

(here we have used commutativity of addition (and, implicitly, associativity) to switch the inner two terms of each sum)

$= \begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{ 12}+a_{12}b_{22}\\a_{21}b_{11}+a_{22}b_{21}&a_{21} b_{12}+a_{22}b_{22} \end{bmatrix} + \begin{bmatrix}a_{11}c_{11}+a_{12}c_{21}&a_{11}c_{ 12}+a_{12}c_{22}\\a_{21}c_{11}+a_{22}c_{21}&a_{21} c_{12}+a_{22}c_{22} \end{bmatrix}$

$= \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}\begin{bmatrix}b_{11}&b_{12}\\b_{21}& b_{22} \end{bmatrix} + \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}\begin{bmatrix}c_{11}&c_{12}\\c_{21}& c_{22} \end{bmatrix}$

$= AB + AC$.