User:J D Bowen/Math735 HW4

7.3.26)

a)

The function defined in the problem satisfies

$$\phi(a+b)=\underbrace{1+\dots+1}_{a \ \text{times}} + \underbrace{1+\dots+1}_{b \ \text{times}} = \underbrace{1+\dots+1}_{a+b \ \text{times}} = \phi(a)+\phi(b) \ $$

and

$$\phi(ab)=\underbrace{1+\dots+1}_{ab \ \text{times}}= \underbrace{\underbrace{1+\dots+1}_{a \ \text{times}}+\dots+\underbrace{1+\dots+1}_{a \ \text{times}}}_{b \ \text{times}} =\phi(a)\phi(b) \ $$.

Similar arguments work for when one or both of $$a,b \ $$ are negative, and so $$\phi \ $$ is a ring homomorphism.

Now to show the kernel is $$n\mathbb{Z} \ $$

Observe that $$\phi(n\mathbb{Z}) = \left\{{\phi(na):a\in\mathbb{Z} }\right\} \ $$. But

$$\phi(na)=\underbrace{1+1+\dots+1}_{an \ \text{times}}=\underbrace{\underbrace{1+\dots+1}_{n \ \text{times}}+\dots+\underbrace{1+\dots+1}_{n \ \text{times}}}_{a \ \text{times}} = \underbrace{0+\dots+0}_{a \ \text{times}} = 0 \ $$,

since $$\text{char}(R)=n \ $$. So, $$n\mathbb{Z}\subseteq\text{ker}(\phi) \ $$.

Now suppose $$x\in\text{ker}(\phi) \ $$. Since $$\text{char}(R)=n \ $$, we can be sure $$(x=0) \or (x>n) \ $$. Obviously $$0=x\in n\mathbb{Z} \ $$, so consider the second case. By the division algorithm, we can write $$x=an+b, \ b<n \ $$. Then $$0=\phi(x)=\phi(an+b)=\phi(an)+\phi(b) \ $$. By the previous paragraph, this becomes $$0=\phi(an)+\phi(b)=\phi(b) \ $$, but then $$(\phi(b)=0)\and(b<n) \ $$ contradicts $$\text{char}(R)=n \ $$. So $$\text{ker}(\phi)\subseteq n\mathbb{Z} \ $$.

We have $$(\text{ker}(\phi)\subseteq n\mathbb{Z}) \and( n\mathbb{Z}\subseteq\text{ker}(\phi) )\implies \text{ker}(\phi)=n\mathbb{Z} \ $$.

b) Let's examine the expression $$ {p \choose k} = \frac{p!}{k!(p-k)!} \ $$.

Since $$k<p, \ p-k<p \ $$, all the factors of $$k!(p-k)! \ $$ are less than $$p \ $$. So the only way $$p \ $$ could divide $$k!(p-k)! \ $$ would be if $$p \ $$ had factors in that expression; but $$p \ $$ is prime.

So $$p \ $$ divides $$\frac{p!}{k!(p-k)!} \ $$, and so $$\frac{p!}{k!(p-k)!} \ \text{mod}(p) = 0 \ $$.

Consider the binomial expansion

$$(a+b)^p = \sum_{k=0}^p {p \choose k} a^k b^{p-k} \ $$. By the above result,

$$(a+b)^p = 1 a^p + 0+\dots+0+1b^p \ $$.

7.3.2) Observe that the subring of constant functions $$\left\{{f\in\mathbb{Q}[x]:f(x)=q \ \forall x}\right\} \cong \mathbb{Q} \ $$.

Suppose there is an isomorphism from $$\mathbb{Q}[x]\to\mathbb{Z}[x] \ $$. Then $$\phi \ $$ carries the division ring $$\mathbb{Q} \ $$ to a division ring in $$\mathbb{Z}[x] \ $$. But since $$\mathbb{Z} \ $$ is not a division ring, there are no division rings in $$\mathbb{Z}[x] \ $$. Hence no such isomorphism exists.

7.3.12) Let $$x\in GR \ $$, so

$$Nx=N(\Sigma a_i g_i) = \Sigma Na_i g_i = \Sigma a_i N g_i \ $$

So, to show $$xN=Nx \ $$, it suffices to show $$g_i N = N g_i \forall g_i\in G \ $$.

Assume $$g_i N \neq N \ $$. Then $$gN=\Sigma gg_i \implies gg_i = gg_j \ $$, since there must be an omitted $$g_j \ $$. But then $$g_i = g_j =e \ $$, a contradiction. Hence $$gN=N \ $$. A similar argument shows $$Ng=N \ $$. Hence, $$N\in Z(RG) \ $$.