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I'm trying to disprove the statement 'Every ring is a subring of a semisimple Artinian ring' where semisimple is defined as having trivial Jacobson radical. The way I approached this was to first consider what would not be a counter example; namely commutative domains. The reason I think that these would not work is because any commutative domain is a subring of its field of fractions (or equiv localise with everything but $0$).

So using this thought I wondered if $\mathbb{Z}_4 = \mathbb{Z}/4\mathbb{Z}$ would be a counter example since it is not a domain so I could not construct a field of fractions.

However, how can I prove that $\mathbb{Z}_4$ is not a subring of $any$ semisimple Artinian ring. At the moment I am struggling to see how it could be a subring of anything else as it is a factor ring. Further, if it is indeed a subring of a semisimple Artinian ring, how might I go about thinking of a suitable counterexample?

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It is not possible to embed $\mathbb{Z}/4\mathbb{Z}$ in a ring with trivial Jacobson radical. I assume you mean, in your post, rings with unity.

Indeed, let $R$ be a ring containing $\mathbb{Z}/4\mathbb{Z}$ as a subring. Let $\mathfrak{m}$ be a maximal ideal of $R$. Then $\mathfrak{m}$ contains the element $\overline{2}$ of $\mathbb{Z}/4\mathbb{Z}$. Indeed, if $\overline{2}$ was not in $\mathfrak{m}$, then its projection in the field $R/\mathfrak{m}$ would be invertible, which is impossible since $\overline{2}^2 = 0$.

Therefore, all maximal ideals of $\mathfrak{m}$ contain $\overline{2}$. Since the Jacobson radical of $R$ is the intersection of its maximal ideals, this means that $\overline{2}$ is in the Jacobson radical.

Summarizing: if $R$ contains $\mathbb{Z}/4\mathbb{Z}$ as a subring, then the Jacobson radical of $R$ is non-trivial.

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    I was just wondering, what if $R$ is noncommutative? Because then $\mathfrak{m}$ is just a one sided ideal so we cannot conclude that $R/\mathfrak{m}$ is a field, only that it is simple? – 2017-02-18