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Let $K$ be a field. Let $R=K[X,Y]$. Observe the ideals $(X), (Y), (X,Y)$ as $R$-modules. Which of them are isomorphic, which are not?

My guess is, that $(X)\cong (Y)$ and $(X)\ncong (X,Y)$, $(Y)\ncong (X,Y)$.

So I want to give a homomorphism of $R$-modules $f: (X)\to (Y)$ and show that it is an isomorphism. How can I give such an homomorphism? It is clear, that $f(X)=Y$, but how can I define $f$ so that it is a homomorphism?

Do I need the universal property of polynomial rings?

Thanks in advance.

Xam
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Cornman
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2 Answers2

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The ideals $(X)$ and $(Y)$ are both free of rank $1$ (isomorphic to $R$) and an explicit isomorphism is the mapping $$P(X,Y)X\longmapsto P(X,Y)Y.$$ On the other hand, the ideal $(X,Y)$ is not even free: $\{X,Y\}$ is a minimal set of generators, and an obvious relation between these generators is $$Y\cdot X- X\cdot Y=0. $$ Actually we have a finite free resolution of length $1$ for the ideal $(X,Y)$: \begin{align} 0\longrightarrow R[X,Y]&\longrightarrow R[X,Y]^2\xrightarrow{\qquad} (X,Y)\longrightarrow 0\\ 1& \longmapsto (Y,-X)\\ &\hspace3.5em(1,0)\longmapsto X,~(0,1)\longmapsto Y \end{align}

Bernard
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$(X)$ and $(Y)$ are principal ideals. Any nonzero principal ideal $(a)$ in an integral domain $R$ is isomorphic to $R$ as an $R$-module. The map $r\mapsto ar$ is an isomorphism from $R$ to $(a)$. Conversely an ideal in an integral domain $R$ which is isomorphic to $R$ as an $R$-module is principal. In this example, $(X,Y)$ is not a principal ideal.

Angina Seng
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  • So we have $(X)\cong R$ and $(Y)\cong R$ but $(X,Y)\ncong R$. Therefore $(X)\cong (Y)$ and $(X)\ncong (X,Y)$ [Same for $(Y)$]. I am still struggeling with R-modules. We observe the map $f: R\to (X)$, $r\mapsto rX$. But this is not a R-homomorphism, is it? – Cornman Sep 03 '17 at 09:14