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1. • CommentRowNumber1.
   • CommentAuthorTobyBartels
   • CommentTimeAug 26th 2015
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   Author: TobyBartels
   Format: MarkdownItexA bit of trivial algebra at [[trivial subalgebra]].

   A bit of trivial algebra at trivial subalgebra.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 27th 2015
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   Author: Todd_Trimble
   Format: MarkdownItexDid you mean to say that the trivial subring of a ring $A$ is $\{\ldots, -2, -1, 0, 1, 2, \ldots\}$? In other words, the image of the initial ring in $A$?

   Did you mean to say that the trivial subring of a ring $A$ is $\{\ldots, -2, -1, 0, 1, 2, \ldots\}$? In other words, the image of the initial ring in $A$?

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeAug 27th 2015
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   Author: TobyBartels
   Format: MarkdownItexYes, you\'re right, it\'s the set of *all* constants, not just the set of the basic constants (as it must be to be a subring). Rookie mistake!

   Yes, you're right, it's the set of all constants, not just the set of the basic constants (as it must be to be a subring). Rookie mistake!