If $g \in I$ for an ideal $I$ over a commutative ring $R$, and $f \not \in I$, what can we say about $f-g$?

Asked Nov 24 '23 at 22:11

Active Nov 24 '23 at 22:18

Viewed 31 times

This is a really basic question:

Let $R$ be a commutative ring and $I \trianglelefteq R$ be an ideal.

Let $g \in I$ and let $f \not \in I$.

Can we say for certain that $f - g \not \in I$?

A classmate points out that you could say that

$$g+I =I$$ so that $$(f+(-g))+I = f+I \neq I.$$

edited Nov 24 '23 at 22:18

asked Nov 24 '23 at 22:11

Ben123

2

If $f-g$ and $g$ are both in $I$, then so is $(f-g)+g$. – Arturo Magidin Nov 24 '23 at 22:13
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Yes since if $f-g\in I$, then we have that $f-g=h$ for some $h\in I$, so then $f=h+g$, and as $g,h\in I$, we then have that $h+g=f\in I$ a contradiction – Steven Creech Nov 24 '23 at 22:13
Makes sense, thanks :) – Ben123 Nov 24 '23 at 22:16

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