Try the Free Math Solver or Scroll down to Tutorials!

 

 

 

 

 

 

 

 
 
 
 
 
 
 
 
 

 

 

 
 
 
 
 
 
 
 
 

Please use this form if you would like
to have this math solver on your website,
free of charge.


Math 6310 Homework 5

Csar Lozano

Problem 1 Show that a free group has no elements of finite order. (except e).

Solution. Consider an element a ∈F(S). This means that

We may assume this is a reduced word. Suppose such a element has order two:

Since a is reduced, then this fact gives relation to the group, as sns1 = e for instance. Then
a ∈ F(S) cannot have order two. We can argue the same for

If we have an = e we got group relations sns1= e and therefore the group F(S) is no longer
free. Therefore, an element a ∈ F(S) may not have finite order.

Problem 2 A Boolean ring is a ring R such that x^2 = x for all x ∈ R. If R is a boolean
ring, show that

•x = -x for all x ∈ R

•R is commutative.

Solution. The firs statement is equivalent to 2x = 0 for all x ∈ R

This show that 2x = 0.

Now, in order to prove the second statement, just look the following computation

Taking into account in the last equation that x = -x, we have xy = yx, as desired.

Problem 3 Show that the only homomorphism from R to R is the identity.

Solution. First, let f : R -> R be a ring homomorphism. Consider n ∈ N, then

Thus the restriction f|Z is the identity map. Now consider n ∈ Q, this means we can write

Then we can conclude that the map is the identity map on the rational number Q.

Observe that if x > 0, there exits positive number y^2 = x such that f(x) = f(y)f(y) > 0.
As a result we have, if x - y > 0 then

In other words, if x < y, implies f(x) < f(y).

Now consider {xn} a sequence of rational numbers converging to r ∈ R,Q.
We can assume this since Q is dense in R. Then for each
there exist N ∈ N such that

however applying f to this inequality, we may have

Therefore, the function f(r) is the identity on R as desired

Problem 4

Solution. From now on, consider an arbitrary element y ∈ R.

•Multiplicative closure. Consider

•Additivity closure. Consider

•Additive Inverses. Consider this implies

Problem 5 Let S be a multiplicative set in a commutative ring R and let I be an ideal
maximal with respect to the property thatShow that I is prime.

Solution. If f,g ∈ R are not in I, then by the maximality of I, both I +(f) and I +(g)
meet S. Thus, there are elements of the form af + r and bg + s in S such that r, s ∈ I. If
fg ∈ I, the the product of af + r and bg + s would be in I, contradicting the fact that I
does not meet S

Problem 6 If R is an integral domain that is not a field, show that R[x] is not a principal
domain.

Solution. Observe that in a principal ideal domain R, each prime ideal is maximal. This
fact is due to if and there is then x = ty. Further ty ∈< x >
where then t = rx. Therefore x = rxy which means 1 = ry because R is integral
domain. Therefore, < y >= R.

Keeping this argument in mind, suppose R an principal domain. Thus we have that < x >
is maximal. However, the quotient is not a field, which is a contradiction.

Problem 7 Let R a commutative ring and let G be a group. Let R[G] be the group ring.
The augmentation ideal of R[G] is the kernel of the homomorphism from R[G] to R that sends

Show that the augmentation ideal is generated by the elements g - e for g ∈ G.

Solution. Thus, certainly the elements of the form {(x - e) : x ∈G} are in the kernel.
Now we want to see the kernel is generated by only those elements. Note that if a1x1 + … +
anxn -> 0 this implies that a1 + … + an = 0. It is clear that a1(x-e) + … + an(xn - e) goes
to zero as desired.

Problem 8 Let p a prime number, and let R be a commutative ring such that pr = 0 for all
r ∈R. Show that the function form R to R that sends r to rp is a ring homomorphism.

Solution. First, observe that p prime number implies that

has always p as a factor. Therefore,

since px = 0. Moreover

since R is a commutative ring. As a result the map x -> xp is a ring homomorphism.