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Let $R_{1}\times R_{2}$ be a direct product of Noetherian rings. Prove the product is Noetherian.

An ideal of $R_{1}\times R_{2}$ is of the form $I_{1}\times I_{2}$ where $I_{1}$, $I_{2}$ are ideals of $R_{1}$, $R_{2}$ respectively. Let $I_{1}$ be finitely generated by $x_{1},..,x_{n}$ and $I_{2}$ by $y_{1},..,y_{m}$. Then given $(a,b)$ in $I_{1}\times I_{2}$, there exist $v_{1},...,v_{n}$ in $R_{1}$ and $w_{1},...,w_{m}$ in $R_{2}$ such that $(a,b)=(v_{1}x_{1}+\cdots+v_{n}x_{n},w_{1}y_{1}+\cdots+w_{m}y_{m})$.

Say $n{\leq}m$.

Then $(a,b)=(v_{1},w_{1})(x_{1},y_{1})+\cdots+(v_{n},w_{n})(x_{n},y_{n})$. Therefore $I_{1}\times I_{2}$ is finitely generated by $(x_{1},y_{1}),..,(x_{n},y_{n})$. Correct?

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Not really!

$(a,b)=(v_1,w_1)(x_1,y_1)+\cdots+(v_n,w_n)(x_n,y_n)+(0,w_{n+1})(0,y_{n+1})+\cdots+(0,w_m)(0,y_m)$

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  • $\begingroup$ So i need to add $(0,y_{n+1}),..(0,y_{m})$ to my generating set? $\endgroup$
    – user137090
    Commented Apr 25, 2014 at 16:35

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