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Bundles that have many sections don't have a special name, but their slightly more useful special case goes under the name of very ample bundles.

Start from the beginnning. Line bundles can be defined over any field $k$ but it has to be the same as the field of definition of your original manifold. Why the field cannot be different? By definition a line bundle is something whose total space is locally $A^1\times B$ for $B$ as base.

Now yes, what you're doing can be written more formally and more invariantly. Take the global sectionsfor a line bundle $L$. By definition, for a given point $x$ we have a linear pairing $H^0(L)\times \{x\} \to L_x$, the fiber of $L$ at $x$ (which is isomorphic to $k$). If nonzero, a linear map to a one-dimensional space (even though the isomorphism with $k$ is not canonical) by definition gives a canonical point in $\mathbb P(H^0(L))^*$. As Anton points out, this map takes a point $x$ to the hyperplane of sections that vanish at $x$ (and it fails iff all sections vanish at $x$).

As you vary $x$, you get the map to $\mathbb P(H^0(L))^*$, a very well-defined and canonical projective space. As the intuition would tell you, for this to be effective there should be a way to tell two points apart, that is if I have points $x, y\in X$ then there should be a section $s$ such that $s(x) = 0$ but $s(y) \ne 0$.

It's hard to construct sections of an arbitrary line bundle, therefore for some bundles the pairings above will be zero at some points. This will mean that the map above will be not defined.

For many other bundles, the map will be defined but it may not have good properties, e.g. it could send the whole variety to a single point — it's not very useful, thus a line bundle is called very ample iff this map is an immersion.

Now if take arbitrary scheme (skip if you're not familiar with schemes), they locally look like $\mathrm{Spec}\\,A$. Now it's easy to define a line bundle: it's something that maps into the scheme and the map locally looks like $$\mathrm{Spec}\\,A[t] \to \mathrm{Spec} \\,A.$$ You can define the cohomology as usual, so the space $\mathbb P(H^0(L))^*$ continues to prove a good target to map into. It's slightly more unusual to draw pictures, but you can start with tensoring everything by $\mathbb F_q$ to gain the intuition.

Return to curves over field $k$. There the situation with sections is quite easy. Every line bundle there has the form $O(D)$ for some divisor $D$. As you add more points to your divisor, you're guaranteed lots of sections. Therefore, there are many very ample lines bundles.

Finally, a line bundle $L$ is ample iff some of its powers $L^{\otimes n}$ is very ample.

Bundles that have many sections don't have a special name, but their slightly more useful special case goes under the name of very ample bundles.

Start from the beginning. Line bundles can be defined over any field $k$ but it has to be the same as the field of definition of your original manifold. Why the field cannot be different? By definition a line bundle is something whose total space is locally $A^1\times B$ for $B$ as base.

Now yes, what you're doing can be written more formally and more invariantly. Take the global sections for a line bundle $L$. By definition, for a given point $x$ we have a linear pairing $H^0(L)\times \{x\} \to L_x$, the fiber of $L$ at $x$ (which is isomorphic to $k$). If nonzero, a linear map to a one-dimensional space (even though the isomorphism with $k$ is not canonical) by definition gives a canonical point in $\mathbb P(H^0(L))^*$. As Anton points out, this map takes a point $x$ to the hyperplane of sections that vanish at $x$ (and it fails iff all sections vanish at $x$).

As you vary $x$, you get the map to $\mathbb P(H^0(L))^*$, a very well-defined and canonical projective space. As the intuition would tell you, for this to be effective there should be a way to tell two points apart, that is if I have points $x, y\in X$ then there should be a section $s$ such that $s(x) = 0$ but $s(y) \ne 0$.

It's hard to construct sections of an arbitrary line bundle, therefore for some bundles the pairings above will be zero at some points. This will mean that the map above will be not defined.

For many other bundles, the map will be defined but it may not have good properties, e.g. it could send the whole variety to a single point — it's not very useful, thus a line bundle is called very ample iff this map is an immersion.

Now if take arbitrary scheme (skip if you're not familiar with schemes), they locally look like $\mathrm{Spec}\,A$. Now it's easy to define a line bundle: it's something that maps into the scheme and the map locally looks like $$\mathrm{Spec}\,A[t] \to \mathrm{Spec}\,A.$$ You can define the cohomology as usual, so the space $\mathbb P(H^0(L))^*$ continues to prove a good target to map into. It's slightly more unusual to draw pictures, but you can start with tensoring everything by $\mathbb F_q$ to gain the intuition.

Return to curves over field $k$. There the situation with sections is quite easy. Every line bundle there has the form $O(D)$ for some divisor $D$. As you add more points to your divisor, you're guaranteed lots of sections. Therefore, there are many very ample lines bundles.

Finally, a line bundle $L$ is ample iff some of its powers $L^{\otimes n}$ is very ample.

Bundles that have many sections don't have a special name, but their slightly more useful special case goes under the name of very ample bundles.

Start from the beginnning. Line bundles can be defined over any field $k$ but it has to be the same as the field of definition of your original manifold. Why the field cannot be different? By definition a line bundle is something whose total space is locally $A^1\times B$ for $B$ as base.

Now yes, what you're doing can be written more formally and more invariantly. Take the global sectionsfor a line bundle $L$. By definition, for a given point $x$ we have a linear pairing $H^0(L)\times \{x\} \to L_x$, the fiber of $L$ at $x$ (which is isomorphic to $k$). If nonzero, a linear map to a one-dimensional space (even though the isomorphism with $k$ is not canonical) by definition gives a canonical point in $\mathbb P(H^0(L))^*$.

As you vary $x$, you get the map to $\mathbb PH^0(L)$, a very well-defined and canonical projective space. As the intuition would tell you, for this to be effective there should be a way to tell two points apart, that is if I have points $x, y\in X$ then there should be a section $s$ such that $s(x) = 0$ but $s(y) \ne 0$.

It's hard to construct sections of an arbitrary line bundle, therefore for some bundles the pairings above will be zero at some points. This will mean that the map above will be not defined.

For many other bundles, the map will be defined but it may not have good properties, e.g. it could send the whole variety to a single point — it's not very useful, thus a line bundle is called very ample iff this map is an immersion.

Now if take arbitrary scheme (skip if you're not familiar with schemes), they locally look like $\mathrm{Spec}\\,A$. Now it's easy to define a line bundle: it's something that maps into the scheme and the map locally looks like $$\mathrm{Spec}\\,A[t] \to \mathrm{Spec} \\,A.$$ You can define the cohomology as usual, so the space $\mathbb PH^0(L)$ continues to prove a good target to map into.

Return to curves over field $k$. There the situation with sections is quite easy. Every line bundle there has the form $O(D)$ for some divisor $D$. As you add more points to your divisor, you're guaranteed lots of sections. Therefore, there are many very ample lines bundles.

A line bundle $L$ is ample iff some of its powers $L^{\otimes n}$ is very ample.

Bundles that have many sections don't have a special name, but their slightly more useful special case goes under the name of very ample bundles.

Start from the beginnning. Line bundles can be defined over any field $k$ but it has to be the same as the field of definition of your original manifold. Why the field cannot be different? By definition a line bundle is something whose total space is locally $A^1\times B$ for $B$ as base.

Now yes, what you're doing can be written more formally and more invariantly. Take the global sectionsfor a line bundle $L$. By definition, for a given point $x$ we have a linear pairing $H^0(L)\times \{x\} \to L_x$, the fiber of $L$ at $x$ (which is isomorphic to $k$). If nonzero, a linear map to a one-dimensional space (even though the isomorphism with $k$ is not canonical) by definition gives a canonical point in $\mathbb P(H^0(L))^*$. As Anton points out, this map takes a point $x$ to the hyperplane of sections that vanish at $x$ (and it fails iff all sections vanish at $x$).

As you vary $x$, you get the map to $\mathbb P(H^0(L))^*$, a very well-defined and canonical projective space. As the intuition would tell you, for this to be effective there should be a way to tell two points apart, that is if I have points $x, y\in X$ then there should be a section $s$ such that $s(x) = 0$ but $s(y) \ne 0$.

It's hard to construct sections of an arbitrary line bundle, therefore for some bundles the pairings above will be zero at some points. This will mean that the map above will be not defined.

For many other bundles, the map will be defined but it may not have good properties, e.g. it could send the whole variety to a single point — it's not very useful, thus a line bundle is called very ample iff this map is an immersion.

Now if take arbitrary scheme (skip if you're not familiar with schemes), they locally look like $\mathrm{Spec}\\,A$. Now it's easy to define a line bundle: it's something that maps into the scheme and the map locally looks like $$\mathrm{Spec}\\,A[t] \to \mathrm{Spec} \\,A.$$ You can define the cohomology as usual, so the space $\mathbb P(H^0(L))^*$ continues to prove a good target to map into. It's slightly more unusual to draw pictures, but you can start with tensoring everything by $\mathbb F_q$ to gain the intuition.

Return to curves over field $k$. There the situation with sections is quite easy. Every line bundle there has the form $O(D)$ for some divisor $D$. As you add more points to your divisor, you're guaranteed lots of sections. Therefore, there are many very ample lines bundles.

Finally, a line bundle $L$ is ample iff some of its powers $L^{\otimes n}$ is very ample.

Bundles that have many sections don't have a special name, but their slightly more useful special case goes under the name of very ample bundles.

Start from the beginnning. Line bundles can be defined over any field $k$ but it has to be the same as the field of definition of your original manifold. Why the field cannot be different? By definition a line bundle is something whose total space is locally $A^1\times B$ for $B$ as base.

Now yes, what you're doing can be written more formally and more invariantly. Take the global sections $V = H^0(L)$ for a line bundle $L$. By definition, for a given point $x$ we have a linear pairing $V \times \{x\} \to L_x$, the fiber of $L$ at $x$ (which is isomorphic to $k$). If nonzero, a linear map to a one-dimensional space (even though the isomorphism with $k$ is not canonical) by definition gives a canonical point in $\mathbb PV^*$.

As you vary $x$, you get the map to $\mathbb PH^0(L)$, a very well-defined and canonical projective space. As the intuition would tell you, for this to be effective there should be a way to tell two points apart, that is if I have points $x, y\in X$ then there should be a section $s$ such that $s(x) = 0$ but $s(y) \ne 0$.

It's hard to construct sections of an arbitrary line bundle, therefore for some bundles the pairings above will be zero at some points. This will mean that the map above will be not defined.

For many other bundles, the map will be defined but it may not have good properties, e.g. it could send the whole variety to a single point — it's not very useful, thus a line bundle is called very ample iff this map is an immersion.

For the curves situation with sections is quite easy. Every line bundle there has the form $O(D)$ for some divisor $D$. As you add more points to your divisor, you're guaranteed lots of sections. Therefore, there are many very ample lines bundles.

A line bundle $L$ is ample iff some of its powers $L^{\otimes n}$ is very ample.

Bundles that have many sections don't have a special name, but their slightly more useful special case goes under the name of very ample bundles.

Start from the beginnning. Line bundles can be defined over any field $k$ but it has to be the same as the field of definition of your original manifold. Why the field cannot be different? By definition a line bundle is something whose total space is locally $A^1\times B$ for $B$ as base.

Now yes, what you're doing can be written more formally and more invariantly. Take the global sectionsfor a line bundle $L$. By definition, for a given point $x$ we have a linear pairing $H^0(L)\times \{x\} \to L_x$, the fiber of $L$ at $x$ (which is isomorphic to $k$). If nonzero, a linear map to a one-dimensional space (even though the isomorphism with $k$ is not canonical) by definition gives a canonical point in $\mathbb P(H^0(L))^*$.

As you vary $x$, you get the map to $\mathbb PH^0(L)$, a very well-defined and canonical projective space. As the intuition would tell you, for this to be effective there should be a way to tell two points apart, that is if I have points $x, y\in X$ then there should be a section $s$ such that $s(x) = 0$ but $s(y) \ne 0$.

It's hard to construct sections of an arbitrary line bundle, therefore for some bundles the pairings above will be zero at some points. This will mean that the map above will be not defined.

For many other bundles, the map will be defined but it may not have good properties, e.g. it could send the whole variety to a single point — it's not very useful, thus a line bundle is called very ample iff this map is an immersion.

Now if take arbitrary scheme (skip if you're not familiar with schemes), they locally look like $\mathrm{Spec}\\,A$. Now it's easy to define a line bundle: it's something that maps into the scheme and the map locally looks like $$\mathrm{Spec}\\,A[t] \to \mathrm{Spec} \\,A.$$ You can define the cohomology as usual, so the space $\mathbb PH^0(L)$ continues to prove a good target to map into.

Return to curves over field $k$. There the situation with sections is quite easy. Every line bundle there has the form $O(D)$ for some divisor $D$. As you add more points to your divisor, you're guaranteed lots of sections. Therefore, there are many very ample lines bundles.

A line bundle $L$ is ample iff some of its powers $L^{\otimes n}$ is very ample.