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Consider the following question asked in my quiz on algebraic geometry:

Let $I$ be an ideal and let $S=1+I =\{ 1+x : x\in I\}$. Prove that $S$ is a multiplicatively closed subset. Prove that $S^{-1} I$ is contained in Jacobson Radical of $S^{-1} A$.

I have proved S to be multiplicatively closed but I am struggling with the other assertion: I took an element $i/{1+i} \in S^{-1} I$, I have to prove that it lies in each maximal ideal of $S^{-1} A$.Let it not lie in some maximal ideal M,which means that <i/i+1> generates $S^{-1} A$. which means that I is not a proper Ideal of A. ( But I was never given to be proper ideal of A).So, should I assume that I is a proper ideal of A?( Is it a typo in the question?)

Kindly shed some light on this?

user26857
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  • $I$ should be a proper ideal of $A$. Almost all the time anyone says "let $I$ be an ideal of the ring $R$", they implicitly mean a proper ideal. I bet if you ask your instructor, they'd confirm this, and if you bug them about it they might be more likely to remember to include the clarifying language next time! – Hank Scorpio Dec 12 '21 at 20:28
  • I'm going to chip in my two cents and say that yes, it should have been phrased "proper ideal" but also add that I disagree with the previous comment: I don't think most people implicitly mean "proper ideal" when they say "ideal." It's a good idea to have your antennae up for things like this, though! – rschwieb Dec 12 '21 at 20:41
  • @rschwieb I have a question in answer of hm2020 but I can't ask him as he has been banned for 1 year. Do , you mind answering that question? I have commented it below his answer. –  Dec 18 '21 at 13:53
  • @No-One because it is the product of the units $1+u_1$ and $\frac{1}{1+w}$ . It is also explained at this duplicate https://math.stackexchange.com/q/4335469/29335 – rschwieb Dec 18 '21 at 14:06
  • @rschwieb I don't think my question is not duplicate, How are you sure that $1+u_1$ and $\frac{1} {1+w}$ are units? Kindly elaborate! –  Dec 19 '21 at 13:44
  • @No-One seems you need to reread your definitions again. – rschwieb Dec 19 '21 at 20:40

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question: "I have proved S to be multiplicatively closed but I am struggling with the other assertion."

Answer: There is a criterion: If $B$ is a commutative ring with Jacobson radical $J(B)$, an element $x\in B$ is in $J(B)$ iff $1-xy\in B^*$ is a unit for all $y\in B$.

Let $S:=1+I$ for an ideal $I \subseteq A$ and let $B:=S^{-1}A, J:=S^{-1}(I) \subseteq B$. let

$$z_1:=\frac{x}{1+y} \in S^{-1}(I) \subseteq B$$

be any element in $J$. For any element $z_2:=u/1+v \in B$ we get

$$1-z_1z_2=\frac{1+w-ux}{1+w}=\frac{1+u_1}{1+w} \in B^*$$

for some $u_1,w\in I$. Hence for any $z_2$ it follows $1-z_1z_2$ is a unit, hence $z_1\in J(B)$.

hm2020
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