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I'm struggling to understand the definition of ideals in ring homomorphisms generated by a set.

If $R$ is commutative and has a $1$, then Ideal of $R$ generated by a subset $A$ of $R$:

$$⟨ A ⟩ = \{r_1a_1+\dotsb+r_na_n\mid r_i\in R, a_i\in A, n\in \mathbb{N}\}.$$

Now if $R$ has a $1$ isn't it sufficient to always use $⟨1⟩$ to express each element in the ideal?

$$⟨ 1 ⟩ = \{1r \mid r \in R\} = R$$

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In fact more is true: For a commutative ring $R$ with unity, an ideal $I$ such that $I \cap R^\times\neq \varnothing$, $I=R$. –  user21436 Apr 6 '12 at 12:16
    
But I don't understand what is your question and how is the last equation related to your question. –  user21436 Apr 6 '12 at 12:17

1 Answer 1

up vote 3 down vote accepted

I'm not sure what you're asking. If $1$ is in your ideal $I$ then yes, as you wrote, $I = R$, so it's not a proper ideal.

Maybe an example of an ideal generated by a set helps: Let $R = \mathbb Z$ and $A = \{7\}$. Then $\langle A \rangle = 7 \mathbb Z$.

If $R = K[x,y]$ for some field $K$ and $A = \{x,y\}$ then $I = \langle x,y \rangle $ is the set of all polynomials with no constant term. On the other hand, if $A = \{2x \}$ then $\langle 2x \rangle$ is the set of all polynomials with no constant term and with only even coefficients.

Hope this helps. If I misunderstood your question just drop me a comment.

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I had the exact same thing, well not the examples, but similar kind of, in my mind. +1. –  user21436 Apr 6 '12 at 12:20
    
@KannappanSampath Thanks : ) –  Rudy the Reindeer Apr 6 '12 at 12:21
    
Thanks! I did not think about the case when $I$ has no $1$. But if $I$ has a $1$, then it is sufficient to use $\langle 1 \rangle$ to generate $I$? –  joachim Apr 6 '12 at 12:33
    
@joachim Yes, it is enough, because if $I$ has $1$, then, $I=R$ and $\langle 1 \rangle=R$. –  user21436 Apr 6 '12 at 12:35
    
@joachim $\langle 1 \rangle = R$, that is, if $1$ is in $I$ then $I$ is the whole ring so you don't have a proper ideal. If you want a proper ideal you have to have $1 \notin I$. Did you mean "...to generate $R$?" in your last sentence? –  Rudy the Reindeer Apr 6 '12 at 12:35

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