# Direct Product cancellation in Hopfian rings

A ring $$R$$ is called to be 'Hopfian' if every ring homomorphism of $$R$$ onto $$R$$ is an automorphism of $$R$$

Question: Given $$R$$, $$S$$, $$T$$, Hopfian rings and $$R \times S \cong T \times S$$ implies that there exists an isomorphism $$R \cong T$$

My approach:

Let $$X1, X2, Y \in C$$. According to product rule in Category theory for every object Y and pair of morphisms

$$f1 : Y \to X1, \space\space f2 : Y → X2$$

there exists a unique morphism $$f : Y → X1 × X2$$ where $$f = $$

Given $$f: R \times S \to T \times S$$, therefore $$f = $$ where $$f_1: R \times S \to T$$ and $$f_2: R \times S \to S$$. Since $$S$$ is Hopfian, therefore $$f_2$$ is an isomorphic function. I wonder if isomorphism of $$f$$ does implies isomorphism of $$f_1$$ and $$f_2$$. Also please let me know if this is the correct approach or should I look at it in some other way.

• Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:R\times S\to S$ an iso? – Berci Dec 10 '18 at 8:27
• Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R \times S \to S$ where $f_2(r,s) \to s$ is a bijective mapping since S is given to be hopfian object. – forcehandler Dec 10 '18 at 8:50