$\DeclareMathOperator\GL{GL}$The definition of $K_1$ is stated in "The K-book" by Charles Weibel as a quotient of $\GL(R)$, where $\GL(R)$ is the union of the sequence $R^{ \times} = \GL_1(R) \subset \GL_2(R) \subset \dotsb\subset \GL_n(R) \subset \GL_{n+1}(R) \subset\dotsb$ and $K_1(R)$ is defined as $\GL(R)/[\GL(R),\GL(R)]$. Right now I am stuck in the computation of $K_1(k[x]/(x^2))$, for a field $k$. The sudden interest to this is due to the example mentioned in the book that for every field $k$, $K_1(k) = k^{\times}$. So out of curiosity I experimented and understood that $K_1((k[x]/(x^2))_\text{red}) \cong k^{\times}$. But I got stuck in the computation of my headline question. Any hints or way to the solution are really appreciated.
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2$\begingroup$ For any artin local ring, $K_1$ is just the group of units. $\endgroup$– MohanCommented Nov 6, 2022 at 17:40
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1$\begingroup$ @Mohan: I don't think we even need Artin for this. $\endgroup$– Steven LandsburgCommented Nov 6, 2022 at 17:58
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3$\begingroup$ For a commutative ring $A$, the determinant splits $K_1(A)\simeq A^{\times}\oplus (\text{SL}(A)/E(A)$, and the second summand is trivial when $\text{SL}(A)$ is generated by elementary matrices. This is the case when $A$ is a local ring, as in your example. I don't have Weibel's textbook, but probably this is discussed there since it is standard for an introduction to algebraic K-theory. $\endgroup$– F ZaldivarCommented Nov 6, 2022 at 18:32
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$\begingroup$ Thank you very much. I will read through it definitely. $\endgroup$– user443060Commented Nov 7, 2022 at 4:48
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$\begingroup$ I wanted to post this as a separate question but in this thread if I get some hint it would be helpful. So $K_1(k[x]/(x^2))$ is properly understood. But if I extend it to $K_1((k[x]/(x^2))[y])$ will the result be same? Here I find the problem at $SK_1$. $\endgroup$– user443060Commented Nov 17, 2022 at 13:08
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