A reduced Noetherian ring injects into a product of fields

I am trying to prove the following:

If a Noetherian ring $R$ is reduced, then there is an injection from $R$ into a product of fields.

What I know is that every associated prime of $R$ is minimal and for a minimal prime $p$ the localization $R_p$ is field. But I cannot proceed further. I need some help.

Many Thanks.

• Then why don't try $R\to\prod R_p$ where $p$ runs over the minimal primes? (Btw, in this case $R_p$ is the field of fractions of $R/p$, and recover the answer below.) – user26857 Sep 22 '16 at 6:49
• yes this is also possible and first i thought to write that, but Noetherian gives a finite direct product of fields, and so went with that, he can use either of them as he pleases. – user114539 Sep 22 '16 at 6:51
• For those who might be interested: the souped-up version of this is called Goldie's theorem. – rschwieb Sep 22 '16 at 10:07
• The statement also holds without the assumption that $R$ is Noetherian. (In this case, the product may be infinite.) See Tag 02LV in the Stacks Project for the precise statement and the proof. – Ingo Blechschmidt Feb 27 '18 at 15:10

Since $R$ is Noetherian, the set of minimal primes is finite, say $p_1,\cdots, p_n$ are the minimal primes of $R$. Since $R$ is reduced the natural map $R\to R/p_{1}\times \cdots \times R/p_{n}$ is injective. Now each $R/p_{i}$ is a domain and hence injects into its field of fractions $Q_i$. Hence we have an injection $R\to Q_1\times \cdots \times Q_n$.