Interesting invariants of a scheme satisfying additivity in the product
I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
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I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
– Minseon Shin
Dec 1 at 20:08
add a comment |
I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
algebraic-geometry schemes
asked Dec 1 at 10:41
user73260
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112
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
– Minseon Shin
Dec 1 at 20:08
add a comment |
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
– Minseon Shin
Dec 1 at 20:08
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
– Minseon Shin
Dec 1 at 20:08
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
– Minseon Shin
Dec 1 at 20:08
add a comment |
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If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
– Minseon Shin
Dec 1 at 20:08