Algebras are defined with respect to an underlying ring. Now it makes sense that the underlying ring is smaller. But I was wondering if it is possible to have an underlying ring which is larger. In this case the homomorphism from the ring $R$ to the algebra $A$ $$\alpha : R \rightarrow A$$ will be a very weird one. Is there any rule for stopping the existence of such a weird homomorphism? Or do such algebras exist?
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4$\begingroup$ $R=\mathbb{Z}$ and $A=\mathbb{Z}/(2)$ $\endgroup$– MihailCommented Apr 28, 2019 at 15:03
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1$\begingroup$ Mihail's example is the one that pops to mind immediately, but there are even weirder ones, $R= C^0(\mathbb{R},\mathbb{R}), A=\mathbb{R}$, for instance $\endgroup$– Maxime RamziCommented Apr 28, 2019 at 15:05
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2 Answers
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Take the coefficient ring to be $\mathbb Z$ and the algebra to be $\mathbb Z/n\mathbb Z$. Any finite ring will also do as an example because any ring is a $\mathbb Z$-algebra.
answered Apr 28, 2019 at 15:36
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You may consider $R=k[X]$ and $A=k$ with $\alpha(X)=c$ for some constant $c\in k$. Though this is not particularly interesting.
answered Apr 28, 2019 at 15:05