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.