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) \lor (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)\land(b<n) \ $ contradicts $\text{char}(R)=n \ $. So $\text{ker}(\phi)\subseteq n\mathbb{Z} \ $.

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