Question

1. Let 0 : Z/9Z+Z/12Z be the map 6(x + 9Z) = 4.+ 12Z (a) Prove that o is a ring homomorphism. Note: You must first show that o is well-defined (b) Is o injective? explain (c) Is o surjective? explain 2. In Z, let I = (3) and J = (18). Show that the group I/J is isomorphic to the group Z6 but that the ring I/J is not ring-isomorphic to the ring Z6. 3. Let R and S be nonzero rings with unity and denote their respective unities by IR and 1s. Let o: R + S be a nonzero homomorphism of rings. Prove that if $(1R) +1s then (1R) is a zero divisor in S. Deduce that if S is an integral domain then every ring homomorphism from R to S send the unity of R to the unity of S. 4. Let o: R + S be a surjective ring homomorphism. Suppose that J is an ideal of S. Let I = 0-'(J). (a) Prove that R/I is ring-isomorphic to S/J (b) Suppose that R and S are commutative rings with unity. Use part (a) to prove that if J is a prime ideal of S, then I is a prime ideal of R (c) Suppose that R and S are commutative rings with unity. Use part (a) to prove that if J is a maximal ideal of S, then I is a maximal ideal of R. 5. Suppose that F is a field and o: F + R is a ring homomorphism. Prove that o is either trivial or o is injective (Hint: Look at the kernel)

Please solve all questions

Answer 1

1) Let   ...(i).

x = y mod 9

Here we need to show $\phi (x+9z)=\phi (y+9z)$ to show that  $\phi$ is well defined.

From (i) we hake
-y=9k

+x=y+9k
  for some integrer k.

So we have  $\phi (x+9z)=\phi (y+9k+9z)$.

Now $9k+9z=9z$ Implies .

(1 +92) = oly +9k +92) = (y +92)

This proves that the function is well defined.

a)Now we show function is a homomorphism.

i) Linearity.

$\phi(x+y+9z)=4(x+y)+12z=4x+4y+12z$

Other side $\phi(x+9z)=4x+12z;~\phi(y+9z)=4y+12z$ .

Then $\phi(x+9z)+\phi(y+9z)=4x+12z+4y+12z=4x+4y+12z$ .

That is $\phi(x+y+9z)=\phi(x+9z)+\phi(y+9z)$ .

ii) Product.

$\phi(xy+9z)=xy+12z$.

$\phi(x+9z)\phi(y+9z)=(x+12z)(y+12z)=xy+12z$.

That is $\phi(x+9z)\phi(y+9z)=\phi(xy+9z)$ .

b) Not Injective:

We see by example.

$\bar 0,\bar 3\in \frac{z}{12z}$.

Then $\phi (\bar 0)=12z=\bar 0~(in~12z)$ and $\phi (\bar 3)=4*3+12z=12z=\bar 0~(in~12z)$ .

That is $\phi (\bar 3)=\phi (\bar 0)$ . This showed that $\phi$ is not injective.

3) Not Surjective:

Here we see by listing the images under $\phi$ .

$\phi (\bar 0)=\bar 0;~\phi (\bar1)=4;\phi (\bar 2)=\bar 8;\phi (\bar 3)=12=\bar 0\\ \phi (\bar 4)= 16=\bar 4;\phi (\bar 5)= 20=\bar 8;\phi (\bar 6)= 24=\bar 0\\ \phi (\bar 7)=28 =\bar4;\phi (\bar 8)=32=\bar 8$

That is set of images is $\left \{ \bar 0,\bar 4,\bar 8 \right \}$ which is a proper subset of $\frac{z}{12z}$ .

So the function is not subjective since except $\left \{ \bar 0,\bar 4,\bar 8 \right \}$ other elements don't have preimage.

Note: For some $\phi (\bar x)=\bar y$ means $\bar x\in \frac{z}{9z}$ and $\bar y\in \frac{z}{12z}$ .