Problems30-32 J.O

30. (Preliminary for T, April 10; finalized Sat, April 21) Let p be prime. Show that for all natural numbers n and elements x,y of the finite field Fp , we have (x+y)pn=xpn+ypn.

Since p prime and $1\leq n\leq p-1$], it follows that n! and (p-n)! do not divide p.

There is such an integer m such that
$\binom{p}{n} =pm \equiv 0$

So $(x+y)^p = x^p +y^p$.

To prove it works for $p^n$, use induction.

$(x+y)^{{p}^{n}}=((x+y)^p)^{{p}^{n-1}} = (x^p+y^p)^{{p}^{n-1}} = x^{{p}^{n}}+y^{{p}^{n}}$

31. (Preliminary for T, April 10; finalized Sat, April 21) a. Let f:R→S be a ring homomorphism, let A and I be a subring and ideal of R, and let B and J be a subring and ideal of S. Is the image f(A) of A in S necessarily a subring? Is the image f(I) necessarily an ideal? Likewise, is the preimage f−1(B) of B in R necessarily a subring? Is the preimage f−1(J) necessarily an ideal? For each question, either prove the assertion or give a counterexample.

The image of $f(A)$ of $A$ in $S$ is not necessarily a subring. The image $f(I)$ is an ideal, however. The preimage $f^{-1}(B) of B in R$ is not necessarily a subring as it could just be a surjective function. The preimage also doesn't have to necessarily be an ideal.

b. Prove the First Isomorphism Theorem for Rings: If φ:R→S is a ring homomorphism, then R/ker (φ)≅φ(R).

Let $\phi :R\to S$ be a ring homomorphism. If $r\in R \wedge r'\in ker(\phi)$, then we have $rr', r'r \in ker(\phi)$.
Since $ker(\phi)$ is also a subring of R, it is an ideal of R. It is then clear that $\phi(R)$ is a subring of S.

32. (Preliminary for T, April 10; finalized Sat, April 21) Let p be a prime. It is a fact that for each k≥1 there is a unique (up to isomorphism) finite field of cardinality pk, denoted Fpk. Prove that there is no surjective ring homomorphism from Z onto Fpk for k>1.

Suppose there is a surjective ring homomorphism from $\mathbb{Z}$ onto $F_{{p}^{k}}$ for $k>1$.
Then each element of $\mathbb{Z}$ would be mapped onto some element of $F_{{p}^{k}}$ for $k>1$. However this is impossible because of the problem of $\mathbb{Z}$ not being able to have a surjective homomorphism from $\mathbb{Z}$ onto $F_{{p}^{k}}$ for $k>1$ for $k>1$.

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