Finalized problems 26-29 J.O.

26. (Preliminary for T, April 3; finalized Sat, April 14) a. Prove that equality modulo an ideal I of a ring R is an equivalence relation. (In our class, you may always assume (unless stated otherwise) that a ring is commutative and has 1.)

Reflexivity:$x-x\in I \Longrightarrow 0\in I$

Symmetric: Want to show: $x\sim y\longleftrightarrow y\sim x$
Thus we get two situations: $x-y\in I$ and $y-x\in I$
If we perform the operation $0-(x-y)\in I$ because of closedness under addition and also that $0\in I$.
Thus we get $-x+y\in I$ which is equal to $y-x\in I$.

Thus it works.

Transitivity: If we know [[x\sim y, y\sim z$]], then we want to show that it implies that $x\sim z$.
By adding both known properties:
$(x-y)+(y-z)$, we get
$x-z\in I$. Thus $x\sim z$ is implied as well.

b. Prove that multiplication on the quotient ring R/I is well defined: (r+I)(s+I)=rs+I.
$r\sim r' \longrightarrow r\cdot r'\in I$
$s\sim s' \longrightarrow s\cdot s'\in I$

If we let $a=r-r'$ and $b=s-s'$,
$\Longrightarrow rs+I=(r'+a)(s'+b) + I$
$\Longrightarrow r's' + r'b+ as' +ab+I=r's'+I$

27. (Preliminary for T, April 3; finalized Sat, April 14) a. Show that every unit of a ring R has a unique inverse.
Let $\mu,v_1,v_2 \in \mathbb{R}$
Let $\mu v_1=1$ and $\mu v_2=1$
$\Longrightarrow \mu v_1=\mu v_2$
$\Longrightarrow \mu v_1v_1=\mu v_1v_2$
$\Longrightarrow v_1=v_2$.

b. Show that the units of Z/nZ are precisely those [a] such that GCD(a,n)=1.

Assume $GCD(a,n)=1$. Then $\exists u,v \in \mathbb{Z}$ such that
$au+nv=1 \longrightarrow a \underline{u} + n\underline{v} = 1$
However $n\underline{v} =0$.
So $a\underline{u} = 1$.

28. (Preliminary for Th, April 5; finalized Sat, April 14) Prove that an ideal I of a ring R is maximal if and only if the quotient ring R/I is a field.

Assume r/I is a field. Let J be an ideal such that $I \subset J$. Then $\exists x\in J x\notin I$. Therefore, $I+x\notin I+0$.
Since R/I is a field, $\exists (I+y) \in R/I \longrightarrow (I+x)(I+y)=I+1$.
Thus $xy-1\in I$.
Since $I\subset J, (xy-1)\in J$.
Then consider $1= xy-(xy-1)$.
All element of the equation are in J, so $1\in J$.
Therefore, J=R and I is maximal.

Now assume that I is maximal.
Let $I+a \in R/I \longrightarrow (I+a)\neq (I+0) \wedge a\notin I$
Consier $J=I+Ra :=\{i+ra:i\in I, r\in R\}.$
Then $\forall i\in I, i=i+0a \longrightarrow I\subset J$.
Since I is maximal, J=I or J=R. However, $a=0+1a$. so $a\in J \wedge a\notin I$. so $J=R$.
Since I+Ra=R and $1\in R$,
1=i+ra for some $i\in I \wedge r\in R$.
$\Longrightarrow 1-ra=i\in I$, so (I+1)=(I+ra) implies (I+1)=(I+r)(I+a).
I+a is the inverse of I+r.
$\forall a\in R$, exists r such that ar=1.
All elements have an inverse, so R/I is afield.

29. (Preliminary for Th, April 5; finalized Sat, April 14) a. Consider the polynomial ring Q[x,y] in two variables with rational coefficients. (One way to think about it is as the polynomial ring in variable y whose coefficients come from the single-variable polynomial ring Q[x].) Find an ideal I⊆Q[x,y] that is prime but not maximal. (You don't have to write out a very formal proof, but explain clearly what your candidate is and why it works. Try to make your example as simple as possible.)

let $I = (x + 1)$ be the ideal that is prime but not maximal. It works because any polynomial of this form is irreducible and the ideal can only be reduced by the polynomail itself but it also can encompass all types of polynomials when they are multiplied together.

b. An ideal I of a ring R is a principal ideal if there exists x∈R such that I=(x); that is, the ideal is generated by
a single element. Prove that every ideal of Z is principal (we say that Z is a principal ideal domain or PID).

Let I be an ideal and Let x be the minimal positive element of I. Then we know that $(x)\in I$. Let $J = I / (x)$. If (x) doesn't equal I, then J is nonempty, so it too has a minimal positive element. Call that y. Note that $x < y$, so $0 < y-x.$
Now, y-x is less than y, so it can't be in J. Both x and y were in I, though, so y-x must also be in I and therefore, in (x).
Thus, we can write
$y-x = ax$ for some integer a
But that means that y is divisible by x, so y is in (a) This is a contradiction.

c. Show that Q[x,y] is not a PID by finding an ideal I that is not principal. Justify your answer. (Again, look for the simplest possible counterexample.)

Consider $I = (x, y)$. The term 1 does not lie inside I because neither x or y have constant terms and would not be able to be equal to 1.

Comments - Andrew Furash
Great work, but I don't quite understand your logic for 29b.

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