**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.**

Pf:

If $I$ is a maximal ideal and $a\notin I$, then $(a)+I$ must equal $R$ so there are elements $r\in R, m\in I$ such that $ra+m=1$.

I.e.: $(r+I)(a+I)= 1+I$ in $R/I$.

Conversely, if $R/I$ is a field and $a\notin I$, then $(r+I)(a+I)=1+I$ for some $r\in R$. Then $ra-1\in I$ so $1\in (a)+I$. And then $R=(a)+I$.

Since this is for every $a\notin I$, $I$ is maximal.

**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.)**

Pf:

If an ideal is prime then that means that $\forall x=ab\in I, a,b\in R: a\in I\vee a\in I$ and for it to also be maximal, the ideal must be such that $I\subset R$.

Suppose an ideal that is of the form $\mathbb{Q}/(X^2Y^2)$. Thus it has parts in it that when multiplied by parts in the ring, then either is in the ideal and it can still be equal to the entire ring.

**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 $d$ be the least positive integer in $I$. All integers are of the form md, where m is an integer in the ideal.

'd' is some integer in I that is not of the form md. We then use the division algorithm $d'=qd+r$. for some integer $q<0<r<d$. But there is a contradiction because whether $r$ is in I or not, as r is positive and less than d. If d is in our ideal I, then $d'-qd$ is also in our ideal. But $d'-qd=qd+r-dq=r$. Hence r is also in our ideal.

**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 the ideal I of the rational polynomial ring such that it is not principal, then it shows that $\mathbb{Q}$ is not a PID. Guess the subring $\mathbb{Q}/(X^2Y^2)$.

Shows that it's not principal.