Prelim Numbers 28,29 J.O

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.

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.)

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.

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