User:Kcbetancourt/AlgebraHW8

9.1.5) Prove $$(x,y), \ (2,x,y) \ $$ are prime ideals of $$\mathbb{Z}[x,y] \ $$, but that only the latter is maximal.

Let $$ ab \in (x,y) $$ such that $$ a,b \in \Z [x,y] $$. Since $$ ab \in (x,y) $$, each term of $$ ab $$ contains an x or y. Assume that $$ a\notin (x,y) $$ and $$ b\notin (x,y) $$. Then there are terms in $$ a $$ and $$ b $$ that do not contain an x or y i.e, there exists at least one term in both $$ a $$ and $$ b $$ that contains neither x nor y. Then, the product $$ ab $$ would have a term that contains neither x nor y, which implies that $$ ab \notin (x,y) $$. Thus we have reached a contradiction. Therefore, $$ ab \in (x,y) $$ implies that either $$ a \in (x,y) $$ or $$ b \in (x,y) $$. Thus $$ (x,y) $$ is a prime ideal of $$\mathbb{Z}[x,y] \ $$.

Now let $$ ab \in (2,x,y) $$ such that $$ a,b \in \Z [x,y] $$. Then each term of $$ ab $$ contains an x or y or is a constant multiple of 2. Assume that $$ a\notin (2,x,y) $$ and $$ b\notin (2,x,y) $$. Then there is a constant term in both $$ a $$ and $$ b $$ which is not a multiple of 2. Then, the product $$ ab $$ would have a constant term that is not a multiple of 2, because any even constant would have an even factor. This implies that $$ ab \notin (2,x,y) $$. Thus we have reached a contradiction. Therefore, $$ ab \in (2,x,y) $$ implies that either $$ a \in (2,x,y) $$ or $$ b \in (2,x,y) $$. Thus $$ (2,x,y) $$ is a prime ideal of $$\mathbb{Z}[x,y] \ $$.

Now assume that $$ I $$ is an ideal of $$\mathbb{Z}[x,y] \ $$ such that $$ (2,x,y) \subset I \subset \Z [x,y] $$. Then $$ I $$ contains some element in $$\mathbb{Z}[x,y] \ $$ that is not in $$ (2,x,y) $$. Therefore, this element has an odd constant term. Therefore, since $$ I $$ must be closed under addition, $$ I $$ contains all the elements of $$ (2,x,y) $$ and elements with odd constant terms. But this generates all of $$\mathbb{Z}[x,y] \ $$. Therefore, the ideal $$ (2,x,y) $$ must be maximal.

To show that $$ (x,y) $$ is not maximal, we need to show that there exists an ideal, $$ J $$ such that $$ (x,y) \subset J \subset \Z [x,y] $$. Let $$ J = (2,x,y) $$. Clearly, $$ (x,y) \subset (2,x,y) $$ but $$ (2,x,y) $$ contains elements with even constant terms, which are not in $$ (x,y) $$. For instance, $$ 2 \in (2,x,y) $$ but $$ 2 \notin (x,y) $$. Thus $$ (x,y) \subset (2,x,y) \subset \Z [x,y] $$, which implies that $$ (x,y) $$ is not maximal.