The post Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field has proved that under the restriction that the algebra admits no zero-divisors, it is a field. I wonder whether this condition is necessary. Can anyone construct a commutative finite-dimensional algebra that is not a field?
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5How about the $2$-dimensional $\mathbb{R}$-algebra $\mathbb{R}\times\mathbb{R}$? Note in this example, $(1,0)$ is a zero-divisor. – morrowmh Jul 24 '22 at 05:00
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How is the multiplication defined? – Tyson Che Jul 24 '22 at 05:05
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Coordinate-wise, also see https://math.stackexchange.com/questions/2638026/product-of-two-algebras-is-still-a-algebra – morrowmh Jul 24 '22 at 05:06
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I see. Thanks a lot. – Tyson Che Jul 24 '22 at 05:09
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1Also keep in mind $\mathbb R[\epsilon]$ where $\epsilon ^2 = 0 $. – Arkady Jul 24 '22 at 05:19