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Prove that the intersection of any set of Ideals of a ring is an Ideal.

I'm looking for hints.

Let A, B both be Ideals of a ring R.

Suppose $I \equiv A\cap B$.

Since A and B are both Ideals of a ring R, A and B are both Subrings of a ring R. In particular, we have that $\left ( A,+ \right ),\left ( A\setminus \left \{ 0 \right \} ,\cdot \right ),\left ( B,+ \right ),\left ( B\setminus \left \{ 0 \right \},\cdot \right )$ are Abelian.

Now, Suppose $x_{1},x_{2} \in I$.

I'm not entirely sure how I can justify $x_{1}+\left ( -x_{2} \right ) \in I.$ Might be overthinking this but I might have to use the fact that I is the intersection.

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  • $\begingroup$ When you say "any set of ideals", do you mean only finite intersections? In that case, you can work with two ideals and the general case follows by induction. If you mean arbitrary intersection, this will not do. $\endgroup$ – M. Vinay May 28 '16 at 5:23
  • $\begingroup$ Yes, assume finite intersection although this is not explicitly mentioned in the text. $\endgroup$ – Mathematicing May 28 '16 at 5:24
  • $\begingroup$ Yes. "A ring with unity" contains the multiplicative identity 1. "A ring" contains no unity. $\endgroup$ – Mathematicing May 28 '16 at 5:25
  • $\begingroup$ If it's not mentioned, you should not assume it. And you don't have to, for the result to be true. Assuming it makes the proof no simpler, so there is no advantage in such an unwarranted assumption. $\endgroup$ – M. Vinay May 28 '16 at 5:28
  • $\begingroup$ You don't need to assume the collection is finite. You need to show that the intersection of all the ideals form an additive subgroup and, given $r\in R$, $r$ times the intersection is in the intersection. $\endgroup$ – John Douma May 28 '16 at 5:28
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Hints, as requested:

Let $I = \bigcap\limits_{j \in J} I_j$ be an intersection of ideals $I_j$ (where $J$ is an indexing set, finite or otherwise).

  1. Show that for any $x, y \in I$, $x - y \in I$. Use the fact that $x, y \in I_j$, $\forall j \in J$.
  2. Show that for any $r \in R$ and $x \in I$, $rx \in I$. Again, use the fact that $x \in I_j$, $\forall j \in J$. [If the ring is not commutative, this works only for left ideals, and the proof is similar for right ideals and two-sided ideals].

Solution (using above hints):

  1. Since $x, y \in I_j$, and $I_j$ is an ideal, $x - y \in I_j$, $\forall j \in J$. Therefore, $x - y \in \bigcap\limits_{j \in J} I_j = I$.
  2. Similarly, since $x \in I_j$, $rx \in I_j$, $\forall j \in J$. Therefore, $rx \in I$. The proof is similar for right ideals and two-sided ideals (or alternatively, a two-sided ideal is both a left ideal and a right ideal).

Note: If rings are defined to have $1$, it is enough to show $x + y \in I$, since $-y = (-1)y$.

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    $\begingroup$ I have solved it. Nonetheless, will give you an up vote for staying with the question. $\endgroup$ – Mathematicing May 28 '16 at 5:32
  • $\begingroup$ @Mathematicing Thanks. I added the full solution. $\endgroup$ – M. Vinay May 28 '16 at 5:48
  • $\begingroup$ @M. Vinay Thanks for a solution. But how do we know, that the intersection is not an ampty set ? So how come, that if x-y belongs to an ideal Ij, than it suddenly belongs to Ij for every j. ? Thanks for clarification in advance. $\endgroup$ – martina May 31 '18 at 9:16
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    $\begingroup$ @martina The statement is that if $x, y \in I = \bigcap\limits_{j \in J} I_j$, then by definition of intersection, $x, y \in I_j$, $\forall j \in J$, and therefore $x - y \in I_j$, $\forall j \in J$ (since each $I_j$ is an ideal). The question of whether $I$ is non-empty does not arise here. If you separately ask whether the intersection of ideals is non-empty, then yes, it is non-empty, since every ideal must contain $0$, so their intersection must also contain $0$. $\endgroup$ – M. Vinay May 31 '18 at 9:48

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