Krull dimension of a direct product of rings Let $R$ and $S$ be two non-zero rings. 

The Krull dimension of $R \times S$ is greater or equal to the maximum of the Krull dimensions of $R$ and $S$. Does anyone have an idea how to show this ? 

I know the definition of the Krull dimension of a ring, but how can I define the Krull dimension of a direct product of rings ?
Thank you for your help. 
 A: 
I know the definition of the Krull dimension of a ring, but how can I define the Krull dimension of a direct product of rings ?

I'm not sure what you mean by "how can I define" the Krull dimension for direct products. The direct product of rings is certainly a ring itself, so the definition of Krull dimension makes sense. In particular, the Krull dimension of a ring is one less than the length of the longest chain of ideals in the poset of prime ideals.
Now one can compute the Krull dimension of the direct product of rings $R,S$: it turns that dim$(R \times S)$ is the maximum of that of $R$ and $S$.
To prove this, first note that all the ideals of $R\times S$ are of the form $I \times J$, where $I\subset R,J \subset S$ are ideals (the key observation here is that if $(a,b)$ is in the ideal, then so is $(a,0)$ and $(b,0)$ by multiplying by $(1,0)$ and $(0,1)$).
Moreover, the prime ideals  of $R\times S$ are of the form $R \times P$ or $Q\times S$, where $P\subset S,Q\subset R$ are primes. One readily sees that these are prime ideals, and indeed if $I \times J$ are primes, then $(1,0)(0,1) = (0,0)$ is in the ideal, so either $I$ or $J$ contains $1$. From here it is easy to see that the other one must be prime.
Thus any chain of prime ideals in $R \times S$ really arises from either a chain of prime ideals in $R$ or one in $S$. Thus the longest chain must come from the longest chain in $R$ or $S$ (which ever one has a longer chain) so dim$(R\times S) = $ max$($dim$(R),$dim$(S))$.
