[xsl] Why do the namespace appears in transformation ?

["Fabien Tillier" <f.tillier@xxxxxxxx>]

Hi List.
I am trying to generate an XSL template from an XML file that describes
some filters.
So, basically, I get
<Results>
   <Row>
      <MTF_NUMERO_TABLEAU>2</MTF_NUMERO_TABLEAU>
      <TA_TITRE_A>Titre 1</TA_TITRE_A>
   </Row>
   <Row>
      <MTF_NUMERO_TABLEAU>3</MTF_NUMERO_TABLEAU>
      <TA_TITRE_A>Titre 2</TA_TITRE_A>
   </Row>
 </Results>

I am parsing it with (not finished, of course)

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
exclude-result-prefixes="xsl"
 xmlns:xso="dummy"
  >
 <xsl:output method = "xml" encoding="UTF-8"/>
 <!--  When transforming, all xso namespace elements will become xsl -->
 <xsl:namespace-alias stylesheet-prefix="xso" result-prefix="xsl"/>
 <xsl:template match="/Results">
 	<xsl:for-each-group select="Row" group-by="MTF_NUMERO_TABLEAU">
 		<xsl:variable name="numtableau">
 			<xsl:value-of select="current-grouping-key()"/>
 		</xsl:variable>
 		<xso:template match="node()[(MTF_NUMERO_TABLEAU =
'{$numtableau}')]">
 			<!--Do something-->
 		</xso:template>
		<xsl:value-of select="current-grouping-key()"/>,

	</xsl:for-each-group>
 </xsl:template>
 <xsl:template match="Row">

 </xsl:template>
</xsl:stylesheet>



The output (with Kernow) is

<?xml version="1.0" encoding="UTF-8"?>
<xsl:template xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
match="node()[(MTF_NUMERO_TABLEAU = '2')]"/>2,
<xsl:template xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
match="node()[(MTF_NUMERO_TABLEAU = '3')]"/>3

Which is almost what I want....
But, how can I get rid of the xmlns declaration in each <xsl:template
section ?
(So that to get
<?xml version="1.0" encoding="UTF-8"?>
<xsl:template match="node()[(MTF_NUMERO_TABLEAU = '2')]"/>2,
<xsl:template match="node()[(MTF_NUMERO_TABLEAU = '3')]"/>3
)

Thanks in advance
Regards,
Fabien