Let $K$ be a field and $R=K\times K$ the product ring. We know this ring is not integral domain. But for all $P \in Spec(R)$, $R_P$ (the localization of $R$ at $P$) is integral domain. I know this but I don't know how can I show that? I need some hints.


1 Answer 1


Hint: for a product of rings $R_1 \times R_2$, all its ideals are of the form $I_1 \times I_2$, where $I_1 \triangleleft R_1$ and $I_2 \triangleleft R_2$. In a field $K$, there are only two ideals: the trivial one and the whole field. Thus, there are four ideals in $K \times K$:

  • $(0) \times (0)$
  • $K \times (0)$
  • $(0) \times K$
  • $K \times K$

Clearly, only $K \times (0)$ and $(0) \times K$ are prime.

Let $P = K \times (0)$. The localization $R_P$ contains inverses for all elements not in $P$, in particular for $(0,1)$. However, $(0,1)$ is a zero divisor in $R$: $(1,0) \cdot (0,1) = (0,0)$. Thus, if $(0,1)$ becomes invertible, then $(1,0)$ becomes zero in $R_P$, and so all elements of the form $(a,0)$, too, and from each element $(a,b)$ only the second coordinate remains.

Thus, we may conjecture that $R_P \cong K$. To prove this, construct a homomorphism $f : R \rightarrow K$ via $f(a,b) = b$. It maps elements from $R \setminus P$ to units in $K$. Combined with the previous paragraph, this proves that $K$ satisfies the universal property of localization, thus it is the localization.

Hence, the localization is $K$, which is an integral domain indeed. The case for the ideal $(0) \times K$ is entirely similar.

  • $\begingroup$ @haziranyagmur If you mean $\mathbb Z$, the ring of integers, - it actually is an integral domain. Nevertheless, I've added a bit on how to compute the localizations. $\endgroup$
    – lisyarus
    Jun 25, 2018 at 22:44

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