如何在没有模式信息的情况下将c#对象序列化为xml

我是这样做的:

可序列化的类:

[Serializable()]
public class Ticket
{
    public string   CitationNumber { get; set; }
    public decimal Amount { get; set; }
}

然后将模型序列化为xml:

var model = cart.Citations
    .Select(c => new Ticket(c.Number, c.Amount)).ToList();
var serializer = new XmlSerializer(typeof (List<Ticket>));
var sw = new StringWriter();
serializer.Serialize(sw, model);
return sw.ToString();

输出sw.ToString()类似于

<?xml version="1.0" encoding="utf-16"?>
<ArrayOfTicket xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Ticket>
<CitationNumber>00092844</CitationNumber>
<Amount>20</Amount>
</Ticket>
</ArrayOfTicket>  

是否有一种方法可以自定义Serialize()输出以删除那些模式信息，如:<?xml version="1.0" encoding="utf-16"?>和xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" ?

我怎么能把根元素ArrayOfTicket变成别的东西?

我可以控制这些输出吗?

var serializer = new XmlSerializer(typeof(List<Ticket>));
var ns = new XmlSerializerNamespaces();
ns.Add("", "");
var sw = new StringWriter();
var xmlWriter = XmlWriter.Create(sw, new XmlWriterSettings() { OmitXmlDeclaration = true });
serializer.Serialize(xmlWriter, model, ns);
string xml = sw.ToString(); 

<ArrayOfTicket>
  <Ticket>
    <CitationNumber>a</CitationNumber>
    <Amount>1</Amount>
  </Ticket>
  <Ticket>
    <CitationNumber>b</CitationNumber>
    <Amount>2</Amount>
  </Ticket>
</ArrayOfTicket>