Absorbing multiplication? More like "Amazing lectures; thanks for all the great information!" I do want to call out that issue with the polynomial ideals in the example at 11:55 (as other commenters have), but other than that, thanks for another great video 👍
11:55 Isn't J = the set of all polynomials divisible by t OR t^2+1? i.e. = { tp + (t^2+1)q : p, q in Z[t]}? And so in an analogous manner to the example of at 6:40, the ideal is NOT just the intersection of and (which would be as you claimed). If I'm understanding correctly, I believe that is actually all of Z[t], since t(-t) + (t^2+1)(1) = 1 is in , and hence, being an ideal, p(1) = p is in for any p in Z[t].
I believe there's something fishy with the ideal generated by t and t^2 + 1. It is not the intersection and not the ideal generated by t^3 + t. For an analogy, refer back to the ideal in Z. It is not the intersection of 4Z and 10Z. In short, = {polynomials divisible by t + polynomials divisible by (t^2 + 1)} != {polynomials divisible by t^3 + t}. An easy counterexample as stated by @Alex is the polynomial t = t + 0. Here t is divisible by t and 0 is a polynomial divisible by (t^2 + 1).
+Ming Hao Quek Yes, he made a mistake since = Z[t] (1 = t(-t) + (t^2 + 1) ) and once we can make 1 we can make anything.) An example of a non-principal ideal in Z[t] is .
Am I misunderstanding, or did you make a mistake? I think you said near the end of the video that the ideal generated by t and t^2 + 1 was equal to the ideal generated by t^3 + t. Isn't this false? For example t is an element of the ideal generated by the two polynomials, while t is not an element of the ideal generated by the one polynomial t^3 + t. Where am I not understanding this correctly? Thanks for the videos!!
This channel is amazing! I have been watching it for years now and I keep coming back. Thank you, Matthew Salomone.
Absorbing multiplication? More like "Amazing lectures; thanks for all the great information!"
I do want to call out that issue with the polynomial ideals in the example at 11:55 (as other commenters have), but other than that, thanks for another great video 👍
11:55 Isn't J = the set of all polynomials divisible by t OR t^2+1? i.e. = { tp + (t^2+1)q : p, q in Z[t]}? And so in an analogous manner to the example of at 6:40, the ideal is NOT just the intersection of and (which would be as you claimed). If I'm understanding correctly, I believe that is actually all of Z[t], since t(-t) + (t^2+1)(1) = 1 is in , and hence, being an ideal, p(1) = p is in for any p in Z[t].
Nice Matt.
I think J= = {tp(t) + (t^2+1)q(t), p(t), q(t) in Z(t)} and not a principal ideal since t is in J but t is not in
Ideals are just the latest incarnation of equivalence classes.
I believe there's something fishy with the ideal generated by t and t^2 + 1. It is not the intersection and not the ideal generated by t^3 + t. For an analogy, refer back to the ideal in Z. It is not the intersection of 4Z and 10Z.
In short, = {polynomials divisible by t + polynomials divisible by (t^2 + 1)} != {polynomials divisible by t^3 + t}. An easy counterexample as stated by @Alex is the polynomial t = t + 0. Here t is divisible by t and 0 is a polynomial divisible by (t^2 + 1).
+Ming Hao Quek Yes, he made a mistake since = Z[t] (1 = t(-t) + (t^2 + 1) ) and once we can make 1 we can make anything.)
An example of a non-principal ideal in Z[t] is .
Am I misunderstanding, or did you make a mistake? I think you said near the end of the video that the ideal generated by t and t^2 + 1 was equal to the ideal generated by t^3 + t. Isn't this false? For example t is an element of the ideal generated by the two polynomials, while t is not an element of the ideal generated by the one polynomial t^3 + t. Where am I not understanding this correctly? Thanks for the videos!!
9:28 Derp, I thought you wrote R, so I was so confused about your reasoning. Then I realized you actually wrote Q.
Pause at 04:47 for some funny moment :)
What was funny about it?