Ring homomorphism $f$ from field to non-trivial ring where $operatorname{Im}(f)$ is not the zero ring












0












$begingroup$


When I looked at the wikipedia page about ring homomorphisms (here) I noticed the following statement:




Let $f:Rto S$ be a ring homomorphism.



If $R$ is a field, $S$ is not the zero ring, and $operatorname{Im}(f)$ is not the zero
ring then $f$ is injective.




My question is why the fact that $operatorname{Im}(f)$ is not the zero ring is required as I don't see where this would be used in a proof.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe $f(x)=0$ is a ring homomorphism?
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:14










  • $begingroup$
    but isn't f(1) = 1 required for f to be a ring morphism?
    $endgroup$
    – oneguy
    Jan 10 at 16:19










  • $begingroup$
    Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring.
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:21
















0












$begingroup$


When I looked at the wikipedia page about ring homomorphisms (here) I noticed the following statement:




Let $f:Rto S$ be a ring homomorphism.



If $R$ is a field, $S$ is not the zero ring, and $operatorname{Im}(f)$ is not the zero
ring then $f$ is injective.




My question is why the fact that $operatorname{Im}(f)$ is not the zero ring is required as I don't see where this would be used in a proof.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe $f(x)=0$ is a ring homomorphism?
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:14










  • $begingroup$
    but isn't f(1) = 1 required for f to be a ring morphism?
    $endgroup$
    – oneguy
    Jan 10 at 16:19










  • $begingroup$
    Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring.
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:21














0












0








0





$begingroup$


When I looked at the wikipedia page about ring homomorphisms (here) I noticed the following statement:




Let $f:Rto S$ be a ring homomorphism.



If $R$ is a field, $S$ is not the zero ring, and $operatorname{Im}(f)$ is not the zero
ring then $f$ is injective.




My question is why the fact that $operatorname{Im}(f)$ is not the zero ring is required as I don't see where this would be used in a proof.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$




When I looked at the wikipedia page about ring homomorphisms (here) I noticed the following statement:




Let $f:Rto S$ be a ring homomorphism.



If $R$ is a field, $S$ is not the zero ring, and $operatorname{Im}(f)$ is not the zero
ring then $f$ is injective.




My question is why the fact that $operatorname{Im}(f)$ is not the zero ring is required as I don't see where this would be used in a proof.






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 16:24









Yuval Gat

572213




572213










asked Jan 10 at 16:11









oneguyoneguy

458




458








  • 1




    $begingroup$
    Maybe $f(x)=0$ is a ring homomorphism?
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:14










  • $begingroup$
    but isn't f(1) = 1 required for f to be a ring morphism?
    $endgroup$
    – oneguy
    Jan 10 at 16:19










  • $begingroup$
    Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring.
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:21














  • 1




    $begingroup$
    Maybe $f(x)=0$ is a ring homomorphism?
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:14










  • $begingroup$
    but isn't f(1) = 1 required for f to be a ring morphism?
    $endgroup$
    – oneguy
    Jan 10 at 16:19










  • $begingroup$
    Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring.
    $endgroup$
    – SmileyCraft
    Jan 10 at 16:21








1




1




$begingroup$
Maybe $f(x)=0$ is a ring homomorphism?
$endgroup$
– SmileyCraft
Jan 10 at 16:14




$begingroup$
Maybe $f(x)=0$ is a ring homomorphism?
$endgroup$
– SmileyCraft
Jan 10 at 16:14












$begingroup$
but isn't f(1) = 1 required for f to be a ring morphism?
$endgroup$
– oneguy
Jan 10 at 16:19




$begingroup$
but isn't f(1) = 1 required for f to be a ring morphism?
$endgroup$
– oneguy
Jan 10 at 16:19












$begingroup$
Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring.
$endgroup$
– SmileyCraft
Jan 10 at 16:21




$begingroup$
Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring.
$endgroup$
– SmileyCraft
Jan 10 at 16:21










1 Answer
1






active

oldest

votes


















3












$begingroup$

This is just an error in the Wikipedia page; the assumption that the image is not the zero ring is unnecessary. If ring homomorphisms are not defined to be unital (that is, $f(1)=1$), then it is necessary, since otherwise $f$ could just map everything to $0$. However, the Wikipedia page does define ring homomorphisms to be unital, so this is not an issue.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The zero homomorphism to the trivial ring ${0}$ is unital. I think that's what the authors had in mind.
    $endgroup$
    – rschwieb
    Jan 10 at 16:27








  • 1




    $begingroup$
    But the statement also includes the requirement that $S$ is not the zero ring.
    $endgroup$
    – Eric Wofsey
    Jan 10 at 16:31










  • $begingroup$
    Oh that part is superfluous. Got crossed up.
    $endgroup$
    – rschwieb
    Jan 10 at 17:01











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

This is just an error in the Wikipedia page; the assumption that the image is not the zero ring is unnecessary. If ring homomorphisms are not defined to be unital (that is, $f(1)=1$), then it is necessary, since otherwise $f$ could just map everything to $0$. However, the Wikipedia page does define ring homomorphisms to be unital, so this is not an issue.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The zero homomorphism to the trivial ring ${0}$ is unital. I think that's what the authors had in mind.
    $endgroup$
    – rschwieb
    Jan 10 at 16:27








  • 1




    $begingroup$
    But the statement also includes the requirement that $S$ is not the zero ring.
    $endgroup$
    – Eric Wofsey
    Jan 10 at 16:31










  • $begingroup$
    Oh that part is superfluous. Got crossed up.
    $endgroup$
    – rschwieb
    Jan 10 at 17:01
















3












$begingroup$

This is just an error in the Wikipedia page; the assumption that the image is not the zero ring is unnecessary. If ring homomorphisms are not defined to be unital (that is, $f(1)=1$), then it is necessary, since otherwise $f$ could just map everything to $0$. However, the Wikipedia page does define ring homomorphisms to be unital, so this is not an issue.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The zero homomorphism to the trivial ring ${0}$ is unital. I think that's what the authors had in mind.
    $endgroup$
    – rschwieb
    Jan 10 at 16:27








  • 1




    $begingroup$
    But the statement also includes the requirement that $S$ is not the zero ring.
    $endgroup$
    – Eric Wofsey
    Jan 10 at 16:31










  • $begingroup$
    Oh that part is superfluous. Got crossed up.
    $endgroup$
    – rschwieb
    Jan 10 at 17:01














3












3








3





$begingroup$

This is just an error in the Wikipedia page; the assumption that the image is not the zero ring is unnecessary. If ring homomorphisms are not defined to be unital (that is, $f(1)=1$), then it is necessary, since otherwise $f$ could just map everything to $0$. However, the Wikipedia page does define ring homomorphisms to be unital, so this is not an issue.






share|cite|improve this answer









$endgroup$



This is just an error in the Wikipedia page; the assumption that the image is not the zero ring is unnecessary. If ring homomorphisms are not defined to be unital (that is, $f(1)=1$), then it is necessary, since otherwise $f$ could just map everything to $0$. However, the Wikipedia page does define ring homomorphisms to be unital, so this is not an issue.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 16:25









Eric WofseyEric Wofsey

188k14216346




188k14216346












  • $begingroup$
    The zero homomorphism to the trivial ring ${0}$ is unital. I think that's what the authors had in mind.
    $endgroup$
    – rschwieb
    Jan 10 at 16:27








  • 1




    $begingroup$
    But the statement also includes the requirement that $S$ is not the zero ring.
    $endgroup$
    – Eric Wofsey
    Jan 10 at 16:31










  • $begingroup$
    Oh that part is superfluous. Got crossed up.
    $endgroup$
    – rschwieb
    Jan 10 at 17:01


















  • $begingroup$
    The zero homomorphism to the trivial ring ${0}$ is unital. I think that's what the authors had in mind.
    $endgroup$
    – rschwieb
    Jan 10 at 16:27








  • 1




    $begingroup$
    But the statement also includes the requirement that $S$ is not the zero ring.
    $endgroup$
    – Eric Wofsey
    Jan 10 at 16:31










  • $begingroup$
    Oh that part is superfluous. Got crossed up.
    $endgroup$
    – rschwieb
    Jan 10 at 17:01
















$begingroup$
The zero homomorphism to the trivial ring ${0}$ is unital. I think that's what the authors had in mind.
$endgroup$
– rschwieb
Jan 10 at 16:27






$begingroup$
The zero homomorphism to the trivial ring ${0}$ is unital. I think that's what the authors had in mind.
$endgroup$
– rschwieb
Jan 10 at 16:27






1




1




$begingroup$
But the statement also includes the requirement that $S$ is not the zero ring.
$endgroup$
– Eric Wofsey
Jan 10 at 16:31




$begingroup$
But the statement also includes the requirement that $S$ is not the zero ring.
$endgroup$
– Eric Wofsey
Jan 10 at 16:31












$begingroup$
Oh that part is superfluous. Got crossed up.
$endgroup$
– rschwieb
Jan 10 at 17:01




$begingroup$
Oh that part is superfluous. Got crossed up.
$endgroup$
– rschwieb
Jan 10 at 17:01


















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