I was doing some practice abstract algebra questions off the internet since I have a quiz coming up soon. However, I am not very skilled at abstract algebra. In fact, I did very average in my group theory class, so I am struggling in my ring theory one. Can someone please help explain what is happening in this proof? I'm very sorry if it's extremely straightforward, I just think I need some time to get used to the way of thinking that's required to solve these questions.Let R1 and R2 be commutative rings with identities and let R=R1×R2. The question asks to show that every ideal I of R is of the form I=I1×I2 with I1 an ideal of R1 and I2 an ideal of R2.

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I was doing some practice abstract algebra questions off the internet since I have a quiz coming up soon. However, I am not very skilled at abstract algebra. In fact, I did very average in my group theory class, so I am struggling in my ring theory one. Can someone please help explain what is happening in this proof? I&#039;m very sorry if it&#039;s extremely straightforward, I just think I need some time to get used to the way of thinking that&#039;s required to solve these questions.Let R1 and R2 be commutative rings with identities and let R=R1×R2. The question asks to show that every ideal I of R is of the form I=I1×I2 with I1 an ideal of R1 and I2 an ideal of R2.

annieljcddj0 · Accepted Answer

Step 1Because&amp;nbsp;I&amp;nbsp;is an ideal, if you left- or right-multiply an element of&amp;nbsp;I&amp;nbsp;by an element of&amp;nbsp;R, then the result is in&amp;nbsp;I.LetI1={x∈R1|there exists&amp;nbsp;y∈R2&amp;nbsp;such that&amp;nbsp;(x,y)∈I}I2={y∈R2|there exists&amp;nbsp;x∈R1&amp;nbsp;such that&amp;nbsp;(x,y)∈I}.Step 2That is,&amp;nbsp;I1&amp;nbsp;is the image of I under the projection&amp;nbsp;π1:R→R1&amp;nbsp;on the first coordinate, and&amp;nbsp;I2&amp;nbsp;is the image of&amp;nbsp;I&amp;nbsp;under the projection&amp;nbsp;π2:R→R2. Because these are the images of an ideal under a surjective group homomorphism, we know that&amp;nbsp;I1&amp;nbsp;is an ideal of&amp;nbsp;R1&amp;nbsp;and&amp;nbsp;I2&amp;nbsp;is an ideal of&amp;nbsp;R2&amp;nbsp;(isomorphism theorems).Step 3Now, by construction,&amp;nbsp;I⊆I1×I2&amp;nbsp;(verify!).To show that&amp;nbsp;I1×I2⊆I, let&amp;nbsp;a∈I2. Then there exists&amp;nbsp;b∈R2&amp;nbsp;such that&amp;nbsp;(a,b)∈I. Then&amp;nbsp;(1R1,0)(a,b)=(a,0)∈I, so&amp;nbsp;I1×{0}⊆I.Step 4Now prove that likewise&amp;nbsp;{0}×I2⊆I.Conclude that&amp;nbsp;I1×I2⊆I. This gives the equality.

Marin Lowe · Accepted Answer

Sometimes you only need to follow the definition and ignore others. In this case, it can be proved easily from the definition.Suppose I is an ideal of R1×R2. Let (a,b),(c,d)∈I and (r1,r2)∈R1×R2. Then (a,b)−(c,d)=(a−c,b−d)∈Iimpliesa−c∈π1(I)andb−d∈π2(I)Where πi(I) is the ith-projection of I. Also (r1,r2)(a,b)=(r1a,r2b)∈π1(I)×π2(I)Let I1=π1(I) and I2=π2(I). Then I1 and I2 are ideals of R1 and R2.

I was doing some practice abstract algebra questions

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