4
$\begingroup$

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?

$$⟨ 1 ⟩ = \{1r \mid r \in R\} = R$$

  • 0
    In fact more is true: For a commutative ring $R$ with unity, an ideal $I$ such that $I \cap R^\times\neq \varnothing$, $I=R$.2012-04-06
  • 1
    But I don't understand what is your question and how is the last equation related to your question.2012-04-06

1 Answers 1

6

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.

  • 0
    I had the exact same thing, well not the examples, but similar kind of, in my mind. +1.2012-04-06
  • 0
    @KannappanSampath Thanks : )2012-04-06
  • 0
    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 $I$?2012-04-06
  • 0
    @joachim Yes, it is enough, because if $I$ has $1$, then, $I=R$ and $\langle 1 \rangle=R$.2012-04-06
  • 0
    @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?2012-04-06
  • 0
    Actually I was not sure what is was asking for up until now. Thanks I think I got it!2012-04-06
  • 0
    @joachim Cool, I'm glad : ) Otherwise feel free to ask further questions.2012-04-06