Ring homomorphism $f$ from field to non-trivial ring where $operatorname{Im}(f)$ is not the zero ring
$begingroup$
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
edited Jan 10 at 16:24
Yuval Gat
572213
asked Jan 10 at 16:11
oneguyoneguy
458
$endgroup$
add a comment |
$begingroup$
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
edited Jan 10 at 16:24
Yuval Gat
572213
asked Jan 10 at 16:11
oneguyoneguy
458
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 10 at 16:24
Yuval Gat
572213
asked Jan 10 at 16:11
oneguyoneguy
458
$endgroup$
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
edited Jan 10 at 16:24
Yuval Gat
572213
asked Jan 10 at 16:11
oneguyoneguy
458
edited Jan 10 at 16:24
Yuval Gat
572213
asked Jan 10 at 16:11
oneguyoneguy
458
edited Jan 10 at 16:24
Yuval Gat
572213
edited Jan 10 at 16:24
Yuval Gat
572213
edited Jan 10 at 16:24
Yuval Gat
572213
572213
asked Jan 10 at 16:11
oneguyoneguy
458
asked Jan 10 at 16:11
oneguyoneguy
458
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 10 at 16:25
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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$begingroup$
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.
answered Jan 10 at 16:25
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 10 at 16:25
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 10 at 16:25
Eric WofseyEric Wofsey
188k14216346
$endgroup$
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.
answered Jan 10 at 16:25
Eric WofseyEric Wofsey
188k14216346
answered Jan 10 at 16:25
Eric WofseyEric Wofsey
188k14216346
answered Jan 10 at 16:25
Eric WofseyEric Wofsey
188k14216346
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
add a comment |
$begingroup$
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
add a comment |
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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