Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Not really!

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

share|improve this answer
    
So i need to add $(0,y_{n+1}),..(0,y_{m})$ to my generating set? –  user137090 Apr 25 at 16:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.