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Let $R$ be a non-zero ring with identity. Is it possible for $R[x]$ to have only a finite number of maximal left ideals !?

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Do you want to exclude $R=0$? – Martin Brandenburg Jan 7 '13 at 11:22
yes, usually when speaking of rings with identity, it means non-zero rings or equivalently $1 \not = 0$. – user30276 Jan 7 '13 at 11:50
If $R$ is commutative, then you can choose a maximal ideal $m$ of $R$, and let $k=R/m$. Since $k[x]$ has infinitely many maximal ideals, the same is true for $R[x]$. – François Brunault Jan 7 '13 at 20:20
tvector: This is not "usually" (and would mess around with ideal theory). – darij grinberg Jan 7 '13 at 21:12
@tvector: No, this would contradict the modern point of view of universal algebra and category theory. I know that there are people assuming that integers are $ \neq 0$, rings $ \neq 0$ (unital or not isn't important, the zero ring is unital with $1=0$), sets $\neq \emptyset$ etc. but this is just nonsense. Anyway, so the question should be: Let $R$ be a unital ring such that $R[x]$ has only finitely many maximal ideals. Does it follow $R=0$? (yes, see shatich's answer) – Martin Brandenburg Jan 8 '13 at 0:05
up vote 22 down vote accepted

Lets try something like Euclid's proof of infiniteness of prime numbers. Let $f_1 = x, f_2 = xf_1 + 1, f_3=xf_1f_2+1, f_4=xf_1f_2f_3 + 1$ and so on. The left ideal generated by each $f_i$ is proper. Thus each $f_i$ is contained in a maximal left ideal. On the other hand for any $i < j$ we have $ f_i | f_j-1$ (note that $f_i$'s are in the center of $R[x]$). This means that $f_i$ and $f_j$ can not be both in one maximal left ideal. Hence we have an infinite number of maximal left ideals.

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