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# 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 s_{n}s_{1} = e for instance. Then

a ∈ F(S) cannot have order two. We can argue the same for

If we have a^{n} = e we got group relations
s_{n}s_{1}= 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 {x_{n}} 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
a_{1}x_{1} + … +

a_{n}x_{n} -> 0 this implies that a_{1} +
… + a_{n} = 0. It is clear that a_{1}(x-e) +
… + a_{n}(x_{n}
- 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 r^{p} 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 -> x^{p}
is a ring homomorphism.