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From Robert Ash's "Basic Abstract Algebra"

Let $A$ be a commutative ring and $F$ a field. Show that $A$ is an algebra over $F$ if and only if $A$ contains (an isomorphic copy of) $F$ as a subring.

Suppose that $A$ is an algebra over $F$. Then we can define a homomorphism $f: F \rightarrow A$ where the image is in the center of $A$ (which is $A$ itself in this case). We need to show that $f$ is injective. But since $F$ is a field, the only ideals are $\{0\}$ and $F$. We know that the kernel cannot equal $F$, because otherwise we would have $f(1) = 1 = 0$ and we are assuming that $0 \not= 1$. So the kernel must be trivial, and we're done.

Now suppose that $A$ contains an isomorphic copy of $F$ as a subring. Then we will have an embedding $f: F \rightarrow A$. Since $A$ is commutative, the image will (of course) be contained in the center. So $A$ must be an $F$-algebra.

Give an example of an $R$-module $M$ with nonzero elements $r \in R$ and $x \in M$ such that $rx=0$.

Any ring is a module over itself. So if we choose a ring with zero divisors $r$ and $x$, then $rx=0$.

Do you think my answers are correct?

Thank you in advance

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    $\begingroup$ The headline claim is false: the trivial ring is an $F$-algebra. But that is the only problem. For the second problem you don't have to take $R = M$, but what you suggested works too. $\endgroup$ – Zhen Lin Jun 13 '13 at 19:16
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    $\begingroup$ Also you want $A$ to have an identity (perhaps this is implicit in your definition of ring, or $F$-algebra, but just to be extra precise). When the identity is equal to $0$, you get the zero ring, which is perfectly acceptable, and it won't contain an isomorphic copy of $F$, as Zhen indicates. $\endgroup$ – Keenan Kidwell Jun 13 '13 at 19:18
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    $\begingroup$ @ZhenLin Do we still run into that problem if $0 \not= 1$? Because my textbook assumes that $0 \not= 1$ when it's defining a ring. $\endgroup$ – user58289 Jun 13 '13 at 19:29
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    $\begingroup$ Then burn this book. Does it also define quotient rings $R/I$ then only for proper ideals? Does it only define localizations $S^{-1} R$ when $S$ doesn't contain nilpotent elements? For which rings $R,S$ is $R \otimes S$ defined? etc. ... it is nonsense to exclude the zero ring. $\endgroup$ – Martin Brandenburg Jun 13 '13 at 20:39
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    $\begingroup$ Dear @Artus, No, your proof is correct under the assumption $0\neq 1$. You just have to exclude the case $0=1$, because then the only ring map $F\rightarrow A$ sends everything to $0=1$, and is not injective. But, as is indicated in Martin's comment, the zero ring should definitely be allowed as an $F$-algebra. Otherwise you have to make special comments about it all the time. $\endgroup$ – Keenan Kidwell Jun 13 '13 at 20:39
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For the second question, the answer can be given in a fairly general way. If $R$ is not a division ring, it has a proper nonzero left ideal $I$; then, for $r\in I$, $r\ne0$, you have $r(1+I)=r+I=0+I$ as elements of $R/I$ and $1+I\ne0+I$ by assumption.

Of course, in a division ring $R$, from $rx=0$ ($r\in R$, $x\in M$) you can deduce that either $r=0$ or $x\ne 0$.

As already commented, the first assertion is valid if and only if $A$ is not the zero ring (that is, $1\ne0$), assuming that ring homomorphisms take $1$ to $1$ (and your proof is correct).

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Your answers are correct, but for me an example is more concrete. So instead of talking just about a ring with zero divisors I would really give an example, e.g. $R=M=\mathbb{Z}/(4)$. Here it is no problem, but later on, sometimes one also has to argue why the example you give exists.

The $0\neq 1$ removes the trivial ring as an $F$-algebra.

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