Show that if $R$ and $S$ are Noetherian rings then so is $R\underset{T}{\times}S$.

Asked Oct 10 '18 at 23:25

Active Oct 10 '18 at 23:25

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Consider homomorphisms of rings,

\begin{figure}[htb!] \vspace{-10pt} \begin{center} \begin{tikzpicture}[scale=1.5,auto] \node (A) at (0,0) {$R$}; \node (B) at (1.5,0) {$T$}; \node (C) at (1.5,1.5) {$S$}; \node (D) at (0,1.5) {$R\times S$};

\path[->] (C) edge node[right] {$\psi$} (B) (A) edge node[above] {$\varphi$} (B) ;

\path[dashed,->] (D) edge node[right] {$\pi_1$} (A) (D) edge node[above] {$\pi_2$} (C) ; \end{tikzpicture} \end{center} \vspace{-20pt} \end{figure}

(Sorry for the code above. I don't know how to produce the diagram on stackexchange.)

The fiber product of $R$ and $S$ over $T$ is the subring $R\underset{T}{\times}S=\{(r,s)|\varphi(r)=\psi(s)\}$ of $R\times S$. Assume $\varphi$ and $\psi$ are surjective. Show that if $R$ and $S$ are Noetherian rings then so is $R\underset{T}{\times}S$.

If $R,S$ are Noetherian rings, then there exists a surjective homomorphism. Is it enough to show that $\pi_1$ and $\pi_2$ are also surjective?

asked Oct 10 '18 at 23:25

Username Unknown

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https://math.stackexchange.com/questions/51697/fibre-product-is-noetherian – Tsemo Aristide Oct 10 '18 at 23:58
But isn't it trivial that $\pi_1$ and $\pi_2$ are surjective? – Username Unknown Oct 11 '18 at 00:02
It's not trivial for fiber products, but now it follows from the condition that $\varphi$ and $\psi$ are surjective. – Berci Oct 11 '18 at 00:10
Ok. Once I show that, am I done? Or is there more to show? – Username Unknown Oct 11 '18 at 00:18

Show that if $R$ and $S$ are Noetherian rings then so is $R\underset{T}{\times}S$.

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