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Trouble with interpreting a question about polynomial rings?
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I trying to do problem $9.3.1$ in Dummut & Foote; I think they use some "abuse of notation" that I don't understand.
Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $colorredtextProve that if a(x) not in textR[x] then R is not a U.F.D.$ Deduce that $mathbbZ[2sqrt2]$ is not a U.F.D.
The part in red is what's confusing me. The reason why I think it's "abuse of notation" is because there is nothing to distinguish $a(x)$ from $b(x)$. So I could see an argument for each of the following interpretations, and I don't know which one is correct:
- Exactly one of $a(x), b(x)$ is not in $R[x]$
- At least one one of $a(x), b(x)$ is not in $R[x]$
- Precisely both of $a(x), b(x)$ are not in $R[x]$
Any help is appreciated.
Edit for Lord Shark the Unknown:
Proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$ then $a, b in mathbbZ$.
Since $(x+a)(x+b) = x^2 + (a+b)x + ab$ we know $ab in mathbbZ$ and $a+b in mathbbZ$. Thus:
$$a^2+2ab+b^2 in mathbbZ$$
$$implies a^2-2ab +b^2in mathbbZ$$
$$implies (a-b)^2 in mathbbZ$$
$$implies a-b in mathbbZ$$
$$implies (a+b)(a-b) in mathbbZ$$
$$implies a^2-b^2 in mathbbZ$$
Now we can also get that $a^2+b^2 in mathbbZ$ using the same trick we used in the first steps. So $(a^2+b^2)-(a^2-b^2) in mathbbZ$,
$$implies 2a^2 in mathbbZ$$
$$implies a^2 in mathbbZ$$
$$implies a in mathbbZ$$
And by a similar argument, $b in mathbbZ$.
abstract-algebra ring-theory polynomial-rings
asked Jul 23 at 1:03
Ovi
11.3k935105
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show 1 more comment
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I trying to do problem $9.3.1$ in Dummut & Foote; I think they use some "abuse of notation" that I don't understand.
Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $colorredtextProve that if a(x) not in textR[x] then R is not a U.F.D.$ Deduce that $mathbbZ[2sqrt2]$ is not a U.F.D.
The part in red is what's confusing me. The reason why I think it's "abuse of notation" is because there is nothing to distinguish $a(x)$ from $b(x)$. So I could see an argument for each of the following interpretations, and I don't know which one is correct:
- Exactly one of $a(x), b(x)$ is not in $R[x]$
- At least one one of $a(x), b(x)$ is not in $R[x]$
- Precisely both of $a(x), b(x)$ are not in $R[x]$
Any help is appreciated.
Edit for Lord Shark the Unknown:
Proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$ then $a, b in mathbbZ$.
Since $(x+a)(x+b) = x^2 + (a+b)x + ab$ we know $ab in mathbbZ$ and $a+b in mathbbZ$. Thus:
$$a^2+2ab+b^2 in mathbbZ$$
$$implies a^2-2ab +b^2in mathbbZ$$
$$implies (a-b)^2 in mathbbZ$$
$$implies a-b in mathbbZ$$
$$implies (a+b)(a-b) in mathbbZ$$
$$implies a^2-b^2 in mathbbZ$$
Now we can also get that $a^2+b^2 in mathbbZ$ using the same trick we used in the first steps. So $(a^2+b^2)-(a^2-b^2) in mathbbZ$,
$$implies 2a^2 in mathbbZ$$
$$implies a^2 in mathbbZ$$
$$implies a in mathbbZ$$
And by a similar argument, $b in mathbbZ$.
abstract-algebra ring-theory polynomial-rings
asked Jul 23 at 1:03
Ovi
11.3k935105
Both factors must be in $R[x]$ if the product is, so at least one must not be if the product isn't. It's not an abuse of notation, but it does rely implicitly on the commutativity of multiplication.
– Ethan Bolker
Jul 23 at 1:05
@EthanBolker Thanks! I can see how going very technical, that would be 100% the correct interpretation. However, I was not sure if that was the case or they were using "abus of notation"; Dummit & Foote sometimes seem to drop certain assumptions which I think they want the reader to infer from the previous paragraph of from the section in general.
– Ovi
Jul 23 at 1:09
@EthanBolker Could you please elaborate a little on your first sentence? Since $(a_1x+a_0)(b_1x+b_0) = a_1b_1x^2 + x(a_1b_0+a_0b_1) + a_0b_0$, I don't see why we cannot have $a_0, a_1, b_0, b_1$ being non-integer rationals, yet they conspire together to make $a_1b_1, a_0b_0$, and $a_1b_0+a_0b_1$ all integers.
– Ovi
Jul 23 at 2:03
@Ovi Recall that $a(x)$ and $b(x)$ are assumed to be monic.
– Lord Shark the Unknown
Jul 23 at 4:29
@LordSharktheUnknown I've edited the question with a proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$, then $a, b in mathbbZ$. However, of course this covers just one case, it requires me to prove a couple lemmas which I didn't include, and I know there must be an abstract algebra way to prove it for the general case. Could you please point me in the right direction?
– Ovi
Jul 23 at 4:50
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
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I trying to do problem $9.3.1$ in Dummut & Foote; I think they use some "abuse of notation" that I don't understand.
Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $colorredtextProve that if a(x) not in textR[x] then R is not a U.F.D.$ Deduce that $mathbbZ[2sqrt2]$ is not a U.F.D.
The part in red is what's confusing me. The reason why I think it's "abuse of notation" is because there is nothing to distinguish $a(x)$ from $b(x)$. So I could see an argument for each of the following interpretations, and I don't know which one is correct:
- Exactly one of $a(x), b(x)$ is not in $R[x]$
- At least one one of $a(x), b(x)$ is not in $R[x]$
- Precisely both of $a(x), b(x)$ are not in $R[x]$
Any help is appreciated.
Edit for Lord Shark the Unknown:
Proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$ then $a, b in mathbbZ$.
Since $(x+a)(x+b) = x^2 + (a+b)x + ab$ we know $ab in mathbbZ$ and $a+b in mathbbZ$. Thus:
$$a^2+2ab+b^2 in mathbbZ$$
$$implies a^2-2ab +b^2in mathbbZ$$
$$implies (a-b)^2 in mathbbZ$$
$$implies a-b in mathbbZ$$
$$implies (a+b)(a-b) in mathbbZ$$
$$implies a^2-b^2 in mathbbZ$$
Now we can also get that $a^2+b^2 in mathbbZ$ using the same trick we used in the first steps. So $(a^2+b^2)-(a^2-b^2) in mathbbZ$,
$$implies 2a^2 in mathbbZ$$
$$implies a^2 in mathbbZ$$
$$implies a in mathbbZ$$
And by a similar argument, $b in mathbbZ$.
abstract-algebra ring-theory polynomial-rings
asked Jul 23 at 1:03
Ovi
11.3k935105
I trying to do problem $9.3.1$ in Dummut & Foote; I think they use some "abuse of notation" that I don't understand.
Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $colorredtextProve that if a(x) not in textR[x] then R is not a U.F.D.$ Deduce that $mathbbZ[2sqrt2]$ is not a U.F.D.
The part in red is what's confusing me. The reason why I think it's "abuse of notation" is because there is nothing to distinguish $a(x)$ from $b(x)$. So I could see an argument for each of the following interpretations, and I don't know which one is correct:
- Exactly one of $a(x), b(x)$ is not in $R[x]$
- At least one one of $a(x), b(x)$ is not in $R[x]$
- Precisely both of $a(x), b(x)$ are not in $R[x]$
Any help is appreciated.
Edit for Lord Shark the Unknown:
Proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$ then $a, b in mathbbZ$.
Since $(x+a)(x+b) = x^2 + (a+b)x + ab$ we know $ab in mathbbZ$ and $a+b in mathbbZ$. Thus:
$$a^2+2ab+b^2 in mathbbZ$$
$$implies a^2-2ab +b^2in mathbbZ$$
$$implies (a-b)^2 in mathbbZ$$
$$implies a-b in mathbbZ$$
$$implies (a+b)(a-b) in mathbbZ$$
$$implies a^2-b^2 in mathbbZ$$
Now we can also get that $a^2+b^2 in mathbbZ$ using the same trick we used in the first steps. So $(a^2+b^2)-(a^2-b^2) in mathbbZ$,
$$implies 2a^2 in mathbbZ$$
$$implies a^2 in mathbbZ$$
$$implies a in mathbbZ$$
And by a similar argument, $b in mathbbZ$.
abstract-algebra ring-theory polynomial-rings
asked Jul 23 at 1:03
Ovi
11.3k935105
edited Jul 23 at 4:47
asked Jul 23 at 1:03
Ovi
11.3k935105
asked Jul 23 at 1:03
Ovi
11.3k935105
asked Jul 23 at 1:03
Ovi
11.3k935105
11.3k935105
Both factors must be in $R[x]$ if the product is, so at least one must not be if the product isn't. It's not an abuse of notation, but it does rely implicitly on the commutativity of multiplication.
– Ethan Bolker
Jul 23 at 1:05
@EthanBolker Thanks! I can see how going very technical, that would be 100% the correct interpretation. However, I was not sure if that was the case or they were using "abus of notation"; Dummit & Foote sometimes seem to drop certain assumptions which I think they want the reader to infer from the previous paragraph of from the section in general.
– Ovi
Jul 23 at 1:09
@EthanBolker Could you please elaborate a little on your first sentence? Since $(a_1x+a_0)(b_1x+b_0) = a_1b_1x^2 + x(a_1b_0+a_0b_1) + a_0b_0$, I don't see why we cannot have $a_0, a_1, b_0, b_1$ being non-integer rationals, yet they conspire together to make $a_1b_1, a_0b_0$, and $a_1b_0+a_0b_1$ all integers.
– Ovi
Jul 23 at 2:03
@Ovi Recall that $a(x)$ and $b(x)$ are assumed to be monic.
– Lord Shark the Unknown
Jul 23 at 4:29
@LordSharktheUnknown I've edited the question with a proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$, then $a, b in mathbbZ$. However, of course this covers just one case, it requires me to prove a couple lemmas which I didn't include, and I know there must be an abstract algebra way to prove it for the general case. Could you please point me in the right direction?
– Ovi
Jul 23 at 4:50
 |Â
show 1 more comment
Both factors must be in $R[x]$ if the product is, so at least one must not be if the product isn't. It's not an abuse of notation, but it does rely implicitly on the commutativity of multiplication.
– Ethan Bolker
Jul 23 at 1:05
@EthanBolker Thanks! I can see how going very technical, that would be 100% the correct interpretation. However, I was not sure if that was the case or they were using "abus of notation"; Dummit & Foote sometimes seem to drop certain assumptions which I think they want the reader to infer from the previous paragraph of from the section in general.
– Ovi
Jul 23 at 1:09
@EthanBolker Could you please elaborate a little on your first sentence? Since $(a_1x+a_0)(b_1x+b_0) = a_1b_1x^2 + x(a_1b_0+a_0b_1) + a_0b_0$, I don't see why we cannot have $a_0, a_1, b_0, b_1$ being non-integer rationals, yet they conspire together to make $a_1b_1, a_0b_0$, and $a_1b_0+a_0b_1$ all integers.
– Ovi
Jul 23 at 2:03
@Ovi Recall that $a(x)$ and $b(x)$ are assumed to be monic.
– Lord Shark the Unknown
Jul 23 at 4:29
@LordSharktheUnknown I've edited the question with a proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$, then $a, b in mathbbZ$. However, of course this covers just one case, it requires me to prove a couple lemmas which I didn't include, and I know there must be an abstract algebra way to prove it for the general case. Could you please point me in the right direction?
– Ovi
Jul 23 at 4:50
Both factors must be in $R[x]$ if the product is, so at least one must not be if the product isn't. It's not an abuse of notation, but it does rely implicitly on the commutativity of multiplication.
– Ethan Bolker
Jul 23 at 1:05
Both factors must be in $R[x]$ if the product is, so at least one must not be if the product isn't. It's not an abuse of notation, but it does rely implicitly on the commutativity of multiplication.
– Ethan Bolker
Jul 23 at 1:05
@EthanBolker Thanks! I can see how going very technical, that would be 100% the correct interpretation. However, I was not sure if that was the case or they were using "abus of notation"; Dummit & Foote sometimes seem to drop certain assumptions which I think they want the reader to infer from the previous paragraph of from the section in general.
– Ovi
Jul 23 at 1:09
@EthanBolker Thanks! I can see how going very technical, that would be 100% the correct interpretation. However, I was not sure if that was the case or they were using "abus of notation"; Dummit & Foote sometimes seem to drop certain assumptions which I think they want the reader to infer from the previous paragraph of from the section in general.
– Ovi
Jul 23 at 1:09
@EthanBolker Could you please elaborate a little on your first sentence? Since $(a_1x+a_0)(b_1x+b_0) = a_1b_1x^2 + x(a_1b_0+a_0b_1) + a_0b_0$, I don't see why we cannot have $a_0, a_1, b_0, b_1$ being non-integer rationals, yet they conspire together to make $a_1b_1, a_0b_0$, and $a_1b_0+a_0b_1$ all integers.
– Ovi
Jul 23 at 2:03
@EthanBolker Could you please elaborate a little on your first sentence? Since $(a_1x+a_0)(b_1x+b_0) = a_1b_1x^2 + x(a_1b_0+a_0b_1) + a_0b_0$, I don't see why we cannot have $a_0, a_1, b_0, b_1$ being non-integer rationals, yet they conspire together to make $a_1b_1, a_0b_0$, and $a_1b_0+a_0b_1$ all integers.
– Ovi
Jul 23 at 2:03
@Ovi Recall that $a(x)$ and $b(x)$ are assumed to be monic.
– Lord Shark the Unknown
Jul 23 at 4:29
@Ovi Recall that $a(x)$ and $b(x)$ are assumed to be monic.
– Lord Shark the Unknown
Jul 23 at 4:29
@LordSharktheUnknown I've edited the question with a proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$, then $a, b in mathbbZ$. However, of course this covers just one case, it requires me to prove a couple lemmas which I didn't include, and I know there must be an abstract algebra way to prove it for the general case. Could you please point me in the right direction?
– Ovi
Jul 23 at 4:50
@LordSharktheUnknown I've edited the question with a proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$, then $a, b in mathbbZ$. However, of course this covers just one case, it requires me to prove a couple lemmas which I didn't include, and I know there must be an abstract algebra way to prove it for the general case. Could you please point me in the right direction?
– Ovi
Jul 23 at 4:50
 |Â
show 1 more comment
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Both factors must be in $R[x]$ if the product is, so at least one must not be if the product isn't. It's not an abuse of notation, but it does rely implicitly on the commutativity of multiplication.
– Ethan Bolker
Jul 23 at 1:05
@EthanBolker Thanks! I can see how going very technical, that would be 100% the correct interpretation. However, I was not sure if that was the case or they were using "abus of notation"; Dummit & Foote sometimes seem to drop certain assumptions which I think they want the reader to infer from the previous paragraph of from the section in general.
– Ovi
Jul 23 at 1:09
@EthanBolker Could you please elaborate a little on your first sentence? Since $(a_1x+a_0)(b_1x+b_0) = a_1b_1x^2 + x(a_1b_0+a_0b_1) + a_0b_0$, I don't see why we cannot have $a_0, a_1, b_0, b_1$ being non-integer rationals, yet they conspire together to make $a_1b_1, a_0b_0$, and $a_1b_0+a_0b_1$ all integers.
– Ovi
Jul 23 at 2:03
@Ovi Recall that $a(x)$ and $b(x)$ are assumed to be monic.
– Lord Shark the Unknown
Jul 23 at 4:29
@LordSharktheUnknown I've edited the question with a proof that if $(x+a)(x+b) in mathbbZ[x]$ and $a, b in mathbbQ$, then $a, b in mathbbZ$. However, of course this covers just one case, it requires me to prove a couple lemmas which I didn't include, and I know there must be an abstract algebra way to prove it for the general case. Could you please point me in the right direction?
– Ovi
Jul 23 at 4:50