What is the reason we can't call I a subring? At 2:41 you explained it I think, you said ideals don't need to contain the multiplicative identity, well neither does rings. So my question persists, why can't we call I a subring?
@@mihirbpi theres a small amount of discrepancy. Some definitions of rings require that a multiplicative identity exists while others don't. Ideals without 1 are non-unital subrings but could still be considered subrings.
The ideal generated by x, denoted (x), is the collection of elements ax where a is in R. To show this is an ideal, we just have to show that it is closed under addition and under multiplication by elements in R. So let y and z be elements of (x). Then, by definition, y = ax for some a in R and z = bx for some b in R. Then y+z = ax+bx = (a+b)x by the distributive property of R. Also a+b is in R by closure under addition, so y+z is an element of R times x. Hence, y+z is in (x). Let y be an element of (x) and let r be any element of R. By definition, y = ax for some element a of R. Then ry = r(ax) = (ra)x by the associative property of multiplication. Also ra is in R by closure under multiplication, so ry is an element of R times x. Hence, ry is in (x). Any collection of elements in R generate an ideal in R.
Fantastically pedagogical and helpful. Thank you for providing such clear explanations.
this playlist probably saved me from flunking, thanks dude!
Your explanations are crystal clear. Thank you for taking the time to make this video.
This video helped me a lot with understanding ideals. Thank you!
What is the reason we can't call I a subring? At 2:41 you explained it I think, you said ideals don't need to contain the multiplicative identity, well neither does rings. So my question persists, why can't we call I a subring?
I think rings do need to contain a multiplicative identity. It's just that all elements in the ring don't need to have multiplicative inverses.
So since ideals don't always need to contain the multiplicative identity they can't always be a subring
@@mihirbpi theres a small amount of discrepancy. Some definitions of rings require that a multiplicative identity exists while others don't. Ideals without 1 are non-unital subrings but could still be considered subrings.
15:31 Sir how you that x must generate an ideal?
The ideal generated by x, denoted (x), is the collection of elements ax where a is in R. To show this is an ideal, we just have to show that it is closed under addition and under multiplication by elements in R.
So let y and z be elements of (x). Then, by definition, y = ax for some a in R and z = bx for some b in R. Then y+z = ax+bx = (a+b)x by the distributive property of R. Also a+b is in R by closure under addition, so y+z is an element of R times x. Hence, y+z is in (x).
Let y be an element of (x) and let r be any element of R. By definition, y = ax for some element a of R. Then ry = r(ax) = (ra)x by the associative property of multiplication. Also ra is in R by closure under multiplication, so ry is an element of R times x. Hence, ry is in (x).
Any collection of elements in R generate an ideal in R.