## Linear equivalence of divisors in smooth algebraic surface

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

Let's assume $X$ is a smooth algebraic surface and $C$ a curve containing a smooth point $p_0\in X$, then there exist divisors $H_1$ and $H_2$ non of which contain $p_0$ such that $C+H_1$ is linearly equivalent to $H_2$.

-

5
This is a fairly straightforward consequence of the definition of ample divisor. It's probably useful to work out the details for yourself. – Artie Prendergast-Smith Feb 1 2012 at 17:23
Besides, it is worded like a homework problem, not as a question. – Angelo Feb 2 2012 at 4:40
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.910675048828125, "perplexity_flag": "head"}