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Prime Ideals

by Artusartos
Tags: ideals, prime
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Artusartos
#1
Jan7-13, 02:16 PM
P: 248
I have a question about the proof that I attached...

1) Since R/I is not the zero ring, we know that [tex]1 \not= 0[/tex]. What is the reason to say [tex]1 + I \not= 0 + I[/tex] instead of [tex]1 \not= 0[/tex]?


2) Also, how do we compute something like (a+I)(b+I)? Isn't this correct [tex](a+I)(b+I) = ab+aI+bI+I^2[/tex]?

3) Finally, if we have something like R/I, how do we know if the elements in R/I are of the form a+I or aI? Or is it both (since it's a ring)?

Thank you in advance
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#2
Jan7-13, 09:01 PM
P: 772
Also, how do we compute something like (a+I)(b+I)? Isn't this correct [tex](a+I)(b+I) = ab+aI+bI+I^2[/tex]?
Strictly, yes; but you didn't finish. How can you simplify it further, using what you know about ideals and multiplication?

Finally, if we have something like R/I, how do we know if the elements in R/I are of the form a+I or aI? Or is it both (since it's a ring)?
R/I is the quotient of R by the normal subgroup I of (R,+) satisfying the additional constraint that RI = I. The cosets are of the form a + I.

What is the reason to say [tex]1 + I \not= 0 + I[/tex]
We know that [itex]0 \in I[/itex] by definition, so [itex]0 + I = I[/itex]. We also know that [itex]I[/itex] is a proper ideal, so it can't contain 1 (what happens if an ideal contains a unit?). It follows that [itex]1 + I \neq I[/itex].


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