Sum of ideals is an ideal

**knguyen2005** · October 26th 2009, 10:39 AM

Let R be a ring
Suppose that I_a is an ideal of R, 1=< a=< n.
Show that: I_1 + I_2+....+ I_n = {i_1 + i_2+...+ i_n: a in I, 1=< a=< n} is an ideal of R.

This is my attempt

Since 1 belongs to R then 1 is also belongs to I_a implies I_a is non empty set
For any i_1, i_2 in I_a, then i_1 + i_2 also belongs to I_a
So, for 1=< a=< n, i_1 +.....+ i_n belongs to I_a

Hence, (I_1 + I_2+...+ I_n) is an ideal of R
Is that correct?

Thank you

**TheEmptySet** · October 26th 2009, 11:06 AM

Originally Posted by knguyen2005

Let R be a ring
Suppose that I_a is an ideal of R, 1=< a=< n.
Show that: I_1 + I_2+....+ I_n = {i_1 + i_2+...+ i_n: a in I, 1=< a=< n} is an ideal of R.

This is my attempt

Since 1 belongs to R then 1 is also belongs to I_a implies I_a is non empty set
For any i_1, i_2 in I_a, then i_1 + i_2 also belongs to I_a
So, for 1=< a=< n, i_1 +.....+ i_n belongs to I_a

Hence, (I_1 + I_2+...+ I_n) is an ideal of R
Is that correct?

Thank you

First I think you mean $0$ is in each $I_a$

If 1 is in your ideal your ideal is the whole ring!!

So what you need to show is that is it is closed uder subtraction and that it absorbs ring elements

i.e

let $x =\sum_{k=1}^{a}i_k$ and $y =\sum_{k=1}^{a}j_k$ where $i_k,j_k \in I_k$

What you need to show is that

$x-y$ is in the ideal and $rx$ is in the ideal for every $r$ in the Ring

I hope this helps

**clic-clac** · October 26th 2009, 11:09 AM

Show that: I_1 + I_2+....+ I_n = {i_1 + i_2+...+ i_n: a in I, 1=< a=< n} is an ideal of R.

I guess it is $i_a\in I_a$ .

Since 1 belongs to R then 1 is also belongs to I_a

If you want to prove the sum of ideals $I_1+...+I_n$ is non empty, try to prove that $0$ is one of its elements, using that 0 belongs to any ideal of $R.$

Then what you have to do is: given two elements of $I_1+...+I_n,$ let's say $i=i_1+...+i_n$ and $j=j_1+...+j_n,$ prove that $i-j$ belongs to $I_1+...+I_n.$ It can be done using that $(R,+)$ and $(I_a,+)$ for $a\in\{1,...,n\}$ are abelian group.

Finally, show that for any element $i_1+...+i_n$ of $I_1+...+I_n$ and for any $r$ in $R,$ then $r(i_1+...+i_n)$ is in $I_1+...+I_n.$

**knguyen2005** · October 26th 2009, 12:13 PM

Thanks you guys alot
Sorry to bother you again.

I know the definition of an ideal of a ring But I don't know exactly what it means, can you explain what the ideal means in the ring theory please

I am really appreciated
Cheers

**clic-clac** · October 27th 2009, 04:02 AM

Take an additive subgroup $(B,+)$ of a ring $(A,+,\times).$ Since $(A,+)$ is commutative, $(A/B,+)$ is an abelian group.

Let us write $\overline{x}$ for $x+B$ the class of some element of $A$ in $A/B$

We want $A/B$ to be a ring when we define naturally the multiplication: i.e. $\overline{a}\times \overline{b}=\overline{ab}.$

But given two elements of $\overline{a}$ and $\overline{b},$ let's say $a+c$ and $b+d$ (i.e. $c,d\in B$ ), observe that $(a+c)(b+d)=ab+ad+cb+cd \in \overline{ab}$ iff $ad+cb+cd\in B.$

In general, the last condition is wrong; but if we ask $B$ to be stable under multiplication by any element of $A,$ then it becomes true. And that's precisely the definition of an ideal.

Ideals can be seen as a sort of "normal subgroups" for rings, when you consider the quotient of a ring by one of its ideals, then it is ring (of course you have to verify that the multiplication as defined higher has the required properties, but it is true). The difference with normal subgroups is that an ideal is not a subring: the only case when this is true is when the ideal is the whole ring, indeed a subring contains $1$ .

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